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A vector carries magnitude and direction
After 3 north, the 4 km leg can end anywhere on this circle. Two of those places are 5 from the start.
You ask the way to the market. Someone says, “It is 5 km from here.” That does not get you there. You also need to know which way to go.
Walk 3 north, then 4 east. You end 5 from the start: Pythagoras on legs 3 and 4.
Say you walk 3 km north, and then 4 km east. You end up 5 km from where you started. Class 10 Pythagoras gives it: $\sqrt{3^2 + 4^2} = 5$.
Now walk 3 km north, and then 4 km west. You are 5 km from the start again. But you are in a different place.
Two arrows from the origin, to 4 across, 3 up and to 4 back, 3 up. Both are 5 long, but they are different vectors.
The figure puts the start at the origin. East is along the $x$-axis, and north is up the page. The first walk ends at $(4, 3)$. The second ends at $(-4, 3)$. Both arrows are 5 long. The number 5 cannot tell them apart. Physics 11 called the 5 km a distance, and the arrow a displacement.
Every arrow on this ring has length 5. Only the direction picks out the bold one, 4î + 3ĵ.
A quantity with a size only is a SCALAR. Mass, time and distance are scalars. A quantity with a size and a direction is a VECTOR. Displacement, velocity and force are vectors. You draw a vector as an arrow.
Four things this chapter does with arrows: add them, stretch them, write them as numbers, and multiply them.
Here is the question to carry to the last page. Can arrows be added, stretched, written as numbers and multiplied? Can lengths, angles, work and areas all come out of adding and multiplying them?
The walk comes back several times. Keep it in mind.
A SCHOLAR INDIA REMEMBERS
Dignaga was born in south India, around the 5th or 6th century CE. Part of his work was a theory of how words mean anything at all. Put his question to “$5$ km from here”. Which place does it pick out? Not one. Every place $5$ km away fits, a whole circle of them. Add the way, $3$ km north and then $4$ km east, and the words pick out one place, $(4, 3)$.
Dignaga · the kingfisher
Check yourself
A scalar carries magnitude only. What does a vector carry in addition?
a formula
a sign
direction
units
Check your answer
a formula — A formula is how a vector’s components get computed; it is not itself the property a vector carries beyond magnitude.
a sign — A sign only distinguishes positive from negative along a single number line, it cannot describe a direction anywhere in three-dimensional space.
✓ direction — (C) A vector carries everything a scalar does (magnitude) plus direction, the one addition this whole chapter builds an entire parallel arithmetic around.
units — Units such as kilograms or degrees attach to scalars too, mass and temperature both carry units without carrying direction.
Why can this chapter not just reuse ordinary number-line rules such as ‘bigger magnitude always wins’ once direction enters?
because direction only matters once there are three dimensions
direction lets quantities cancel or reinforce by angle
because vectors are always larger in value than scalars
because vectors cannot be added at all
Check your answer
because direction only matters once there are three dimensions — Direction changes how quantities combine as soon as there is more than one dimension to point in, the effect is not held back for three dimensions specifically.
✓ direction lets quantities cancel or reinforce by angle — (B) Direction means two quantities of any size can point toward or against each other, so a simple size comparison stops deciding the outcome the way it does for plain numbers.
because vectors are always larger in value than scalars — Magnitude values from a vector and a scalar are not comparable in the first place , ‘larger’ is not a claim this chapter makes.
because vectors cannot be added at all — Vectors can be added, that is exactly what the triangle and parallelogram laws build, so this denies that very starting point.
Which of these is a vector quantity, not a scalar?
speed
mass
velocity
temperature
Check your answer
speed — Speed is a scalar: it is velocity, not speed, that adds a direction to it.
mass — Mass is named directly as a scalar example, it has a size but no direction attached to it.
✓ velocity — (C) Velocity is the listed example of a vector, it carries both magnitude and direction.
temperature — Temperature is a scalar: a reading of 30 degrees has a size and points nowhere.
Speed and velocity both describe how fast something moves. Why does this chapter treat them as fundamentally different kinds of quantity?
speed only applies to cars, velocity applies to everything else
velocity is always a larger number than speed
velocity and speed are actually the same quantity under two different names
velocity adds direction on top of the same magnitude speed reports
Check your answer
speed only applies to cars, velocity applies to everything else — Speed and velocity both apply to any moving object at all, the distinction is about direction, not what kind of object is moving.
velocity is always a larger number than speed — Nothing sets velocity’s number above speed’s, a directed reading of the same motion is not automatically a bigger one.
velocity and speed are actually the same quantity under two different names — This is exactly the distinction at stake, direction is what turns a scalar quantity into a vector one, and speed versus velocity is its clearest example.
✓ velocity adds direction on top of the same magnitude speed reports — (D) Velocity is a vector, so it adds direction on top of the same magnitude that speed already reports on its own.
On a grid of streets, any corner is so many blocks east of the start and so many blocks north.
Go back to the walk, and draw it on a map grid. Put the start at the origin. Each square is 1 km. East is to the right, and north is up the page. The first walk ends 4 squares east and 3 squares north of the start.
The point (4, 3) is 4 steps east and 3 steps north. The arrow from the origin carries both numbers.
So the coordinates of a point tell you how to walk there from the origin: 4 east, then 3 north. The arrow from the origin to the point carries the same two numbers. On the grid that point is $(4, 3)$.
In space a third number says how far up. The point (4, 3, 2) is reached by three steps.
Space has a third direction, up. Call the origin $O$. You reach a point $P(x, y, z)$ by $x$ steps along the $x$-axis, $y$ steps along the $y$-axis and $z$ steps up.
An arrow from initial point A to terminal point B. Its length, 5 here, is the magnitude.
Now name the parts. A vector is a line segment with a direction, an arrow. This book writes a vector as $\vec{a}$, with an arrow on top, as NCERT does. Some books print the letter in thick type instead. The length of $\vec{a}$ is its MAGNITUDE, written $\left|\vec{a}\right|$. A magnitude is a scalar.
î, ĵ and k̂ are one step along each axis. Every other vector is built from these three.
A step of length 1 along each axis gets its own name. It is $\hat{i}$ along the $x$-axis, $\hat{j}$ along the $y$-axis and $\hat{k}$ along the $z$-axis. The arrow from $O$ to $P(x, y, z)$ is the POSITION VECTOR of $P$:
$\vec{O P} = x \hat{i} + y \hat{j} + z \hat{k}$.
The length of the arrow
Class 11 gave you the distance of $P$ from $O$. So $\left|\vec{O P}\right| = \sqrt{x^2 + y^2 + z^2}$.
Components of 2, 6 and 3 reach P. The arrow from O to P is 2i plus 6j plus 3k, length 7.
Let us check it on the figure. The point is $P(2, 6, 3)$. Walk 2 along the $x$-axis, 6 along the $y$-axis and 3 up. The arrow is $\vec{O P} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k}$. Its length is $\sqrt{4 + 36 + 9} = \sqrt{49} = 7$.
For the first walk, $\vec{O P} = 4 \hat{i} + 3 \hat{j}$. Its length is $\sqrt{16 + 9} = 5$, the 5 km.
Check yourself
The position vector of a point $P(x, y, z)$ relative to the origin $O$, $\vec{O P}$, equals \_\_\_\_\_\_.
$y \hat{i} + x \hat{j} + z \hat{k}$
$(x+y+z) \hat{i}$
$\sqrt{x^2+y^2+z^2}$
$x \hat{i} + y \hat{j} + z \hat{k}$
Check your answer
$y \hat{i} + x \hat{j} + z \hat{k}$ — The x-coordinate belongs on $\hat{i}$ and the y-coordinate on $\hat{j}$, swapping them mislabels two of the three components.
$(x+y+z) \hat{i}$ — Collapsing the three coordinates into a single sum along $\hat{i}$ discards the $\hat{j}$- and $\hat{k}$-directions entirely.
$\sqrt{x^2+y^2+z^2}$ — That expression is $\left|\vec{O P}\right|$, the position vector’s magnitude, a number, not the directed vector itself.
✓ $x \hat{i} + y \hat{j} + z \hat{k}$ — (D) The position vector reads the point’s three coordinates straight onto the three basis vectors, in order.
Point $P$ has coordinates $(3, -1, 2)$. What is $\vec{O P}$?
$-3 \hat{i} + \hat{j} - 2 \hat{k}$
$3 \hat{i} - \hat{j} + 2 \hat{k}$
$2 \hat{i} - \hat{j} + 3 \hat{k}$
$3 \hat{i} + \hat{j} + 2 \hat{k}$
Check your answer
$-3 \hat{i} + \hat{j} - 2 \hat{k}$ — Flipping every sign gives the vector from $P$ back to $O$, the reverse of what was asked for.
$2 \hat{i} - \hat{j} + 3 \hat{k}$ — Swapping the x- and z-coordinates puts $3$ on $\hat{k}$ and $2$ on $\hat{i}$, the wrong way round.
$3 \hat{i} + \hat{j} + 2 \hat{k}$ — The y-coordinate is $-1$, not $1$, dropping its negative sign changes the vector.
A position vector is given as $\vec{O P} = 4 \hat{i} + 0 \hat{j} - 2 \hat{k}$. What are the coordinates of $P$?
$(4, 0, -2)$
$(-4, 0, 2)$
$(4, -2, 0)$
$(0, 4, -2)$
Check your answer
✓ $(4, 0, -2)$ — (A) Reading the coefficients of $\hat{i}, \hat{j}, \hat{k}$ in order gives the coordinates $(4, 0, -2)$ directly.
$(-4, 0, 2)$ — Flipping every sign describes $\vec{P O}$’s coordinates read backward, not $P$’s actual coordinates from $\vec{O P}$.
$(4, -2, 0)$ — The $\hat{j}$-coefficient is $0$ and the $\hat{k}$-coefficient is $-2$, swapping them misplaces both.
$(0, 4, -2)$ — The $\hat{i}$-coefficient is $4$, not $0$, swapping the first two coordinates misreads the vector.
▲CONCEPT
Some vectors come up often enough to get names.
Shrink 4î + 3ĵ to half, then a quarter, then nothing. The zero vector is a point with no direction.
The ZERO VECTOR $\vec{0}$ has magnitude 0. Its start and its end are the same point, so it has no direction.
Three arrows of lengths 5, 2 and about 2.24. Divide each by its length and it lands on the unit circle.
A UNIT VECTOR has magnitude 1. Take any vector $\vec{a}$ with $\vec{a} \neq \vec{0}$. Divide it by its own length. You get the unit vector in the direction of $\vec{a}$:
$\hat{a} = \vec{a} / \left|\vec{a}\right|$.
The arrow 3 across, 4 up is 5 long. Left: the unit vector along it, length 1. Right: its negative, also 5 long.
Check it on the figure. The arrow $\vec{a} = 3 \hat{i} + 4 \hat{j}$ has length 5. So $\hat{a} = 3/5 \hat{i} + 4/5 \hat{j}$. Its length, $\sqrt{9/25 + 16/25}$, is 1.
The NEGATIVE of $\vec{a}$ is $-\vec{a}$. It has the same length as $\vec{a}$, and the opposite direction. In the figure it is $-3 \hat{i} - 4 \hat{j}$, and it is still 5 long.
Other names describe two or more vectors together.
Coinitial: three arrows from one point. Collinear: arrows on three parallel lines. Equal: one arrow, 3 across and 2 up, drawn from two places.
COINITIAL vectors start at the same point.
COLLINEAR vectors are parallel to the same line. Their lengths can differ, and they can point opposite ways.
EQUAL vectors have the same magnitude and the same direction. Where they start does not matter.
All three kinds are in the figure above. Three arrows leave one point. Three more lie on three parallel lines. The arrow $3 \hat{i} + 2 \hat{j}$ is drawn twice, from two different points. It is one vector.
Your turn: is $\hat{i} + \hat{j}$ a unit vector? If not, what is the unit vector in its direction?
A SCHOLAR INDIA REMEMBERS
Sort four arrows into kinds, and draw the lines first. Take $\vec{a} = 3 \hat{i} + 2 \hat{j}$ from the origin, the same steps from $(1, 1)$, then $-6 \hat{i} - 4 \hat{j}$ and $2 \hat{i} + 3 \hat{j}$. Equal means the same length and the same direction, wherever it starts. So the first two are one vector. The third is $-2 \vec{a}$, collinear with $\vec{a}$ but not equal. The fourth swaps the steps, so it is neither.
Dignaga · the kingfisher
Check yourself
Two vectors are called EQUAL when \_\_\_\_\_\_.
same magnitude and direction, any position
they lie along the same or parallel lines
they have the same magnitude but opposite direction
they start at the same point
Check your answer
✓ same magnitude and direction, any position — (A) Equal vectors match in both magnitude and direction; where either one happens to be drawn makes no difference at all.
they lie along the same or parallel lines — Lying along the same or parallel lines is the definition of COLLINEAR vectors , two collinear vectors can point opposite ways or differ in length.
they have the same magnitude but opposite direction — Same magnitude with opposite direction is exactly the NEGATIVE vector relationship, not equality.
they start at the same point — Sharing a starting point is the definition of COINITIAL vectors, two coinitial vectors can still differ completely in magnitude and direction.
What is the magnitude of the zero vector $\vec{0}$, and what direction does it have?
magnitude $0$, pointing toward the origin
magnitude $1$, no defined direction
magnitude $0$, direction along $\hat{i}$ by convention
magnitude $0$, no defined direction
Check your answer
magnitude $0$, pointing toward the origin — “Pointing toward the origin” assigns a direction where the definition explicitly assigns none.
magnitude $1$, no defined direction — Magnitude $1$ belongs to a UNIT vector, a different category entirely, the zero vector’s magnitude is $0$.
magnitude $0$, direction along $\hat{i}$ by convention — No default direction such as $\hat{i}$ is assigned anywhere, the zero vector’s direction is explicitly left undefined.
✓ magnitude $0$, no defined direction — (D) The zero vector’s magnitude is $0$, and with no length to point along, it carries no defined direction at all.
A vector $\vec{a}$ has magnitude $5$. What is the magnitude of the unit vector $\hat{a} = \vec{a} / \left|\vec{a}\right|$ built from it?
$5$
$1/5$
$1$
$0$
Check your answer
$5$ — $5$ is $\vec{a}$’s own magnitude, dividing by that magnitude is exactly what brings the length down to $1$.
$1/5$ — $1/5$ is $1/\left|\vec{a}\right|$, the number $\vec{a}$ is multiplied by, not the length of the result. Scaling a length of $5$ by $1/5$ gives a length of $1$.
✓ $1$ — (C) Dividing any nonzero vector by its own magnitude always produces a unit vector, magnitude exactly $1$.
$0$ — A unit vector and a zero vector sit at opposite ends of the scale, magnitude $1$ against magnitude $0$.
The walk as arrows: 3ĵ, then 4î from where it ended. The arrow from start to finish is the sum.
Go back to the walk. The first leg is 3 km north. Call it $\vec{a}$. The second leg is 4 km east, and it starts where the first one ended. Call it $\vec{b}$. The whole trip takes you from the start to the end, and one straight arrow does the same.
So one leg and then the next add up to one arrow, from the first start to the last end. Physics 11 adds forces and velocities the same way.
Triangle law: put b’s tail on a’s head. The third side, from a’s tail to b’s head, is a + b.
This is the TRIANGLE LAW of vector addition. Put the tail of $\vec{b}$ at the head of $\vec{a}$. The sum $\vec{a} + \vec{b}$ runs from the tail of $\vec{a}$ to the head of $\vec{b}$.
For the walk, $\vec{a} = 3 \hat{j}$ and $\vec{b} = 4 \hat{i}$. The sum is $4 \hat{i} + 3 \hat{j}$, the 5 km arrow from the start.
Parallelogram law: both arrows from one corner. The diagonal from that corner is the same sum, here 4î + 3ĵ.
There is a second way to draw it. Put both arrows at one point, and complete the parallelogram on them. The diagonal from that point is the sum. This is the PARALLELOGRAM LAW, and it gives the same arrow.
Arrows 4 across, 1 up and 1 across, 3 up, from the origin. Either order, or the diagonal, reaches 5 across, 4 up.
Test both laws on the figure above. Here $\vec{a} = 4 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 3 \hat{j}$. Go along $\vec{a}$ and then $\vec{b}$, and you reach $(5, 4)$. Go along $\vec{b}$ and then $\vec{a}$, and you reach $(5, 4)$ again. The diagonal ends there too. So $\vec{a} + \vec{b} = \vec{b} + \vec{a}$. Addition is COMMUTATIVE.
Three arrows walked one after another end at 6 across, 5 up. Bracket the first two or the last two: the end is the same.
Add three arrows, as the figure above does: $4 \hat{i} + \hat{j}$, then $-\hat{i} + 3 \hat{j}$, then $3 \hat{i} + \hat{j}$. The walk ends at $(6, 5)$. You can add the first two first, or the last two first. The end is the same. So $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$. Addition is ASSOCIATIVE.
Walk a, then walk −a. You are back where you started: a + (−a) is the zero vector.
The zero vector changes nothing: $\vec{a} + \vec{0} = \vec{a}$. And $\vec{a} + (-\vec{a}) = \vec{0}$. Walk somewhere and back, and you are at the start.
Check yourself
The TRIANGLE LAW of vector addition places $\vec{b}$’s tail at $\vec{a}$’s \_\_\_\_\_\_.
tail, so both vectors share a starting point
head, so $\vec{a} + \vec{b}$ runs tail-to-head
head, but the sum then runs head-to-tail instead
midpoint, so the sum runs through the middle of $\vec{a}$
Check your answer
tail, so both vectors share a starting point — Sharing a starting point is how the PARALLELOGRAM law is drawn, the triangle law instead joins one vector’s tail to the other’s head.
✓ head, so $\vec{a} + \vec{b}$ runs tail-to-head — (B) The triangle law joins $\vec{b}$’s tail to $\vec{a}$’s head, and the sum runs from $\vec{a}$’s own tail to $\vec{b}$’s own head.
head, but the sum then runs head-to-tail instead — The starting point is right, but the sum runs from $\vec{a}$’s tail to $\vec{b}$’s head, tail-to-head, not the other way round.
midpoint, so the sum runs through the middle of $\vec{a}$ — Neither addition law ever places a vector’s tail at another vector’s midpoint , the triangle law joins tail to head.
The PARALLELOGRAM LAW gives the sum $\vec{a} + \vec{b}$ as \_\_\_\_\_\_.
the shorter of the two diagonals, always
the diagonal from their shared starting point
the side of the parallelogram opposite $\vec{a}$
the average of the two vectors’ magnitudes
Check your answer
the shorter of the two diagonals, always — Which diagonal is shorter depends on the angle between $\vec{a}$ and $\vec{b}$ , the sum is always the one from their COMMON point, not whichever is shorter.
✓ the diagonal from their shared starting point — (B) Drawing $\vec{a}$ and $\vec{b}$ from a common point and completing the parallelogram, the diagonal from that same point is exactly $\vec{a} + \vec{b}$.
the side of the parallelogram opposite $\vec{a}$ — The side opposite $\vec{a}$ is a parallel copy of $\vec{a}$. It is not a sum of the two vectors at all.
the average of the two vectors’ magnitudes — Averaging magnitudes throws away direction entirely, the sum is a vector, not a single averaged number.
Vector addition is both commutative and associative. What does associativity mean here?
adding $\vec{0}$ to any vector changes it
the order of any three vectors may always be freely rearranged, just as with two
adding $\vec{0}$ to any vector changes it — Adding $\vec{0}$ leaves any vector unchanged, that is exactly what makes it the identity element, the opposite of this claim.
the order of any three vectors may always be freely rearranged, just as with two — Rearranging ORDER is commutativity’s job; associativity only says the GROUPING can change, not that the order can too.
$\vec{a} + \vec{b} = \vec{b} + \vec{a}$ — That equation is COMMUTATIVITY, swapping the ORDER of two vectors. Associativity is about grouping three additions, a different property.
✓ $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$ — (D) Associativity says the GROUPING of three additions makes no difference, bracket the first two or the last two, the total is the same.
$\vec{a} + \vec{0}$ equals \_\_\_\_\_\_.
a vector with $\vec{a}$’s magnitude but reversed direction
$\vec{0}$
$\vec{a}$, unchanged
$2 \vec{a}$
Check your answer
a vector with $\vec{a}$’s magnitude but reversed direction — Reversing the direction is what multiplying by $-1$ does; adding $-\vec{a}$ cancels $\vec{a}$ and leaves $\vec{0}$. Adding the zero vector changes nothing at all.
$\vec{0}$ — An identity element leaves a vector UNCHANGED by definition, claiming it erases the vector to $\vec{0}$ is the opposite of that role.
✓ $\vec{a}$, unchanged — (C) $\vec{0}$ is the identity element for vector addition, adding it to any vector leaves that vector exactly as it was.
$2 \vec{a}$ — Doubling would come from $\vec{a} + \vec{a}$, not from adding the zero vector , $\vec{0}$ contributes nothing to the sum.
▲CONCEPT
Twice 3ĵ is 6ĵ, the same direction and twice as long. Minus one times 3ĵ points the other way.
Take the first leg of the walk, 3 km north, and walk it twice. You go 6 km north. Walk it backwards, and you go 3 km south.
Multiplying by λ multiplies the length by the size of λ. A negative λ also reverses the arrow.
So a number can stretch a vector, or turn it round. For a real number $\lambda$, the vector $\lambda \vec{a}$ has length $\left|\lambda\right| \left|\vec{a}\right|$. If $\lambda > 0$, it points the same way as $\vec{a}$. If $\lambda < 0$, it points the opposite way. If $\lambda = 0$, it is the zero vector. This is SCALAR MULTIPLICATION.
The arrow 2 across, 1 up, then twice it, then minus it. All three lie on one line through the origin.
In the figure above, $\vec{a} = 2 \hat{i} + \hat{j}$ has length $\sqrt{5}$. Then $2 \vec{a} = 4 \hat{i} + 2 \hat{j}$ is twice as long, and points the same way. And $-\vec{a} = -2 \hat{i} - \hat{j}$ is just as long, and points the other way. All three lie on one line through $O$.
−4î − 2ĵ is −2 times 2î + ĵ, so they are collinear. 3î + 2ĵ is no multiple of it.
So a multiple never leaves the line. Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear exactly when $\vec{b} = \lambda \vec{a}$ for some number $\lambda$.
Every vector in components
Picture the map grid again. You reach any point by so many steps east, so many north and so many up. So every vector can be written as
Scalar components are the numbers 1, 2, 2. Vector components are the arrows î, 2ĵ and 2k̂, walked tail to head.
$a_1$, $a_2$, $a_3$ are the SCALAR COMPONENTS of $\vec{a}$, and $a_1 \hat{i}$, $a_2 \hat{j}$, $a_3 \hat{k}$ its VECTOR COMPONENTS. The length is Pythagoras used twice:
A box with edges 1, 2 and 2. The floor diagonal is the square root of 5. The long diagonal is 3.
Use the box above to check it. Take $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$. On the floor of the box, 1 and 2 give a diagonal of $\sqrt{1 + 4} = \sqrt{5}$. That diagonal and the height 2 give $\sqrt{5 + 4} = 3$.
In components, adding and scaling work number by number: (2, 1) + (1, 3) = (3, 4), and 3(2, 1) = (6, 3).
In components, every rule turns into adding or scaling numbers. Take $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ as well.
Your turn: is $6 \hat{i} - 3 \hat{j} + 3 \hat{k}$ collinear with $2 \hat{i} - \hat{j} + \hat{k}$? If it is, what is $\lambda$?
Check yourself
Multiplying $\vec{a}$ by a negative scalar $\lambda$ \_\_\_\_\_\_.
has no effect on the vector at all
keeps the direction the same and scales the magnitude by $\lambda$
reverses the direction and scales the magnitude by $\left|\lambda\right|$
reverses the direction only, leaving the magnitude unchanged
Check your answer
has no effect on the vector at all — A negative $\lambda$ always reverses the direction, so the vector always changes. Only $\lambda = 1$ leaves a vector exactly as it is.
keeps the direction the same and scales the magnitude by $\lambda$ — Direction stays the same only when $\lambda > 0$, a negative $\lambda$ reverses it.
✓ reverses the direction and scales the magnitude by $\left|\lambda\right|$ — (C) A negative $\lambda$ does two things at once: it reverses the direction, and it scales the magnitude by $\left|\lambda\right|$.
reverses the direction only, leaving the magnitude unchanged — The magnitude is scaled by $\left|\lambda\right|$ as well; it stays the same only when $\lambda = -1$.
For $\vec{a} = 3 \hat{i} - 4 \hat{k}$, what is the magnitude $\left|\vec{a}\right|$?
$7$ — Adding $3+4=7$ skips the squaring and square-rooting the magnitude formula actually calls for.
$1$ — Nothing about this vector’s components makes it a unit vector, its magnitude has to be computed, and it comes out to $5$, not $1$.
$\sqrt{7}$ — The formula needs the sum of the SQUARES, $9+16=25$, before the square root, not the square root of the components’ plain sum.
For $\vec{a} = \hat{i} + 2 \hat{j} - 2 \hat{k}$, what is $2 \vec{a}$?
$4 \hat{i} + 2 \hat{j} - 2 \hat{k}$
$2 \hat{i} + 4 \hat{j} - 4 \hat{k}$
$\hat{i} + 4 \hat{j} - 4 \hat{k}$
$2 \hat{i} + 2 \hat{j} - 2 \hat{k}$
Check your answer
$4 \hat{i} + 2 \hat{j} - 2 \hat{k}$ — The doubled value $4$ belongs on $\hat{j}$ (from $2$), not on $\hat{i}$ (from $1$) , the scaled values have been misplaced across components.
✓ $2 \hat{i} + 4 \hat{j} - 4 \hat{k}$ — (B) Scaling by $2$ multiplies every component by $2$: $2 \hat{i} + 4 \hat{j} - 4 \hat{k}$.
$\hat{i} + 4 \hat{j} - 4 \hat{k}$ — The $\hat{i}$-component must be doubled too, from $1$ to $2$, leaving it at $1$ skips exactly one of the three terms.
$2 \hat{i} + 2 \hat{j} - 2 \hat{k}$ — Scaling must apply to EVERY component, leaving the $\hat{j}$- and $\hat{k}$-components at their original values skips two of the three terms.
For $\vec{a} = \hat{i} + 2 \hat{j} - 2 \hat{k}$, compare $\left|2 \vec{a}\right|$ with $2 \left|\vec{a}\right|$.
they are equal; magnitude also scales by $2$
they can only be compared once a direction is chosen
$\left|2 \vec{a}\right|$ is twice $2 \left|\vec{a}\right|$
$\left|2 \vec{a}\right|$ is half of $2 \left|\vec{a}\right|$
Check your answer
✓ they are equal; magnitude also scales by $2$ — (A) $\left|\vec{a}\right| = \sqrt{1+4+4} = 3$, so $\left|2 \vec{a}\right| = 6$ and $2 \left|\vec{a}\right| = 6$, the same number either way.
they can only be compared once a direction is chosen — Magnitude is a single number by definition, with no direction attached, comparing two magnitudes never needs a direction to be chosen first.
$\left|2 \vec{a}\right|$ is twice $2 \left|\vec{a}\right|$ — $\left|\lambda \vec{a}\right| = \left|\lambda\right| \left|\vec{a}\right|$ scales by $\left|\lambda\right|$ exactly ONCE, there is no second doubling anywhere in the rule.
$\left|2 \vec{a}\right|$ is half of $2 \left|\vec{a}\right|$ — The two quantities come out equal, not one half of the other, this reverses a relationship that does not exist in the first place.
The angles α and β that 4î + 3ĵ makes with the axes fix its direction.
Take a vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$, with $\vec{a} \neq \vec{0}$. It makes an angle with each axis. Call them $\alpha$ with the $x$-axis, $\beta$ with the $y$-axis and $\gamma$ with the $z$-axis. Each angle runs from 0 to $\pi$.
Their cosines are the DIRECTION COSINES of $\vec{a}$:
Edges 2, minus 1 and 2 make a box with diagonal 3. The cosines are two thirds, minus one third and two thirds.
Check it on the figure. The vector is $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$. It is the long diagonal of a box with edges 2, $-1$ and 2. Its length is $\sqrt{4 + 1 + 4} = 3$. So $l = 2/3$, $m = -1/3$ and $n = 2/3$.
2î − ĵ + 2k̂ leans to the −y side, so β is obtuse and cos β = −1/3.
The middle one is negative, because $\vec{a}$ leans to the $-y$ side. So its angle with the $y$-axis is obtuse.
The unit vector of 4î + 3ĵ ends at (4/5, 3/5). Squares of the cosines add to 1.
Now square the three and add: $4/9 + 1/9 + 4/9 = 1$. This always happens. It is Pythagoras on the box, divided through by the long diagonal squared. So for every non-zero vector,
$l^2 + m^2 + n^2 = 1$.
Doubling the arrow doubles the direction ratios but not the direction. The cosines stay 2/3, −1/3, 2/3.
Any three numbers in the same ratio as $l$, $m$, $n$ are DIRECTION RATIOS of $\vec{a}$. The components 2, $-1$, 2 are one set. So are 4, $-2$, 4. Direction ratios do not have to square to 1.
Your turn: can $1/2$, $1/2$, $1/2$ be the direction cosines of a vector? Square them and add.
Check yourself
For a vector making angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes, the direction cosines $l, m, n$ must always satisfy \_\_\_\_\_\_.
$l + m + n = 1$
$l^2 + m^2 + n^2 = 0$
$l \cdot m \cdot n = 1$
$l^2 + m^2 + n^2 = 1$
Check your answer
$l + m + n = 1$ — Summing the cosines directly, without squaring first, does not equal $1$ in general , only the sum of their SQUARES does.
$l^2 + m^2 + n^2 = 0$ — A sum of squares equal to zero would force $l=m=n=0$ at once, meaning the vector makes a right angle with all three axes simultaneously, impossible.
$l \cdot m \cdot n = 1$ — Multiplying the three cosines together is a different quantity altogether from the identity, which sums their squares.
✓ $l^2 + m^2 + n^2 = 1$ — (D) The sum of the squares of the three direction cosines always equals $1$, whatever the vector’s own magnitude is.
How do direction ratios $a, b, c$ relate to the direction cosines $l, m, n$ of a vector?
direction ratios and direction cosines are two names for the same three numbers
direction ratios are only defined when the vector is already a unit vector
direction ratios always equal the cosines multiplied by $100$
$l = a/\sqrt{a^2 + b^2 + c^2}$: proportional to the cosines, not equal
Check your answer
direction ratios and direction cosines are two names for the same three numbers — Direction ratios need not be unit-length like cosines are, that is exactly what being merely proportional allows; they are not the same three numbers.
direction ratios are only defined when the vector is already a unit vector — Direction ratios are read straight off any vector’s components, unit-length or not, the restriction to unit vectors is not part of the definition.
direction ratios always equal the cosines multiplied by $100$ — No fixed factor like $100$ appears in the definition. Any three numbers proportional to the cosines are direction ratios, whatever the factor.
✓ $l = a/\sqrt{a^2 + b^2 + c^2}$: proportional to the cosines, not equal — (D) Direction ratios are any numbers proportional to the cosines. Dividing each by $\sqrt{a^2 + b^2 + c^2}$, the square root of the sum of their squares, recovers the cosines exactly.
A vector has direction ratios $2, -3, 6$. The ratios’ own length, $\sqrt{2^2 + (-3)^2 + 6^2}$, is $7$. What is its direction cosine $l$?
$2/7$
$7/2$
$2/49$
$-3/7$
Check your answer
✓ $2/7$ — (A) $l = 2/7$: the first direction ratio divided by $\sqrt{2^2 + (-3)^2 + 6^2} = 7$.
$7/2$ — The rule divides the RATIO by the root, $2/7$. Inverting it gives $7/2$ instead.
$2/49$ — Dividing by $49$, the sum of the squares, instead of its square root $7$ uses the wrong power.
$-3/7$ — $-3/7$ is $m$, built from the second direction ratio $-3$. $l$ uses the first ratio, $2$.
Worked example
Direction cosines of a given vector
$\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ Find the length first. Then divide each component by it.
$\left|\vec{a}\right| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$ Square the components, add them and take the root.
$l = 2/3$, $m = -1/3$, $n = 2/3$ Each direction cosine is a component divided by the length, 3.
$l^2 + m^2 + n^2 = 4/9 + 1/9 + 4/9 = 9/9 = 1$ The squares add to 1, as they must for any non-zero vector.
Check: $l \hat{i} + m \hat{j} + n \hat{k} = 2/3 \hat{i} - 1/3 \hat{j} + 2/3 \hat{k} = \vec{a}/3 = \hat{a}$ The direction cosines are the components of the unit vector along $\vec{a}$. Its length is 1.
The worked example’s answer drawn: a has length 3, and â, one third of it, has length 1.
Check yourself
For $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$, what is $\left|\vec{a}\right|$?
$9$
$\sqrt{5}$
$5$
$3$
Check your answer
$9$ — $9$ is the sum of squares BEFORE the square root, the magnitude itself needs that final square root, giving $3$.
$\sqrt{5}$ — This sum leaves out the third component’s square, $2^2=4$, all three squared components, $4+1+4=9$, must be summed first.
$5$ — Adding $\left|2\right|+\left|-1\right|+\left|2\right|=5$ skips the squaring and square-rooting the magnitude formula actually requires.
For $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$, with $\left|\vec{a}\right| = 3$, what is the direction cosine $m$?
$1/3$
$-1/3$
$-1/9$
$-3$
Check your answer
$1/3$ — The $\hat{j}$-component is $-1$, not $1$, dropping its negative sign gives the wrong cosine.
✓ $-1/3$ — (B) $m = b/r = (-1)/3 = -1/3$, using the $\hat{j}$-component over the magnitude.
$-1/9$ — Direction cosines divide by the magnitude itself, $3$, not by its square, $9$.
$-3$ — The definition divides the COMPONENT by the magnitude, $-1/3$, inverting it gives $-3$ instead.
For $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}$, the direction cosines are $l=2/3$, $m=-1/3$, $n=2/3$. What does checking $l^2+m^2+n^2$ confirm?
it confirms the direction ratios equal the direction cosines exactly
it confirms that $\vec{a}$ is a unit vector
it confirms the identity $l^2+m^2+n^2=1$ every cosine set obeys
it recomputes $\left|\vec{a}\right|$ from scratch
Check your answer
it confirms the direction ratios equal the direction cosines exactly — Here $\left|\vec{a}\right|=3 \neq 1$, so the direction ratios ($2,-1,2$) and direction cosines ($2/3,-1/3,2/3$) are proportional, not equal, the check says nothing about them being identical.
it confirms that $\vec{a}$ is a unit vector — $\vec{a}$’s own magnitude is $3$, not $1$, it is the COSINES, already divided by that magnitude, that satisfy the sum-of-squares-equals-one identity, not $\vec{a}$ itself.
✓ it confirms the identity $l^2+m^2+n^2=1$ every cosine set obeys — (C) $4/9 + 1/9 + 4/9 = 9/9 = 1$, confirming the identity every valid set of direction cosines must satisfy.
it recomputes $\left|\vec{a}\right|$ from scratch — The cosines were already built by dividing by $\left|\vec{a}\right|$, checking their squared sum confirms the identity, it does not recompute the magnitude from scratch.
Cut AB into three equal pieces. The point two pieces from A divides AB in the ratio 2 : 1.
In Class 10 you found the point that divides a line segment in a given ratio. Say the ratio is 2 to 1. Then the point is two thirds of the way from the first end to the second. You found its two coordinates one at a time, with the same rule for each.
Each coordinate divides the same way: x goes 1, 5, 7 and y goes 4, 2, 1.
The Class 10 formula
The ends were $A(x_1, y_1)$ and $B(x_2, y_2)$, and the ratio was $m : n$. The point was
$((m x_2 + n x_1)/(m + n), (m y_2 + n y_1)/(m + n))$.
r = (1/3)a + (2/3)b: a third of A’s position vector, then two thirds of B’s, reaches R(5, 2).
Vectors give the same point in one line. Let $A$ and $B$ have position vectors $\vec{a}$ and $\vec{b}$. Let $R$ be on the line $A B$, with $A R : R B = m : n$.
If $R$ lies between $A$ and $B$, this is INTERNAL DIVISION. The position vector of $R$ is
$\vec{r} = (m \vec{b} + n \vec{a})/(m + n)$.
Order matters. 2 : 1 puts R nearer B at (5, 2); 1 : 2 puts it nearer A at (3, 3).
Look at where each ratio number goes. $m$ is the part of the ratio on $A$’s side ($A R$), and it multiplies $\vec{b}$. So a bigger $m$ pulls $R$ towards $B$.
Ends (1, 4) and (7, 1). Internal 2 to 1: (5, 2). External 3 to 1: (10, minus 0.5). Midpoint: (4, 2.5).
Let us check it on the first panel of the figure:
$A$ is $(1, 4)$, $B$ is $(7, 1)$, ratio $2 : 1$.
$\vec{r} = (2(7 \hat{i} + \hat{j}) + (\hat{i} + 4 \hat{j}))/3 = 5 \hat{i} + 2 \hat{j}$, the point $(5, 2)$. It sits nearer $B$, as the larger share says.
Check: the Class 10 formula gives $((14 + 1)/3, (2 + 4)/3) = (5, 2)$, the same point.
Beyond B: AB is two lengths of RB, so AR is three. That is external division in the ratio 3 : 1.
If $R$ lies on the line $A B$ but outside the segment, this is EXTERNAL DIVISION. Then, with $m \neq n$,
$\vec{r} = (m \vec{b} - n \vec{a})/(m - n)$.
A point past B
The second panel of the three-panel figure above divides $A B$ externally in the ratio $3 : 1$:
$\vec{r} = (3 \vec{b} - \vec{a})/2$.
$= (20 \hat{i} - \hat{j})/2 = 10 \hat{i} - 1/2 \hat{j}$, the point $(10, -0.5)$, just past $B$.
Halfway between A and B
The MIDPOINT is internal division with $m = n = 1$. So $\vec{r} = (\vec{a} + \vec{b})/2$. In the third panel of the three-panel figure, it is $(4, 2.5)$.
Your turn: divide the same $A B$ externally in the ratio $2 : 1$. Where does the point land, and which end is it past?
A SCHOLAR INDIA REMEMBERS
The external formula comes with a condition, $m \neq n$. Do you need it, or is it there for no reason? Try $m = n = 1$. The formula gives $(\vec{b} - \vec{a})/0$, which is no point at all. Now look along the line. A point outside $A B$ is always nearer one end than the other. So no outside point splits $A B$ in the ratio $1 : 1$. The condition says exactly that.
Dignaga · the kingfisher
Check yourself
Dividing $A B$ INTERNALLY in ratio $m:n$ gives the point $R$ at \_\_\_\_\_\_.
$\vec{r} = (m \vec{a} + n \vec{b})/(m+n)$
$\vec{r} = (m \vec{b} + n \vec{a})/(m-n)$
$\vec{r} = (m \vec{b} - n \vec{a})/(m+n)$
$\vec{r} = (m \vec{b} + n \vec{a})/(m+n)$
Check your answer
$\vec{r} = (m \vec{a} + n \vec{b})/(m+n)$ — $m$ weights $\vec{b}$ and $n$ weights $\vec{a}$, swapping them puts each weight on the wrong point.
$\vec{r} = (m \vec{b} + n \vec{a})/(m-n)$ — Dividing by $m-n$ belongs to the EXTERNAL case, internal division always divides by $m+n$.
$\vec{r} = (m \vec{b} - n \vec{a})/(m+n)$ — Subtracting $n \vec{a}$ is the EXTERNAL division formula’s numerator, the internal case adds the two weighted terms.
✓ $\vec{r} = (m \vec{b} + n \vec{a})/(m+n)$ — (D) Internal division weights $\vec{b}$ by $m$ and $\vec{a}$ by $n$, adds them, and divides by $m+n$.
How does the EXTERNAL division formula for ratio $m:n$ differ from the internal one?
both the numerator’s sign and the denominator switch from $+$ to $-$
the two formulas are identical; ‘internal’ and ‘external’ only describe where $R$ ends up
only the denominator switches from $+$ to $-$; the numerator stays additive
only the numerator switches to subtraction; the denominator stays $m+n$
Check your answer
✓ both the numerator’s sign and the denominator switch from $+$ to $-$ — (A) External division switches BOTH the numerator’s addition to subtraction and the denominator from $m+n$ to $m-n$.
the two formulas are identical; ‘internal’ and ‘external’ only describe where $R$ ends up — The formulas genuinely differ. Both the numerator’s sign and the denominator switch between internal and external division, not just the resulting point’s location.
only the denominator switches from $+$ to $-$; the numerator stays additive — The numerator also switches. External division subtracts $n \vec{a}$, it does not add it.
only the numerator switches to subtraction; the denominator stays $m+n$ — The denominator also switches, to $m-n$. It does not stay at $m+n$ for external division.
The midpoint of segment $A B$ is the section-formula result for which ratio $m:n$?
$m = 2, n = 1$
$m = 0, n = 1$
$m = 1, n = 0$
$m = n = 1$
Check your answer
$m = 2, n = 1$ — Ratio $2:1$ weights $B$ more heavily, landing closer to $B$ than to $A$, not the equally-weighted midpoint.
$m = 0, n = 1$ — With $m=0$ the formula reduces to $\vec{r} = \vec{a}$, that is point $A$ itself, not the midpoint.
$m = 1, n = 0$ — With $n=0$ the formula reduces to $\vec{r} = \vec{b}$, that is point $B$ itself, not the point halfway between $A$ and $B$.
✓ $m = n = 1$ — (D) The midpoint is exactly the internal-division case with $m = n = 1$, weighting $\vec{a}$ and $\vec{b}$ equally.
The worked example in space: R at (−1/3, 4/3, 1/3) is twice as far from A as from B.
Worked example
Internal division by the section formula
$A$ at $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$, $B$ at $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ Two points are given by their position vectors. Find the point $R$ that divides $A B$ internally in the ratio $2 : 1$.
$\vec{r} = (2 \vec{b} + 1 \vec{a})/3$ Use internal division with $m = 2$ and $n = 1$. $B$ is multiplied by $m$, the part of the ratio next to $A$, and $A$ by $n$, the part next to $B$.
$2 \vec{b} = -2 \hat{i} + 2 \hat{j} + 2 \hat{k}$, $1 \vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ Multiply each position vector by its weight.
$\vec{r} = (-\hat{i} + 4 \hat{j} + \hat{k})/3$ Add the two vectors, component by component.
*$\vec{r} = (-1/3) \hat{i} + (4/3) \hat{j} + (1/3) \hat{k}$* Divide each component by 3.
Check: $\vec{r} - \vec{a} = -4/3 \hat{i} - 2/3 \hat{j} + 4/3 \hat{k} = 2(\vec{b} - \vec{r})$ The arrow from $A$ to $R$ is twice the arrow from $R$ to $B$, the same way. So $R$ is between them, with $A R : R B = 2 : 1$.
Check yourself
For $A$ at $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ and $B$ at $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$, dividing $A B$ internally in ratio $2:1$, what is the numerator of $\vec{r}$ before dividing by $3$?
$-2 \hat{i} + 2 \hat{j} + 2 \hat{k}$
$-\hat{i} + 4 \hat{j} + \hat{k}$
$-3 \hat{i} + 0 \hat{j} + 3 \hat{k}$
$\hat{i} + 4 \hat{j} - \hat{k}$
Check your answer
$-2 \hat{i} + 2 \hat{j} + 2 \hat{k}$ — This is exactly $2 \vec{b}$ alone, the numerator still needs $\vec{a}$ added on top of it.
$-3 \hat{i} + 0 \hat{j} + 3 \hat{k}$ — Internal division ADDS the two weighted terms, subtracting $\vec{a}$ instead of adding it belongs to a different computation entirely.
$\hat{i} + 4 \hat{j} - \hat{k}$ — The first and third components come from combining BOTH vectors’ contributions , using only $\vec{a}$’s signs there skips $2 \vec{b}$’s actual effect.
Continuing the same division, what is $\vec{r}$ once the numerator $-\hat{i} + 4 \hat{j} + \hat{k}$ is divided by $3$?
$(-1/3) \hat{i} + (2/3) \hat{j} + (1/3) \hat{k}$
$(-1/3) \hat{i} + (4/3) \hat{j} + (1/3) \hat{k}$
$-1 \hat{i} + 4 \hat{j} + 1 \hat{k}$
$(-1/3) \hat{i} + (4/3) \hat{j} - (1/3) \hat{k}$
Check your answer
$(-1/3) \hat{i} + (2/3) \hat{j} + (1/3) \hat{k}$ — $4$ divided by $3$ is $4/3$, not $2/3$, this understates the $\hat{j}$-component’s actual value.
✓ $(-1/3) \hat{i} + (4/3) \hat{j} + (1/3) \hat{k}$ — (B) Dividing each component of the numerator by $3$ gives $(-1/3) \hat{i} + (4/3) \hat{j} + (1/3) \hat{k}$.
$-1 \hat{i} + 4 \hat{j} + 1 \hat{k}$ — This is the numerator BEFORE dividing by $3$, the division step still needs to be applied to every component.
$(-1/3) \hat{i} + (4/3) \hat{j} - (1/3) \hat{k}$ — The numerator’s $\hat{k}$-component is $+1$, giving $+1/3$ after dividing, flipping just that one sign misreads the division.
If instead $A B$ were divided internally in ratio $1:2$ (reversing which weight goes with which point) using the same $\vec{a}, \vec{b}$, would $\vec{r}$ change from the $2:1$ case?
yes, but only the sign of $\vec{r}$ flips, nothing else changes
no, ratio order only matters for external division, not internal
yes, reversing the ratio gives a different, closer-to-the-other-point $\vec{r}$
no, the section formula gives the same point regardless of how the ratio is written
Check your answer
yes, but only the sign of $\vec{r}$ flips, nothing else changes — The new point is a genuinely different weighted combination, not simply the old one with its sign flipped, its individual components come out different, not just negated.
no, ratio order only matters for external division, not internal — Internal division is just as order-sensitive as external, both formulas assign the FIRST ratio term to $\vec{b}$ and the second to $\vec{a}$, so swapping them changes the result either way.
✓ yes, reversing the ratio gives a different, closer-to-the-other-point $\vec{r}$ — (C) Ratio $1:2$ weights $\vec{a}$ more heavily instead of $\vec{b}$, giving $\vec{r} = (\vec{b} + 2 \vec{a})/3$, a genuinely different point from the $2:1$ case.
no, the section formula gives the same point regardless of how the ratio is written — $m:n$ and $n:m$ weight $\vec{a}$ and $\vec{b}$ differently, they describe different points on the segment, one closer to $A$, one closer to $B$.
A bag pulled by its strap. Only the part of the pull along the floor moves it forward.
Think of pulling a school bag along the floor by its strap. The strap slants up. Only part of your pull drags the bag forward. The rest pulls it up, and that part does not move it along the floor.
A pull of 6 at 60 degrees to the floor: 3 of it goes forward, about 5.2 goes up.
In Physics 11 you met the work done by such a pull. A force of size $F$ moves an object a distance $d$. If the force makes an angle $\theta$ with the motion, the work done is $F d \cos \theta$.
So the work needs two lengths and the cosine of the angle between them. The same mix of two vectors turns up all through geometry, so it gets a name.
To read the angle between two vectors, slide them until their tails meet. Here θ is 60 degrees.
Let $\vec{a}$ and $\vec{b}$ be non-zero vectors with angle $\theta$ between them, where $0 \leq \theta \leq \pi$. Their SCALAR PRODUCT, or DOT PRODUCT, is
If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \cdot \vec{b} = 0$. The answer is always a number, not a vector.
Lengths 4 and 5, at 60 degrees. The shadow of the first along the second is 2. Dot product: 2 times 5, or 10.
Look at the figure. A line drops at right angles from the tip of $\vec{a}$ onto the line of $\vec{b}$. The piece it cuts off is $\left|\vec{a}\right| \cos \theta$, the shadow of $\vec{a}$ along $\vec{b}$. This is the PROJECTION of $\vec{a}$ on $\vec{b}$, the number $(\vec{a} \cdot \vec{b})/\left|\vec{b}\right|$.
The projection has a vector version too. Multiply the number by the unit vector $\hat{b}$, and you get the projection vector, an arrow along $\vec{b}$: $((\vec{a} \cdot \vec{b})/\left|\vec{b}\right|^2) \vec{b}$.
In an exam, when a question says only “projection”, it means the number, not the vector.
So the dot product is the projection times the length of $\vec{b}$. Check it on the figure:
$\vec{a} \cdot \vec{a} = \left|\vec{a}\right|^2$, since the angle is 0.
Put the first two rules together and you get the square of a sum: $\left|\vec{a} + \vec{b}\right|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \left|\vec{a}\right|^2 + 2 \vec{a} \cdot \vec{b} + \left|\vec{b}\right|^2$. Multiply out the right side and the cross terms give $2 \vec{a} \cdot \vec{b}$, using the distributive rule twice.
î · î = 1. Two different axis vectors meet at a right angle, so î · ĵ = 0.
Combine the fourth rule ($\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$) with the DISTRIBUTIVE rule, and multiply out $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k})$: every cross term vanishes, and only the matching ones survive. That gives the dot product in components:
The sign reads the angle: a · b is 8 when acute, 0 at a right angle, and −8 when obtuse.
For non-zero vectors, the sign also tells you the kind of angle. A positive dot product means $0 \leq \theta < \pi/2$. Zero means a right angle. A negative dot product means $\pi/2 < \theta \leq \pi$.
Your turn: take $\vec{a} = \hat{i} + 2 \hat{j}$ and $\vec{b} = 3 \hat{i} - 2 \hat{j}$. Find $\vec{a} \cdot \vec{b}$. Is the angle between them acute or obtuse?
A SCHOLAR INDIA REMEMBERS
Ask what kind of thing $\vec{a} \cdot \vec{b}$ is, before you ask what it equals. Read the definition. $\left|\vec{a}\right| \left|\vec{b}\right| \cos \theta$ is three numbers multiplied, so the answer is a number, with no direction. With lengths 4 and 5 at 60 degrees it is $10$, not an arrow of length $10$. The work done, $F d \cos \theta$, is the same kind. The cross product, next, is the other kind. It ends in $\hat{n}$, so it points.
Dignaga · the kingfisher
Check yourself
The scalar (dot) product $\vec{a} \cdot \vec{b}$ is \_\_\_\_\_\_.
a number, equal to $\left|\vec{a}\right| + \left|\vec{b}\right|$
a number $\left|\vec{a}\right| \cdot \left|\vec{b}\right| \cdot \cos \theta$
a vector along $\vec{a}$’s own direction
a vector perpendicular to both $\vec{a}$ and $\vec{b}$
Check your answer
a number, equal to $\left|\vec{a}\right| + \left|\vec{b}\right|$ — Adding the magnitudes drops the angle entirely, the actual formula multiplies them together and scales by $\cos \theta$.
✓ a number $\left|\vec{a}\right| \cdot \left|\vec{b}\right| \cdot \cos \theta$ — (B) The dot product multiplies the two magnitudes by $\cos \theta$, and the result is a single number.
a vector along $\vec{a}$’s own direction — A number has no direction attached to it, assigning it “$\vec{a}$’s own direction” treats a scalar result as if it were a vector.
a vector perpendicular to both $\vec{a}$ and $\vec{b}$ — A vector perpendicular to both inputs is exactly what the CROSS product returns , the dot product returns a number instead.
The PROJECTION of $\vec{a}$ on $\vec{b}$ is defined as \_\_\_\_\_\_.
$(\vec{a} \cdot \vec{b}) / \left|\vec{b}\right|$
$\vec{a} \cdot \vec{b}$, with no division at all
$\left|\vec{a}\right| \cdot \left|\vec{b}\right|$
$(\vec{a} \cdot \vec{b}) / \left|\vec{a}\right|$
Check your answer
✓ $(\vec{a} \cdot \vec{b}) / \left|\vec{b}\right|$ — (A) The projection divides the dot product by the magnitude of the vector being projected ONTO, here $\vec{b}$.
$\vec{a} \cdot \vec{b}$, with no division at all — Skipping the division leaves the raw dot product, whose units are length-squared , the projection needs one more division by a length to become a length itself.
$\left|\vec{a}\right| \cdot \left|\vec{b}\right|$ — Multiplying only the magnitudes drops $\cos \theta$ entirely, the projection depends on the angle between the two vectors, not just their lengths.
$(\vec{a} \cdot \vec{b}) / \left|\vec{a}\right|$ — Dividing by $\left|\vec{a}\right|$ projects $\vec{b}$ onto $\vec{a}$ instead, the question asks for $\vec{a}$’s projection ONTO $\vec{b}$, which divides by $\left|\vec{b}\right|$.
Solving the dot-product formula for the angle between $\vec{a}$ and $\vec{b}$ gives $\cos \theta$ equal to \_\_\_\_\_\_.
$\vec{a} \cdot \vec{b}$, with no division by either magnitude at all
Check your answer
✓ $(\vec{a} \cdot \vec{b}) / (\left|\vec{a}\right| \cdot \left|\vec{b}\right|)$ — (A) Dividing the dot product by the product of both magnitudes isolates $\cos \theta$ on its own.
$(\vec{a} \cdot \vec{b}) / \left|\vec{b}\right|$ — Dividing by only $\left|\vec{b}\right|$ gives the PROJECTION of $\vec{a}$ on $\vec{b}$ , the angle formula divides by BOTH magnitudes.
$\left|\vec{a}\right| \cdot \left|\vec{b}\right| / (\vec{a} \cdot \vec{b})$ — This flips numerator and denominator, the dot product belongs on top, the product of magnitudes on the bottom.
$\vec{a} \cdot \vec{b}$, with no division by either magnitude at all — The raw dot product is not $\cos \theta$ on its own, it still needs dividing by both magnitudes to isolate the angle.
$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ always holds for the dot product. Does the same order-independence hold for a PROJECTION? Is the projection of $\vec{a}$ on $\vec{b}$ the same as the projection of $\vec{b}$ on $\vec{a}$?
no, dividing by only one magnitude breaks the swap symmetry
yes, projection is symmetric because it measures the angle between the two vectors
yes, since the dot product commutes, every formula built from it commutes too
no, but only because the dot product itself is not commutative
Check your answer
✓ no, dividing by only one magnitude breaks the swap symmetry — (A) The projection’s extra division by only ONE magnitude breaks the symmetry that the dot product alone has, swapping which vector gets divided by changes the answer unless the two magnitudes match or the vectors are perpendicular (then both projections are 0).
yes, projection is symmetric because it measures the angle between the two vectors — The angle between two vectors IS symmetric, but that is not what determines projection’s value, the asymmetric division by one vector’s magnitude is what breaks it.
yes, since the dot product commutes, every formula built from it commutes too — Commutativity of the dot product does not carry over automatically, the projection’s extra division by one specific magnitude is exactly the step that breaks the symmetry.
no, but only because the dot product itself is not commutative — The dot product IS commutative, always, the asymmetry comes from the division step in the projection formula, not from $\vec{a} \cdot \vec{b}$ itself.
For $\vec{a} = 2 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$, what is $\vec{a} \cdot \vec{b}$?
$3$
$2$
$-1$
$1$
Check your answer
$3$ — Adding $2+1=3$ drops the subtraction on the $\hat{j}$-term entirely, the second product is $-1$, not $+1$.
$2$ — The second component pair, $1$ and $-1$, also contributes $-1$ to the sum, stopping after the first term misses it.
$-1$ — Flipping the sign of $1$ gives $-1$, the two terms $2$ and $-1$ genuinely sum to a positive $1$.
$\left|\vec{a}\right| = \left|\vec{b}\right| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ Every component has size 1, so both lengths are $\sqrt{3}$.
$\cos \theta = (\vec{a} \cdot \vec{b})/(\left|\vec{a}\right| \cdot \left|\vec{b}\right|) = 1/(\sqrt{3} \cdot \sqrt{3}) = 1/3$ Put the dot product and the two lengths into the angle formula.
*$\theta = \cos^{-1}(1/3)$* The angle is the one whose cosine is $1/3$, with $0 \leq \theta \leq \pi$.
Check: $\left|\vec{a}\right| \left|\vec{b}\right| \cos \theta = (\sqrt{3})(\sqrt{3})(1/3) = 1 = \vec{a} \cdot \vec{b}$ The angle and the dot product agree. Since $\cos \theta > 0$, the angle is acute, about 70.5 degrees.
Check yourself
For $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$, what is $\vec{a} \cdot \vec{b}$?
$-1$ — The three products $1, 1, -1$ genuinely sum to a positive $1$, flipping the sign of that total does not match the arithmetic.
$-3$ — Only $\vec{b}$’s $\hat{k}$-component is negative, treating all three of its components as negative overstates how the two vectors actually differ.
$3$ — $3 = (1)(1) + (1)(1) + (1)(1)$ drops the minus sign on $\vec{b}$’s $\hat{k}$-component. The third product is $(1)(-1) = -1$.
For the same $\vec{a}$ and $\vec{b}$, what are $\left|\vec{a}\right|$ and $\left|\vec{b}\right|$?
✓ both equal $\sqrt{3}$ — (A) $\left|\vec{a}\right| = \left|\vec{b}\right| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$, since squaring erases the sign on $\vec{b}$’s $\hat{k}$-component.
$\left|\vec{a}\right| = \sqrt{3}$, $\left|\vec{b}\right| = 1$ — Squaring erases the sign on $\vec{b}$’s $\hat{k}$-component entirely, $(-1)^2=1$, exactly the same contribution a positive $1$ would give.
both equal $\sqrt{2}$ — Dropping one component leaves only $1+1=2$ under the square root, all three components’ squares, $1+1+1=3$, must be included.
both equal $3$ — $3$ is the sum of the squared components, BEFORE the square root, the magnitude itself is $\sqrt{3}$.
Using $\vec{a} \cdot \vec{b} = 1$ and $\left|\vec{a}\right| = \left|\vec{b}\right| = \sqrt{3}$, what is $\cos \theta$ between them?
$\sqrt{3}$
$3$
$1/3$
$1/\sqrt{3}$
Check your answer
$\sqrt{3}$ — $\sqrt{3}$ is just one of the two magnitudes on its own, the dot product, $1$, still needs to enter the formula as the numerator.
$3$ — The dot product belongs on TOP of the fraction, inverting it, dividing the magnitude-product by the dot product instead, gives the reciprocal of the right answer.
$1/\sqrt{3}$ — The formula divides by BOTH magnitudes multiplied together, $\sqrt{3} \cdot \sqrt{3} = 3$, dividing by only one of them leaves the answer too large.
▲CONCEPT
Force 6 over a distance of 4: work 24 straight along, 12 at 60 degrees, and 0 at a right angle.
Now go back to the bag. Let your pull be the vector $\vec{F}$, and let the move along the floor be $\vec{d}$. The WORK DONE is
The dot product keeps only the part of the pull along the floor. That part is the projection of $\vec{F}$ on $\vec{d}$.
A force of 6 moves the object 4. At 60 degrees, 3 of it acts forward: work 12. At right angles: work 0.
Put in the numbers from the figure:
Pull $= 6$, distance $= 4$, angle $= 60$ degrees, in the left panel.
Forward part of the pull: $6 \cos(\pi/3) = 3$.
Work $= 3 \times 4 = 12$.
In the right panel the pull is at right angles to the motion. Then $\cos \theta = 0$, so $W = 0$. A pull at right angles to the motion does no work along the floor, however hard it is.
Check yourself
The work done by a constant force $\vec{F}$ over displacement $\vec{d}$ is \_\_\_\_\_\_.
$W = \vec{F} \times \vec{d}$
$W = \left|\vec{F}\right| \cdot \left|\vec{d}\right|$, regardless of angle
$W = \left|\vec{F}\right| + \left|\vec{d}\right|$
$W = \vec{F} \cdot \vec{d}$
Check your answer
$W = \vec{F} \times \vec{d}$ — The cross product returns a vector, work is a number, which is exactly what the DOT product, not the cross product, produces.
$W = \left|\vec{F}\right| \cdot \left|\vec{d}\right|$, regardless of angle — Dropping $\cos \theta$ ignores the angle between force and displacement, the exact factor that makes a perpendicular force do zero work.
$W = \left|\vec{F}\right| + \left|\vec{d}\right|$ — Adding the magnitudes ignores the angle between force and displacement entirely , the dot product’s $\cos \theta$ factor is exactly what work depends on.
✓ $W = \vec{F} \cdot \vec{d}$ — (D) Work is the dot product of the force and the displacement it acts through, $W = \vec{F} \cdot \vec{d}$.
Why does a force applied exactly perpendicular to the displacement do zero work?
because work is only ever defined for forces parallel to the displacement
because perpendicular forces have zero magnitude by definition
$\cos \theta = 0$ at $90^\circ$, so the dot product collapses regardless
because $\vec{d}$ itself becomes zero whenever a force is perpendicular to it
Check your answer
because work is only ever defined for forces parallel to the displacement — Work is defined for a force at ANY angle to the displacement. It just happens to come out to zero at exactly $90^\circ$.
because perpendicular forces have zero magnitude by definition — A perpendicular force can be arbitrarily large and still do zero work. Its magnitude is unrelated to its direction relative to $\vec{d}$.
✓ $\cos \theta = 0$ at $90^\circ$, so the dot product collapses regardless — (C) At $\theta = 90^\circ$, $\cos \theta = 0$. The whole product collapses to zero no matter how large $\vec{F}$ or $\vec{d}$ are.
because $\vec{d}$ itself becomes zero whenever a force is perpendicular to it — The displacement’s own size is set independently. A force being perpendicular to it does not shrink $\vec{d}$ to zero.
A force $\vec{F} = 3 \hat{i} + 4 \hat{j}$ moves an object through $\vec{d} = 2 \hat{i}$. What work is done?
$8$
$0$
$6$
$10$
Check your answer
$8$ — $8 = (4)(2)$ pairs the $\hat{j}$-component of $\vec{F}$ with the $\hat{i}$-component of $\vec{d}$. Only matching components multiply, and $\vec{d}$ has no $\hat{j}$-component, so that term is $(4)(0) = 0$.
$0$ — Nothing about $\vec{d}$ pointing along a single axis makes $\vec{F}$ perpendicular to it, $\vec{F}$’s $\hat{i}$-component of $3$ is exactly aligned with $\vec{d}$.
$10$ — $10 = \left|\vec{F}\right| \left|\vec{d}\right| = (5)(2)$ drops the angle. Work is $\left|\vec{F}\right| \left|\vec{d}\right| \cos \theta$, and here $\cos \theta = 3/5$, which gives $6$.
a = 2î and b = î + 2ĵ lie flat on the floor. Their cross product, 4k̂, stands straight up.
There is a second product of two vectors. Its answer is a vector.
Let $\vec{a}$ and $\vec{b}$ be non-zero vectors that are not parallel, with angle $\theta$ between them, where $0 \leq \theta \leq \pi$. Their VECTOR PRODUCT, or CROSS PRODUCT, is
Curl the fingers of the right hand from a to b; the thumb gives a × b. The dashed arrow is the wrong side.
Here $\hat{n}$ is a unit vector at right angles to both $\vec{a}$ and $\vec{b}$. There are two such unit vectors, pointing opposite ways. The RIGHT-HAND RULE picks one. Curl the fingers of your right hand from $\vec{a}$ round to $\vec{b}$. Your thumb points along $\hat{n}$.
Same lengths, three angles: the cross product’s length is 4 at 90 degrees, 2 at 30 degrees, and 0 when they lie along one line.
If $\vec{a}$ and $\vec{b}$ are parallel, then $\sin \theta = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. If either vector is $\vec{0}$, the cross product is $\vec{0}$ too.
Going round in order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
And $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$, since each is parallel to itself.
The cross product as a determinant. Expanding along the top row gives each component; the middle one carries a minus sign.
In components, the cross product is a determinant. Put $\hat{i}$, $\hat{j}$, $\hat{k}$ in row 1. Put the components of $\vec{a}$ in row 2, and those of $\vec{b}$ in row 3. Then expand along row 1, as in Chapter 4:
Check the determinant on î × ĵ: rows 1, 0, 0 and 0, 1, 0 leave only k̂.
Test the determinant on $\hat{i} \times \hat{j}$. Row 2 is $1, 0, 0$, and row 3 is $0, 1, 0$. The expansion gives $0 \hat{i} - 0 \hat{j} + 1 \hat{k} = \hat{k}$, as the right-hand rule says.
Two arrows span a shaded plane. Their cross product, minus i plus 3j plus 5k, stands at right angles to it.
The figure draws $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$ on a shaded plane. Their cross product, $-\hat{i} + 3 \hat{j} + 5 \hat{k}$, stands at right angles to that plane. The worked example below finds it.
The cross product is DISTRIBUTIVE too: it spreads over a sum: $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
Check yourself
The vector (cross) product $\vec{a} \times \vec{b}$ is \_\_\_\_\_\_.
a vector, perpendicular to both $\vec{a}$ and $\vec{b}$
a vector lying in the same plane as $\vec{a}$ and $\vec{b}$
a number, equal to $\left|\vec{a}\right| \cdot \left|\vec{b}\right| \cdot \sin \theta$
a vector along $\vec{a}$’s own direction
Check your answer
✓ a vector, perpendicular to both $\vec{a}$ and $\vec{b}$ — (A) The cross product’s result is itself a vector, perpendicular to both of the two vectors that produced it.
a vector lying in the same plane as $\vec{a}$ and $\vec{b}$ — The cross product points PERPENDICULAR to the plane containing $\vec{a}$ and $\vec{b}$, it does not lie within that plane at all.
a number, equal to $\left|\vec{a}\right| \cdot \left|\vec{b}\right| \cdot \sin \theta$ — That expression is only the cross product’s MAGNITUDE, the cross product itself is a full vector, not the number describing its length.
a vector along $\vec{a}$’s own direction — The cross product is perpendicular to $\vec{a}$, not along it, assigning $\vec{a}$’s own direction contradicts the definition directly.
In the determinant form of $\vec{a} \times \vec{b}$, what goes in the first row?
the components of $\vec{a}$
the components of $\vec{b}$
$\hat{i}, \hat{j}, \hat{k}$
the numbers $1, 1, 1$
Check your answer
the components of $\vec{a}$ — $\vec{a}$’s components go in the SECOND row of the determinant, the first row is reserved for $\hat{i}, \hat{j}, \hat{k}$.
the components of $\vec{b}$ — $\vec{b}$’s components go in the THIRD row, the first row carries the basis vectors, not either input vector.
✓ $\hat{i}, \hat{j}, \hat{k}$ — (C) The first row of the determinant carries the three basis vectors $\hat{i}, \hat{j}, \hat{k}$.
the numbers $1, 1, 1$ — No row of plain ones appears anywhere in the construction, the first row holds the three basis vectors themselves.
Expanding the cross-product determinant along its first row, the $\hat{j}$ term carries which sign?
a plus sign, matching the $\hat{i}$ term
a minus sign, giving $-(a_1 b_3 - a_3 b_1) \hat{j}$
the $\hat{j}$ term is always zero
it depends on the specific numbers in $\vec{a}$ and $\vec{b}$
Check your answer
a plus sign, matching the $\hat{i}$ term — A cofactor expansion alternates sign term by term, the middle ($\hat{j}$) term is always the one that flips to minus.
✓ a minus sign, giving $-(a_1 b_3 - a_3 b_1) \hat{j}$ — (B) The middle term of a cofactor expansion always carries a minus sign, here, the $\hat{j}$ term.
the $\hat{j}$ term is always zero — The $\hat{j}$ term vanishes only for specific choices of $\vec{a}$ and $\vec{b}$, not as a general rule of the expansion itself.
it depends on the specific numbers in $\vec{a}$ and $\vec{b}$ — The alternating sign pattern comes from the term’s POSITION in the expansion, not from whatever numbers $\vec{a}$ and $\vec{b}$ happen to have.
For $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = \hat{j}$, what is $\vec{a} \times \vec{b}$?
$\hat{j}$
$-\hat{i} + \hat{k}$
$\hat{i} - \hat{k}$
$\hat{i} + \hat{k}$
Check your answer
$\hat{j}$ — $\hat{j}$ is $\vec{b}$ itself. The cross product is perpendicular to $\vec{b}$, so it cannot be $\hat{j}$; expanding the determinant gives $\hat{i} + \hat{k}$.
$-\hat{i} + \hat{k}$ — The $\hat{i}$-component works out to $+1$, not $-1$, flipping just that one sign misreads the expansion.
$\hat{i} - \hat{k}$ — Redo the $\hat{k}$ entry of the expansion, it comes out to $+1$ with only that one sign reversed here.
For $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$, use the determinant form to find $\vec{a} \times \vec{b}$.
$\vec{0}$
$-2 \hat{k}$
$-2 \hat{i}$
$2 \hat{k}$
Check your answer
$\vec{0}$ — Looking different is not what a zero cross product requires, only truly parallel vectors give $\vec{0}$, and $\vec{a}$, $\vec{b}$ here are not parallel.
This is called ANTI-COMMUTATIVITY (the dot product, by contrast, is commutative). The length stays the same. Only the direction flips.
Swap the two rows and the determinant changes sign: 3k̂ becomes −3k̂. That is why b × a = −a × b.
The determinant shows why. Swapping $\vec{a}$ and $\vec{b}$ swaps rows 2 and 3. Chapter 4 showed that swapping two rows of a determinant changes its sign. The right-hand rule says the same. Curl your fingers from $\vec{b}$ to $\vec{a}$, and your thumb points the other way.
In the other order, $\vec{b} \times \vec{a} = ((1)(1) - (2)(2)) \hat{k} = -3 \hat{k}$.
The dot product of the same two vectors does not turn round: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = 4$.
Your turn: use $\hat{i} \times \hat{j} = \hat{k}$. What is $\hat{j} \times \hat{i}$?
Check yourself
For two non-parallel vectors $\vec{a}$ and $\vec{b}$, swapping the order of a cross product, comparing $\vec{b} \times \vec{a}$ with $\vec{a} \times \vec{b}$, gives \_\_\_\_\_\_.
the zero vector, always
the same magnitude, opposite direction
the same vector, unchanged
a vector of different magnitude, same direction
Check your answer
the zero vector, always — Swapping order never forces a zero result by itself, the cross product only vanishes when the two vectors are parallel, which a swap does not change.
✓ the same magnitude, opposite direction — (B) Swapping the order reverses the cross product’s direction while leaving its magnitude exactly the same.
the same vector, unchanged — Order matters here, that is exactly what anti-commutativity means; claiming no change at all contradicts it directly.
a vector of different magnitude, same direction — The magnitude stays exactly the same on a swap, it is the direction, not the magnitude, that reverses.
Why does swapping the order of a cross product reverse its sign?
because $\vec{a}$ and $\vec{b}$ must be perpendicular for the cross product to be defined
because vector magnitudes are always positive, forcing a sign flip somewhere
swapping rows 2 and 3 of the determinant flips its sign, and that is the whole mechanism
because the dot product is commutative, so the cross product must be the opposite
Check your answer
because $\vec{a}$ and $\vec{b}$ must be perpendicular for the cross product to be defined — The cross product is defined for ANY two vectors, perpendicularity is never a requirement for it to exist.
because vector magnitudes are always positive, forcing a sign flip somewhere — Magnitudes being positive numbers has no bearing on the sign of a determinant , the sign flip comes from the row swap, not from any magnitude.
✓ swapping rows 2 and 3 of the determinant flips its sign, and that is the whole mechanism — (C) Swapping $\vec{a}$ and $\vec{b}$ swaps rows 2 and 3 of the determinant, and swapping any two rows of a determinant always flips its sign.
because the dot product is commutative, so the cross product must be the opposite — The dot product’s commutativity is a separate fact about a separate operation , the cross product’s sign flip comes from the determinant’s row-swap rule, not from contrasting with the dot product.
If $\vec{a} \times \vec{b} = 3 \hat{i} - 2 \hat{j} + \hat{k}$, what is $\vec{b} \times \vec{a}$?
$-3 \hat{i} - 2 \hat{j} - \hat{k}$
$3 \hat{i} + 2 \hat{j} + \hat{k}$
$3 \hat{i} - 2 \hat{j} + \hat{k}$
$-3 \hat{i} + 2 \hat{j} - \hat{k}$
Check your answer
$-3 \hat{i} - 2 \hat{j} - \hat{k}$ — All THREE components flip sign on a swap, flipping only two of them, and leaving the $\hat{j}$-component untouched, is incomplete.
$3 \hat{i} + 2 \hat{j} + \hat{k}$ — Only the $\hat{j}$-component’s sign changed here, the $\hat{i}$- and $\hat{k}$-components must flip too.
$3 \hat{i} - 2 \hat{j} + \hat{k}$ — Anti-commutativity means the swap DOES change the vector, reporting it unchanged denies the whole property.
✓ $-3 \hat{i} + 2 \hat{j} - \hat{k}$ — (D) Swapping the order flips every one of the three component signs: $-3 \hat{i} + 2 \hat{j} - \hat{k}$.
The worked example by determinant: −î + 3ĵ + 5k̂. Swap the rows and every sign flips.
Worked example
Cross product both ways, determinant form
$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$ Two vectors are given. Find the cross product in both orders.
$\vec{a} \times \vec{b} = -\hat{i} + 3 \hat{j} + 5 \hat{k}$ Work out each bracket. The middle bracket is $-3$, and the minus sign in front makes it $+3$.
$\vec{b} \times \vec{a} = \hat{i} - 3 \hat{j} - 5 \hat{k}$ Swap rows 2 and 3, so that $\vec{b}$ comes first. Expand the same way.
*$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$* Every component has changed sign, as a swap of two rows says.
Check: $(-\hat{i} + 3 \hat{j} + 5 \hat{k}) \cdot \vec{a} = -2 - 3 + 5 = 0$ and $(-\hat{i} + 3 \hat{j} + 5 \hat{k}) \cdot \vec{b} = -1 + 6 - 5 = 0$ Both dot products are 0. So $\vec{a} \times \vec{b}$ is at right angles to both vectors, as it must be.
Check yourself
For $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$, what is the $\hat{i}$-component of $\vec{a} \times \vec{b}$?
$1$
$-1$
$-3$
$3$
Check your answer
$1$ — $1 - 2 = -1$, not $1$. Dropping the negative sign after the subtraction misreads the arithmetic.
$\vec{0}$, since reversing the order cancels the vector — Reversing order changes the SIGN of each component, it does not cancel them, the result is zero only if the original cross product already was.
$\hat{i} + 3 \hat{j} + 5 \hat{k}$, confirming only the first sign flips — Reversing the order flips ALL three signs, not just the first, leaving the $\hat{j}$- and $\hat{k}$-components positive misses two of the three flips.
the same vector, $-\hat{i} + 3 \hat{j} + 5 \hat{k}$, confirming the cross product commutes here — The cross product is never commutative for non-parallel vectors, and this pair is not parallel, the reversed order genuinely gives a different vector.
✓ $\hat{i} - 3 \hat{j} - 5 \hat{k}$, exactly $-(\vec{a} \times \vec{b})$ — (D) Reversing the order flips every one of the three signs, giving $\hat{i} - 3 \hat{j} - 5 \hat{k}$, exactly $-(\vec{a} \times \vec{b})$, confirming anti-commutativity.
✕MISCONCEPTION
Step 1
The determinant gives a × b = k̂
?
Step 2
The order of a dot product does not matter, so the order of a cross product does not matter either
?
Step 3
So b × a = k̂
?
Three steps of a student’s working. Which step is wrong?
Here is a working for $\vec{b} \times \vec{a}$, with $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j}$. Which step is wrong?
The determinant gives $\vec{a} \times \vec{b} = \hat{k}$.
The order of a dot product does not matter, so the order of a cross product does not matter either.
So $\vec{b} \times \vec{a} = \hat{k}$.
Keep your answer, then work it out. Put $\vec{b}$ in row 2 and $\vec{a}$ in row 3. The expansion gives $((0)(1) - (1)(1)) \hat{k} = -\hat{k}$.
a = î + ĵ and b = ĵ: a × b = k̂ points up, and b × a = −k̂ points down.
The second step is wrong. The two products follow different rules. Here $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = 1$, but $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) = -\hat{k}$.
So when you swap the order of a cross product, put in a minus sign.
Check yourself
A student argues: ‘the dot product is commutative, so the cross product must be too.’ What is wrong with this reasoning?
the cross product is commutative too, just with a different formula
nothing is wrong; both products are commutative
commutativity does not carry over between different operations
the dot product is not actually commutative either
Check your answer
the cross product is commutative too, just with a different formula — Anti-commutativity means the SIGN reverses on a swap, hedging with “a different formula” still wrongly claims the two orders give the same result.
nothing is wrong; both products are commutative — The cross product is anti-commutative by its own definition, $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ always, so this denial contradicts the definition directly.
✓ commutativity does not carry over between different operations — (C) Each operation is defined by its own rule, and one operation’s property does not automatically transfer to another, the cross product is anti-commutative precisely because its own determinant-based rule reverses sign on a swap.
the dot product is not actually commutative either — $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ holds for the dot product, always , this option denies a fact stated as true from the start.
For $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = \hat{j}$: which of these is TRUE?
$\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$, but the dot product changes when reversed
neither product is order-independent for these vectors
cross flips sign on swap; the dot product agrees both ways here
both $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ hold
Check your answer
$\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$, but the dot product changes when reversed — This has it backward, the CROSS product is the one that flips sign on a swap; the dot product is the one that stays the same.
neither product is order-independent for these vectors — $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = 1$ genuinely holds here, the dot product IS order-independent; only the cross product is order-sensitive.
✓ cross flips sign on swap; the dot product agrees both ways here — (C) The cross product flips sign on a swap here ($\hat{k}$ versus $-\hat{k}$), while the dot product genuinely agrees both ways, $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = 1$.
both $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ hold — For these vectors the cross product gives $\hat{k}$ one way and $-\hat{k}$ the other, opposite results, not equal ones.
For which of the following pairs of vectors would $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ actually hold?
for any two vectors, since it is a general property
only when $\vec{a} \times \vec{b} = \vec{0}$ already
only when $\vec{a}$ and $\vec{b}$ are perpendicular to each other
only when $\left|\vec{a}\right| = \left|\vec{b}\right|$
Check your answer
for any two vectors, since it is a general property — This is exactly the false generalization at issue, the cross product reverses sign on a swap for almost every pair, not agreeing “for any two vectors.”
✓ only when $\vec{a} \times \vec{b} = \vec{0}$ already — (B) Anti-commutativity forces $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ always; the two sides can only be EQUAL when both are already the zero vector.
only when $\vec{a}$ and $\vec{b}$ are perpendicular to each other — Perpendicular vectors give the LARGEST-magnitude cross product, in two exactly opposite directions on a swap, that is the furthest thing from the two orders agreeing.
only when $\left|\vec{a}\right| = \left|\vec{b}\right|$ — Equal magnitudes say nothing about DIRECTION, whether the cross product vanishes depends on the vectors being parallel, not on matching lengths.
The triangle on (0, 0), (5, 1) and (1, 3), with Chapter 4’s determinant: 14, so the area is 7.
Chapter 4 found the area of a triangle from three points on graph paper, with a determinant. The cross product does the same job for vectors, in the plane or in space. You only need two sides from one corner.
Base times height: about 5.10 times about 2.75 is 14, the same as the length of a × b.
Draw $\vec{a}$ and $\vec{b}$ from one corner. They are two sides of a parallelogram. Its base is $\left|\vec{a}\right|$, and its height is $\left|\vec{b}\right| \sin \theta$. So its area is $\left|\vec{a}\right| \left|\vec{b}\right| \sin \theta$. That is the length of $\vec{a} \times \vec{b}$.
So the AREA OF A PARALLELOGRAM with adjacent sides $\vec{a}$ and $\vec{b}$ is $\left|\vec{a} \times \vec{b}\right|$.
Arrows 5 across, 1 up and 1 across, 3 up make a parallelogram of area 14. Its diagonal cuts two triangles of 7.
A diagonal cuts the parallelogram into two equal triangles. So the AREA OF A TRIANGLE with adjacent sides $\vec{a}$ and $\vec{b}$ is $(1/2) \left|\vec{a} \times \vec{b}\right|$.
So the parallelogram is 14, and each triangle is 7. The determinant of Chapter 4 gives 7 too, for the triangle with corners $(0, 0)$, $(5, 1)$ and $(1, 3)$.
Your turn: find $\left|\vec{a} \times \vec{b}\right|$ for $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 2 \hat{i} + 2 \hat{j}$. What shape do these two vectors make?
Check yourself
For $\vec{a}, \vec{b}$ forming two sides of a parallelogram from a common vertex, its area equals \_\_\_\_\_\_.
$(1/2) \cdot \left|\vec{a} \times \vec{b}\right|$
$\left|\vec{a}\right| \cdot \left|\vec{b}\right|$
$\vec{a} \cdot \vec{b}$
$\left|\vec{a} \times \vec{b}\right|$
Check your answer
$(1/2) \cdot \left|\vec{a} \times \vec{b}\right|$ — Half the cross product’s magnitude is the TRIANGLE’s area, the parallelogram itself keeps the full, undivided value.
$\left|\vec{a}\right| \cdot \left|\vec{b}\right|$ — Multiplying just the two magnitudes ignores the angle between them, a parallelogram squashed nearly flat has far less area than that product suggests.
$\vec{a} \cdot \vec{b}$ — The dot product measures ALIGNMENT between two vectors, not area, area comes from the cross product’s magnitude instead.
✓ $\left|\vec{a} \times \vec{b}\right|$ — (D) The parallelogram’s area is exactly the magnitude of the cross product of its two sides.
Why is the triangle’s area exactly half the parallelogram’s, when both share the same two sides $\vec{a}$, $\vec{b}$?
the parallelogram’s own diagonal splits it into two congruent, equal-area triangles
triangles always have half the area of any shape built from the same two vectors
the cross product itself is halved when computing a triangle instead of a parallelogram
the formula for a triangle uses different vectors than the parallelogram does
Check your answer
✓ the parallelogram’s own diagonal splits it into two congruent, equal-area triangles — (A) The parallelogram’s own diagonal splits it into two congruent triangles, so each one carries exactly half the total area.
triangles always have half the area of any shape built from the same two vectors — This half-area relationship is specific to a PARALLELOGRAM split by its own diagonal , it does not extend to “any shape” built from the same two vectors.
the cross product itself is halved when computing a triangle instead of a parallelogram — $\vec{a} \times \vec{b}$ is identical in both the triangle and parallelogram calculations, the $1/2$ is applied to the resulting AREA, not to the cross product itself.
the formula for a triangle uses different vectors than the parallelogram does — Both the triangle and the parallelogram use the identical two vectors $\vec{a}$, $\vec{b}$, only the $1/2$ factor distinguishes the two areas.
Two sides of a triangle from a common vertex are $\vec{a}$ and $\vec{b}$, with $\left|\vec{a} \times \vec{b}\right| = 10$. What is the triangle’s area?
$10$
$20$
$5$
$1/10$
Check your answer
$10$ — $10$ is the PARALLELOGRAM’s area, the triangle sharing the same two sides carries exactly half of that.
$20$ — Doubling goes the wrong way, the triangle’s area is HALF the cross product’s magnitude, not double it.
✓ $5$ — (C) The triangle’s area is half the cross product’s magnitude: $(1/2)(10) = 5$.
$1/10$ — Inverting $10$ to $1/10$ has no basis in the halving rule, the triangle’s area is simply $10$ divided by $2$.
▲CONCEPT
A door half open. The same push barely turns it near the hinge, and swings it easily at the far edge.
A long spanner on a nut. The long handle is there to put the push far from the turning point.
Push a door near its hinge, and it barely turns. Push it at the far edge, and it swings easily. Push along the door, straight at the hinge, and it does not turn at all. A spanner works the same way. Its long handle lets a small push turn a tight nut. Physics 11 calls this turning effect torque.
The same push of 1 at distances 1, 2 and 4 from the hinge gives a turning effect of 1, 2 and 4.
So the torque depends on how far from the pivot the force acts — that is, on $\vec{r}$. It also depends on the angle between the force and $\vec{r}$.
r = 2ĵ and F = −2î give τ = r × F = 4k̂, an arrow along the hinge.
The cross product holds both. Let a force $\vec{F}$ act at a point whose position vector from the pivot $O$ is $\vec{r}$. The MOMENT OF FORCE, or TORQUE, about $O$ is
$\vec{\tau} = \vec{r} \times \vec{F}$.
With r and the force fixed, the turning effect is 4 sin θ: 2 at π/6, 4 at π/2, and back to 0 at π.
Size and direction
Its size is $\left|\vec{r}\right| \left|\vec{F}\right| \sin \theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. Its direction is along the axis of the torque, by the right-hand rule. The size is largest when the force is perpendicular to $\vec{r}$. It is 0 when the force is along $\vec{r}$.
The same r and the same force. Perpendicular to r, the torque is 4. At 30 degrees it is 2, half as much.
Read the numbers off the figure. The arm and the push are the same in both panels, so $\left|\vec{r}\right| \left|\vec{F}\right| = 4$. In the left panel the push is at 90 degrees, so the torque is $4 \sin(\pi/2) = 4$. In the right panel it is at 30 degrees, so the torque is $4 \sin(\pi/6) = 2$. Only the angle has changed.
Check yourself
The moment of force (torque) about pivot $O$ is defined as \_\_\_\_\_\_.
$\vec{\tau} = \vec{r} \cdot \vec{F}$ — The dot product returns a number, torque is a vector by definition, which is exactly what the cross product, not the dot product, gives.
✓ $\vec{\tau} = \vec{r} \times \vec{F}$ — (B) Torque is the cross product of the position vector from the pivot and the force, $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \vec{F} - \vec{r}$ — Subtracting $\vec{r}$ from $\vec{F}$ has no basis in the definition, torque is built from a cross product, not a subtraction.
$\vec{\tau} = \left|\vec{r}\right| + \left|\vec{F}\right|$ — Adding the magnitudes discards the angle between $\vec{r}$ and $\vec{F}$ entirely, and also discards torque’s direction as a vector.
For a given force and a given distance $\left|\vec{r}\right|$ from $O$ to the point where it acts, the magnitude of the torque is largest when \_\_\_\_\_\_.
$\vec{r}$ and $\vec{F}$ have equal magnitudes
the force acts along the same line as $\vec{r}$, so $\theta = 0^\circ$
perpendicular force, $\theta=90^\circ$, where $\sin \theta$ peaks at $1$
the force points along $\vec{r}$, back towards $O$, so $\theta = 180^\circ$
Check your answer
$\vec{r}$ and $\vec{F}$ have equal magnitudes — Equal magnitudes leave the ANGLE between $\vec{r}$ and $\vec{F}$ completely unspecified, it is the angle, through $\sin \theta$, that controls the maximum.
the force acts along the same line as $\vec{r}$, so $\theta = 0^\circ$ — $\sin 0^\circ = 0$, so a force along the same line as $\vec{r}$ gives ZERO torque , exactly the opposite of the largest possible case.
✓ perpendicular force, $\theta=90^\circ$, where $\sin \theta$ peaks at $1$ — (C) $\sin \theta$ reaches its largest possible value, $1$, exactly at $\theta = 90^\circ$, a force perpendicular to $\vec{r}$.
the force points along $\vec{r}$, back towards $O$, so $\theta = 180^\circ$ — $\sin 180^\circ = 0$, so a force pointing back along $\vec{r}$ towards $O$ gives ZERO torque, just as at $\theta = 0^\circ$.
The scalar triple product (beyond the board syllabus)
▲CONCEPT
This product is beyond the board syllabus. Chapter 11 uses it to find the shortest distance between two lines that are not parallel and never meet.
Cross b with c, then dot with a: b × c = 4î − 2ĵ + k̂, and the result is 8.
Take three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$. Cross two of them, then dot the answer with the third. The number you get is the SCALAR TRIPLE PRODUCT:
Left: base 4, height 2, volume 8. Right: c lies in the base plane, so the volume is 0 and the three are coplanar.
Draw the three vectors from one corner. They are three edges of a slanted box, a PARALLELEPIPED. Its VOLUME is $\left|[\vec{a} \vec{b} \vec{c}]\right|$.
Base area 4, the length of a × b; height 2, the projection of c on a × b. Volume 4 × 2 = 8.
The figure shows why. Its base is the parallelogram on $\vec{a}$ and $\vec{b}$, with area $\left|\vec{a} \times \vec{b}\right|$. The height is how far the tip of $\vec{c}$ rises above that base. Base times height is $\left|\vec{c} \cdot (\vec{a} \times \vec{b})\right|$, and that equals $\left|[\vec{a} \vec{b} \vec{c}]\right|$. Moving the three round in order, $\vec{a} \vec{b} \vec{c}$ to $\vec{c} \vec{a} \vec{b}$, swaps two rows twice and leaves the determinant unchanged.
Determinant $= 2((2)(2) - (0)(1)) = 8$, and base times height is $4 \times 2 = 8$.
The next figure lowers $\vec{c}$ until $\vec{c} = \hat{j}$ lies flat in the base. Its height is then 0, so the volume is 0.
Check yourself
The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is defined as \_\_\_\_\_\_.
$\vec{a} \times (\vec{b} \times \vec{c})$, a vector
$\vec{a} \cdot (\vec{b} \times \vec{c})$, a number
$\vec{a} \cdot \vec{b} \cdot \vec{c}$, multiplying all three dot products together
$(\vec{a} \times \vec{b}) \times \vec{c}$, a vector
Check your answer
$\vec{a} \times (\vec{b} \times \vec{c})$, a vector — That expression is the VECTOR triple product, a different object entirely, returning a vector, not the number this definition describes.
✓ $\vec{a} \cdot (\vec{b} \times \vec{c})$, a number — (B) The scalar triple product dots $\vec{a}$ into the cross product of $\vec{b}$ and $\vec{c}$, giving a single number.
$\vec{a} \cdot \vec{b} \cdot \vec{c}$, multiplying all three dot products together — The dot product only ever combines two vectors at a time, there is no three-way dot product anywhere in the definition.
$(\vec{a} \times \vec{b}) \times \vec{c}$, a vector — Grouping the cross products this way still returns a VECTOR, the scalar triple product is built from one cross product followed by a dot product, not two cross products.
What does the scalar triple product equal, in terms of a single computation over the three vectors’ components?
the product of the three vectors’ magnitudes, with no angle dependence
the sum of the three vectors’ magnitudes
the determinant with the rows in the order $\vec{b}$, $\vec{a}$, $\vec{c}$
the determinant of the three vectors’ components, stacked as rows
Check your answer
the product of the three vectors’ magnitudes, with no angle dependence — The underlying dot-and-cross construction is built entirely from angles, a formula with no angle dependence cannot reproduce it.
the sum of the three vectors’ magnitudes — Summing magnitudes throws away every direction and angle, the actual value depends on how the three vectors are oriented relative to each other, not just their lengths.
the determinant with the rows in the order $\vec{b}$, $\vec{a}$, $\vec{c}$ — Swapping two rows of a determinant changes its sign, so the rows in the order $\vec{b}$, $\vec{a}$, $\vec{c}$ give $-[\vec{a} \vec{b} \vec{c}]$, not the triple product.
✓ the determinant of the three vectors’ components, stacked as rows — (D) The scalar triple product equals the determinant formed by stacking $\vec{a}$, $\vec{b}$, $\vec{c}$’s components as rows, in that order.
What does $\left|[\vec{a} \vec{b} \vec{c}]\right|$ measure, geometrically?
the area of the parallelogram formed by any two of the three vectors
the sum of the three vectors’ individual magnitudes
the volume of the parallelepiped with the three vectors as edges
the perimeter of the parallelepiped
Check your answer
the area of the parallelogram formed by any two of the three vectors — A single cross product’s magnitude gives an AREA, from just two vectors, the full scalar triple product uses all three and gives a VOLUME.
the sum of the three vectors’ individual magnitudes — Volume depends on the ANGLES between all three vectors, not just how long each one is, a magnitude sum ignores that entirely.
✓ the volume of the parallelepiped with the three vectors as edges — (C) The absolute value of the scalar triple product is exactly the volume of the parallelepiped built from the three vectors as edges.
the perimeter of the parallelepiped — Perimeter measures a length around the edges, the scalar triple product measures the enclosed three-dimensional VOLUME instead.
▲CONCEPT
Lower c and the box flattens: volume 8, then 4, then 0 when c lies in the floor with a and b.
When the box is flat, the three vectors lie in one plane. They are COPLANAR.
Three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar exactly when $[\vec{a} \vec{b} \vec{c}] = 0$.
With c = ĵ the determinant is 0. The three vectors lie in one plane: they are coplanar.
So one determinant tests for coplanar vectors. If its value is 0, the three lie in one plane. If it is not 0, they do not.
The determinant above takes $\vec{c} = \hat{j}$, which lies in the plane of $\vec{a} = 2 \hat{i}$ and $\vec{b} = \hat{i} + 2 \hat{j}$. Its value is $2((2)(0) - (0)(1)) = 0$.
Check yourself
Three vectors $\vec{a}, \vec{b}, \vec{c}$ are COPLANAR exactly when \_\_\_\_\_\_.
$\vec{a} \times \vec{b} = \vec{0}$ — $\vec{a} \times \vec{b} = \vec{0}$ makes $\vec{a}$, $\vec{b}$, $\vec{c}$ coplanar for every $\vec{c}$, so it is enough, but it is not needed: the test that holds exactly when they are coplanar is $[\vec{a} \vec{b} \vec{c}] = 0$.
$[\vec{a} \vec{b} \vec{c}] = 1$ — A triple product of $1$ is nonzero, that is exactly the NON-coplanar case; the coplanar test needs the value to be exactly zero.
✓ $[\vec{a} \vec{b} \vec{c}] = 0$ — (C) Coplanarity is exactly the condition that the scalar triple product of the three vectors equals zero.
$\left|\vec{a}\right| = \left|\vec{b}\right| = \left|\vec{c}\right|$ — Three vectors can share the same length and still point in three completely different directions out of any common plane, magnitude says nothing about coplanarity.
Why does a zero scalar triple product mean the three vectors are coplanar?
zero volume means the parallelepiped is flat, its edges sharing one plane
a zero triple product means all three vectors are equal to each other
volume and coplanarity are unrelated; the zero result is a coincidence of the formula
a zero triple product means one of the three vectors is the zero vector
Check your answer
✓ zero volume means the parallelepiped is flat, its edges sharing one plane — (A) The triple product’s absolute value is the parallelepiped’s volume; a volume of exactly zero means the shape has collapsed flat, with all three edges lying in one plane.
a zero triple product means all three vectors are equal to each other — Equal vectors are only ONE special coplanar arrangement, a zero triple product covers every coplanar configuration, not just that one case.
volume and coplanarity are unrelated; the zero result is a coincidence of the formula — Volume and coplanarity are directly linked, not coincidental, a shape with zero enclosed volume has, by definition, collapsed into a flat plane.
a zero triple product means one of the three vectors is the zero vector — Three genuinely nonzero vectors can still give a zero triple product, simply by being dependent (lying in one plane), no single vector needs to vanish.
For $\vec{a} = \hat{i}$, $\vec{b} = \hat{j}$, $\vec{c} = \hat{i} + \hat{j}$, are the three vectors coplanar?
yes, but only because $\vec{a}$ and $\vec{b}$ are perpendicular
yes, the triple product is zero, since $\vec{c}$ already lies in their plane
no, the three vectors point in three different directions, so they cannot be coplanar
no, $\left|[\vec{a} \vec{b} \vec{c}]\right| = 1$, a nonzero volume
Check your answer
yes, but only because $\vec{a}$ and $\vec{b}$ are perpendicular — Whether $\vec{a}$ and $\vec{b}$ happen to be perpendicular plays no role in the coplanarity test, only the full triple product of all three vectors decides it.
✓ yes, the triple product is zero, since $\vec{c}$ already lies in their plane — (B) $\vec{b} \times \vec{c} = \hat{j} \times (\hat{i}+\hat{j}) = -\hat{k}$, and $\vec{a} \cdot (-\hat{k}) = 0$, confirming coplanarity, since $\vec{c}$ is already a combination of $\vec{a}$ and $\vec{b}$.
no, the three vectors point in three different directions, so they cannot be coplanar — Vectors within one shared plane can still point in several different directions , “different directions” is not what the coplanarity test actually checks.
no, $\left|[\vec{a} \vec{b} \vec{c}]\right| = 1$, a nonzero volume — The actual computation here gives $0$, not $1$, $\vec{c}$ is exactly $\vec{a} + \vec{b}$, so it already sits in their shared plane.
Eight idea cards, from scalars and vectors to the triple product. An arrow means the later card uses the earlier one.
The first page asked a question. Can arrows be added, stretched, written as numbers and multiplied? Can lengths, angles, work and areas all come out of adding and multiplying them?
a + b = 5î + 4ĵ and a × b = 11k̂ are vectors; a · b = 7 is just a number.
A VECTOR has a size and a direction. Write it by its components: $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$, with $\left|\vec{a}\right| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
ADD tail to head, or by the parallelogram. In components, add component by component.
SCALE by $\lambda$: the length is multiplied by $\left|\lambda\right|$, and the direction turns round when $\lambda < 0$.
DIRECTION COSINES of $\vec{a} \neq \vec{0}$: $l = a_1 / \left|\vec{a}\right|$, and so on, with $l^2 + m^2 + n^2 = 1$.
SECTION FORMULA: internal, $(m \vec{b} + n \vec{a})/(m + n)$. External, with $m \neq n$, $(m \vec{b} - n \vec{a})/(m - n)$.
DOT PRODUCT: $\vec{a} \cdot \vec{b} = \left|\vec{a}\right| \left|\vec{b}\right| \cos \theta$, with $0 \leq \theta \leq \pi$. It is a number.
For non-zero vectors, $\vec{a} \cdot \vec{b} = 0$ exactly when they are perpendicular. Work is $\vec{F} \cdot \vec{d}$.
CROSS PRODUCT: $\vec{a} \times \vec{b} = \left|\vec{a}\right| \left|\vec{b}\right| \sin \theta \hat{n}$, at right angles to both. The right-hand rule fixes its direction.
Order matters: $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. The parallelogram’s area is $\left|\vec{a} \times \vec{b}\right|$, and torque is $\vec{r} \times \vec{F}$.
SCALAR TRIPLE PRODUCT (beyond the board syllabus): $\left|[\vec{a} \vec{b} \vec{c}]\right|$ is the volume of the box. It is 0 exactly when the three are coplanar.
Here is the answer. Yes. Write an arrow by its components along three axes, and add and stretch it component by component. The dot product gives a number that says how far two arrows point the same way: an angle, a projection, the work done. The cross product gives an arrow at right angles to both, and its length is the area they span.
The opening walk again: 4î + 3ĵ has length 5 and direction cosines 4/5, 3/5 and 0.
Write the opening walk in vectors. Going 3 km north and then 4 km east is $4 \hat{i} + 3 \hat{j}$. Its length, $\sqrt{16 + 9} = 5$, says how far. Its direction cosines, $4/5$, $3/5$ and 0, say which way.
Check yourself
A problem asks ‘how much of one force acts along a given direction?’ Which operation from this chapter answers that, and which would be the wrong one to reach for?
the dot product measures alignment; the cross product measures spread
vector ADDITION answers it, by combining the force with the direction
the CROSS product answers it, since it involves both vectors together
the scalar triple product answers it, since it combines three vectors into one number
Check your answer
✓ the dot product measures alignment; the cross product measures spread — (A) The dot product is built to measure alignment, exactly the ‘how much lies along’ question, the cross product instead measures perpendicular spread, a different question entirely.
vector ADDITION answers it, by combining the force with the direction — Addition combines two vectors into a resultant sum, it never measures how much of one already lies along another; that comparison is the dot product’s role.
the CROSS product answers it, since it involves both vectors together — The cross product measures how much the two vectors spread apart PERPENDICULAR to each other, the opposite of measuring how much one lies along the other.
the scalar triple product answers it, since it combines three vectors into one number — The scalar triple product needs THREE vectors and answers a volume or coplanarity question, alignment between two vectors is the dot product’s job.
(Beyond the board syllabus.) A problem gives three vectors and asks whether they lie in one plane. Which single operation from this chapter answers that directly?
the cross product of any two of them: it always points out of the plane the three would form
the scalar triple product: it is exactly zero when the three vectors are coplanar
vector addition of all three: a zero sum would mean they are coplanar
the dot product taken pairwise between all three: matching signs would mean they are coplanar
Check your answer
the cross product of any two of them: it always points out of the plane the three would form — A single cross product uses only two of the three vectors, it has no way to test where the THIRD vector sits relative to them.
✓ the scalar triple product: it is exactly zero when the three vectors are coplanar — (B) The scalar triple product is built exactly for this question, it equals zero precisely when the three vectors are coplanar.
vector addition of all three: a zero sum would mean they are coplanar — A zero sum tests whether the three vectors close into a triangle-like loop, an entirely different question from whether they share a plane.
the dot product taken pairwise between all three: matching signs would mean they are coplanar — Pairwise dot products describe angles between PAIRS, no combination of matching signs among them directly tests whether all three share one plane.
Work and torque both come from multiplying a force by a displacement or position vector. Why is work a NUMBER while torque is a VECTOR?
work and torque actually use the same product; only the units differ
torque is a number too, once its magnitude is taken; the ‘vector’ description is just informal
work is a vector because force and displacement are both vectors, and combining two vectors always gives a vector
work uses the dot product (alignment, a number); torque uses the cross product (spread, a vector)
Check your answer
work and torque actually use the same product; only the units differ — Work is $\vec{F} \cdot \vec{d}$ and torque is $\vec{r} \times \vec{F}$, genuinely different operations, not the same computation wearing different units.
torque is a number too, once its magnitude is taken; the ‘vector’ description is just informal — Torque is a vector BY DEFINITION, $\vec{r} \times \vec{F}$, taking its magnitude afterward gives one property of it, not a reason to call the ‘vector’ description merely informal.
work is a vector because force and displacement are both vectors, and combining two vectors always gives a vector — Combining two vectors does not always give a vector, the DOT product of two vectors is a number, which is exactly the case work falls into.
✓ work uses the dot product (alignment, a number); torque uses the cross product (spread, a vector) — (D) Work uses the dot product, which returns a number measuring alignment; torque uses the cross product, which returns a vector measuring perpendicular spread, different operations, different kinds of result.
Both vector addition and the section formula combine two vectors into a third. What is the key difference in what each one produces?
addition requires a ratio $m:n$; the section formula does not
they produce the same result; the section formula is just addition written differently
addition gives the full resultant; the section formula gives one weighted point
the section formula only works when $\vec{a}$ and $\vec{b}$ are equal in magnitude
Check your answer
addition requires a ratio $m:n$; the section formula does not — This reverses the roles, ADDITION needs no ratio at all; the SECTION FORMULA is the one built entirely around a chosen ratio $m:n$.
they produce the same result; the section formula is just addition written differently — A weighted average controlled by a ratio $m:n$ is a different construction from a plain sum, the two do not coincide in general, only at specific ratios.
✓ addition gives the full resultant; the section formula gives one weighted point — (C) Addition produces the full resultant of two vectors, while the section formula produces one specific point on segment $A B$, its position controlled by a chosen ratio $m:n$.
the section formula only works when $\vec{a}$ and $\vec{b}$ are equal in magnitude — The section formula places no requirement on the two magnitudes being equal, it works for any $\vec{a}$, $\vec{b}$, and any chosen ratio $m:n$.
Check yourself: the whole chapter
Speed and velocity both carry the same units, such as m/s. Which statement correctly tells them apart?
Velocity is a vector because it has direction; speed lacks one.
Both are vectors, since both carry a magnitude and a matching unit.
Both are scalars; direction is just extra detail added to the number.
Velocity is a vector only while its direction stays fixed and unchanging.
Check your answer
✓ Velocity is a vector because it has direction; speed lacks one. — (A) Speed reports magnitude only. Velocity reports magnitude and direction together, and that direction is exactly what makes it a vector.
Both are vectors, since both carry a magnitude and a matching unit. — A matching unit does not make a quantity a vector. What decides it is whether direction is part of the quantity itself.
Both are scalars; direction is just extra detail added to the number. — Direction is not optional detail. A vector is defined by carrying magnitude and direction together, always.
Velocity is a vector only while its direction stays fixed and unchanging. — Direction need not stay constant. Velocity is a vector at every instant, whether or not that direction later changes.
Vectors $\vec{a}$ and $\vec{b}$ share the same magnitude and point the same way, but are drawn starting from two different points in space. Are they equal vectors?
No, equal vectors must share the same starting point.
Yes, position does not matter; magnitude and direction decide it.
No, they are only parallel, since they start at different points.
Yes, but only because both happen to be unit vectors.
Check your answer
No, equal vectors must share the same starting point. — Starting point plays no role in equality. Two vectors are equal once magnitude and direction match, wherever each is drawn.
✓ Yes, position does not matter; magnitude and direction decide it. — (B) Position never enters the definition. Same magnitude and same direction, drawn from anywhere, makes two vectors equal.
No, they are only parallel, since they start at different points. — Parallel vectors need not share a magnitude. Equal vectors are the stricter case: same magnitude, same direction, any position.
Yes, but only because both happen to be unit vectors. — Equality never needs magnitude one. Any two vectors with matching magnitude and direction are equal, unit or not.
By the triangle law, $\vec{a} + \vec{b}$ places the tail of $\vec{b}$ at the head of $\vec{a}$. If that order is reversed, placing the tail of $\vec{a}$ at the head of $\vec{b}$, what happens to the sum?
The sum reverses direction, since the drawing order was swapped.
The sum is undefined; the triangle law only holds in one fixed order.
The sum becomes $\vec{a} - \vec{b}$ instead of a true sum.
The sum stays the same, since vector addition is commutative.
Check your answer
The sum reverses direction, since the drawing order was swapped. — Reversing the drawing order changes nothing about the result. $\vec{a} + \vec{b}$ and $\vec{b} + \vec{a}$ are the same arrow.
The sum is undefined; the triangle law only holds in one fixed order. — The triangle law holds in either order. It is one way of drawing a sum that was already commutative.
The sum becomes $\vec{a} - \vec{b}$ instead of a true sum. — Reordering two vectors being added is not subtraction. Subtracting means adding the negative of one vector, a separate step.
✓ The sum stays the same, since vector addition is commutative. — (D) Vector addition is commutative. Whichever order the triangle law is drawn in, the resulting arrow stays exactly the same.
For nonzero vectors $\vec{a}$ and $\vec{b}$, what kind of object does $\vec{a} \cdot \vec{b}$ give?
A vector perpendicular to both $\vec{a}$ and $\vec{b}$.
A vector along $\vec{a}$, of length $\left|\vec{b}\right| \cos \theta$.
A number; the dot product never produces a vector.
A vector along $\vec{b}$, whose sign depends on $\theta$.
Check your answer
A vector perpendicular to both $\vec{a}$ and $\vec{b}$. — A perpendicular vector is the cross product’s signature, not the dot product’s. The dot product never returns a vector at all.
A vector along $\vec{a}$, of length $\left|\vec{b}\right| \cos \theta$. — The dot product is the number $\left|\vec{a}\right| \left|\vec{b}\right| \cos \theta$. Building an actual projection vector is a further, separate step.
✓ A number; the dot product never produces a vector. — (C) The dot product is defined as $\left|\vec{a}\right| \left|\vec{b}\right| \cos \theta$, a single number. It never comes out as a vector.
A vector along $\vec{b}$, whose sign depends on $\theta$. — The dot product itself hands no direction to $\vec{b}$ either. It is a single number, nothing more.
To compute $\vec{a} \times \vec{b}$ by the determinant recipe, which row order is correct?
Basis row $\hat{i}, \hat{j}, \hat{k}$, then $\vec{a}$, then $\vec{b}$.
The parts of $\vec{a}$ first, then the basis row, then the parts of $\vec{b}$.
Row order makes no difference, since a determinant never depends on it.
The parts of $\vec{b}$ first, then $\vec{a}$, then the basis row last.
Check your answer
✓ Basis row $\hat{i}, \hat{j}, \hat{k}$, then $\vec{a}$, then $\vec{b}$. — (A) The recipe is fixed: basis row, then the parts of $\vec{a}$, then of $\vec{b}$, in that order.
The parts of $\vec{a}$ first, then the basis row, then the parts of $\vec{b}$. — Putting the basis row in the middle is one row swap from the recipe, so the determinant changes sign: this order gives $-(\vec{a} \times \vec{b})$.
Row order makes no difference, since a determinant never depends on it. — Row order matters here. Swapping any two rows flips the determinant’s sign, which flips the resulting vector.
The parts of $\vec{b}$ first, then $\vec{a}$, then the basis row last. — That order computes $\vec{b} \times \vec{a}$, the negative of what was asked. Row order fixes which vector comes first.
Exercise 10.1 practises the names and the lengths of vectors. You tell scalars from vectors, find unit vectors and test for collinear vectors. You also find direction cosines from components, from two points and from direction ratios.
Exercise 10.1: Vector basics
practice Classify each of the following as a scalar or a vector quantity: (i) mass (ii) displacement (iii) time period (iv) force (v) speed.
practice State whether $\vec{a} = 3\hat{i} - 6\hat{j} + 9\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} - 3\hat{k}$ are collinear.
practice Find the magnitude of the vector $\vec{a} = 6\hat{i} - 2\hat{j} + 3\hat{k}$.
practice Find the unit vector in the direction of $\vec{a} = 8\hat{i} - 6\hat{j}$.
practice Find the direction cosines of $\vec{a} = \hat{i} - 2\hat{j} - 2\hat{k}$.
practice Find the direction cosines of the line joining $P(1,2,-3)$ and $Q(-1,-2,1)$, directed from $P$ to $Q$.
practice A line has direction ratios $(12,-4,3)$. Find its direction cosines.
practice Let $\vec{O A} = 2\hat{i}+3\hat{j}$, let $\vec{B E}$ run from $B(1, 1)$ to $E(3, 4)$, and let $\vec{O C} = -2\hat{i}-3\hat{j}$, $\vec{O D} = 3/5 \hat{i}+4/5 \hat{j}$ and $\vec{0}$ be the zero vector. (i) Which two vectors are EQUAL? (ii) Which vector is the NEGATIVE of $\vec{O A}$? (iii) $\vec{O A}$ and $\vec{O C}$ share the starting point $O$. What does this make them? (iv) Which listed vector is a UNIT vector, and which is the zero vector?
Answers
(i) scalar (ii) vector (iii) scalar (iv) vector (v) scalar
Yes: $\vec{a} = -3 \vec{b}$, so the two are collinear.
$7$
$4/5 \hat{i} - 3/5 \hat{j}$
$(1/3, -2/3, -2/3)$
$(-1/3, -2/3, 2/3)$
$(12/13, -4/13, 3/13)$
$\vec{B E}$ and $\vec{O A}$ are equal; $\vec{O C}$ is the negative of $\vec{O A}$; $\vec{O A}$ and $\vec{O C}$ are coinitial; $\vec{O D}$ is a unit vector and $\vec{0}$ is the zero vector.
Exercise 10.2 practises adding and scaling vectors in components. You find lengths, unit vectors and direction cosines. You also use the section formula, inside the segment and outside it.
Exercise 10.2: Addition and components
practice Find the sum of the vectors $\vec{a} = \hat{i}-2\hat{j}+\hat{k}$, $\vec{b} = -2\hat{i}+4\hat{j}+5\hat{k}$ and $\vec{c} = \hat{i}-6\hat{j}-7\hat{k}$.
practice Find the magnitude of $\vec{a} = 3\hat{i}+4\hat{j}+12\hat{k}$.
practice Find the scalar components and the magnitude of the vector with initial point $(2,3)$ and terminal point $(7,15)$.
practice Find the values of $x$ and $y$ so that the vectors $2\hat{i}+3\hat{j}$ and $x \hat{i}+y \hat{j}$ are equal.
practice Multiply $\vec{a} = 2\hat{i}-3\hat{j}+5\hat{k}$ by the scalar $\lambda = -3$, and find the magnitude of the result.
practice Find the unit vector in the direction of $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$.
practice Find the unit vector in the direction of the vector from $P(1,2,3)$ to $Q(4,6,3)$.
practice For $\vec{a} = 2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b} = -\hat{i}+\hat{j}-\hat{k}$, find the unit vector in the direction of $\vec{a}+\vec{b}$.
practice Find a vector of magnitude $8$ in the direction of $\vec{a} = 5\hat{i}-\hat{j}+2\hat{k}$.
practice Show that $\vec{a} = 2\hat{i}-3\hat{j}+4\hat{k}$ and $\vec{b} = -4\hat{i}+6\hat{j}-8\hat{k}$ are collinear.
practice Find the direction cosines of $\vec{a} = 2\hat{i}+3\hat{j}-6\hat{k}$.
practice Find the direction cosines of the line joining $A(4,4,-2)$ and $B(1,0,-2)$, directed from $A$ to $B$.
practice Show that $\vec{a} = \hat{i}+\hat{j}+\hat{k}$ is equally inclined to the three coordinate axes.
practice $P$ has position vector $\vec{p} = 2\hat{i}-\hat{j}+3\hat{k}$ and $Q$ has position vector $\vec{q} = 4\hat{i}+3\hat{j}-\hat{k}$. Find the position vector of $R$ dividing $P Q$ internally in the ratio $3:1$.
practice For the same $P$ and $Q$ as above, find the position vector of $R$ dividing $P Q$ externally in the ratio $3:1$.
practice Find the position vector of the midpoint of $P Q$, where $P$ is at $3\hat{i}+2\hat{j}-\hat{k}$ and $Q$ is at $\hat{i}+4\hat{j}+3\hat{k}$.
practice Show that the points $A(1,1,1)$, $B(2,3,3)$ and $C(3,-1,2)$, given by their position vectors, are the vertices of a right-angled triangle.
practice In triangle $A B C$, which of the following is NOT necessarily true?
$\vec{A B}+\vec{B C}+\vec{C A} = \vec{0}$
$\vec{A B}+\vec{B C}-\vec{A C} = \vec{0}$
$\vec{A B}+\vec{B C}-\vec{C A} = \vec{0}$
$\vec{A B}-\vec{C B}+\vec{C A} = \vec{0}$
practice If $\vec{a}$ and $\vec{b}$ are collinear vectors, which of the following statements is INCORRECT?
$\vec{a} = \lambda \vec{b}$ for some scalar $\lambda$
The respective components of $\vec{a}$ and $\vec{b}$ are proportional
Both vectors necessarily point in the same direction
Both vectors lie along the same or parallel lines
practice Find $\lambda$ for which $\vec{a} = 2\hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b} = 4\hat{i}-6\hat{j}+2\hat{k}$ are collinear.
Answers
$-4\hat{j}-\hat{k}$
$13$
$5$ and $12$; magnitude $13$
$x=2$, $y=3$
$3\sqrt{38}$
$(\hat{i}+\hat{j}+2\hat{k})/\sqrt{6}$
$3/5 \hat{i}+4/5 \hat{j}$
$(\hat{i}+\hat{k})/\sqrt{2}$
$(40\hat{i}-8\hat{j}+16\hat{k})/\sqrt{30}$
$\vec{b} = -2 \vec{a}$, so the two vectors are collinear.
$(2/7, 3/7, -6/7)$
$(-3/5, -4/5, 0)$
$1/\sqrt{3}$: equal in all three, so equally inclined to the axes.
$7/2 \hat{i}+2\hat{j}$
$5\hat{i}+5\hat{j}-3\hat{k}$
$2\hat{i}+3\hat{j}+\hat{k}$
$0$: the dot product vanishes, so the angle at $A$ is a right angle and the triangle is right-angled.
(C) $\vec{A B}+\vec{B C}-\vec{C A} = \vec{0}$.
(C) Both vectors necessarily point in the same direction.
Exercise 10.3 practises both products. The first questions use the dot product for angles, projections and work. The later ones use the cross product for areas, perpendicular vectors and torque.
Exercise 10.3: Product of vectors, dot and cross
practice Find $\vec{a} \cdot \vec{b}$ for $\vec{a} = 2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b} = \hat{i}-2\hat{j}+4\hat{k}$.
practice Find $\vec{a} \cdot \vec{b}$ for $\vec{a} = 5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b} = \hat{i}+3\hat{j}-5\hat{k}$.
practice Find the angle between $\vec{a} = \hat{i}-\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+\hat{j}-\hat{k}$.
practice Find the angle between $\vec{a} = 3\hat{i}+4\hat{j}$ and $\vec{b} = 4\hat{i}+3\hat{j}$.
practice Find the projection of $\vec{a} = 3\hat{i}-2\hat{j}+\hat{k}$ on $\vec{b} = \hat{i}+2\hat{j}+2\hat{k}$.
practice Find the projection of $\vec{a} = \hat{i}-\hat{j}$ on $\vec{b} = \hat{i}+\hat{j}$.
practice A constant force $\vec{F} = 3\hat{i}+2\hat{j}-4\hat{k}$ (in newtons) displaces a body from $(1,2,3)$ to $(5,4,1)$ (in metres). Find the work done.
practice A force $\vec{F} = 5\hat{i}-3\hat{j}$ acts on a body moving along displacement $\vec{d} = 3\hat{i}+5\hat{j}$. Find the work done.
practice For what value of $\lambda$ are $\vec{a} = 2\hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b} = 4\hat{i}-2\hat{j}-2\hat{k}$ perpendicular?
practice Show that $\vec{a} = 6\hat{i}+2\hat{j}-8\hat{k}$ and $\vec{b} = 4\hat{i}-4\hat{j}+2\hat{k}$ are perpendicular.
practice If $\left|\vec{a}\right| = 3$, $\left|\vec{b}\right| = 4$ and $\vec{a} \cdot \vec{b} = 6$, find $\left|\vec{a}+\vec{b}\right|$.
practice If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 15$, find $\left|\vec{x}\right|$.
practice Two vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and $\vec{a} \cdot \vec{b} = 1/2$. Find $\left|\vec{a}\right|$.
practice If $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$, is it always true that $\vec{b} = \vec{c}$? Justify your answer.
practice If $\vec{a}$ and $\vec{b}$ are unit vectors with the angle between them $60^\circ$, find $\left|\vec{a}-\vec{b}\right|$.
practice If $\left|\vec{a}\right| = 2$, $\left|\vec{b}\right| = 3$ and $\left|\vec{a}+\vec{b}\right| = \sqrt{19}$, find $\vec{a} \cdot \vec{b}$.
practice Verify the triangle inequality $\left|\vec{a}+\vec{b}\right| \leq \left|\vec{a}\right|+\left|\vec{b}\right|$ for $\vec{a} = \hat{i}+\hat{j}$ and $\vec{b} = \hat{j}+\hat{k}$.
practice If $\vec{a}$ is a nonzero vector of magnitude $a$ and $\lambda$ is a nonzero scalar such that $\lambda \vec{a}$ is a unit vector, then:
$\lambda = 1$
$\lambda = -1$
$a = \left|\lambda\right|$
$a = 1/\left|\lambda\right|$
practice Find $\vec{a} \times \vec{b}$ for $\vec{a} = \hat{i}-2\hat{j}+3\hat{k}$ and $\vec{b} = 3\hat{i}+2\hat{j}-\hat{k}$.
practice Find $\vec{a} \times \vec{b}$ for $\vec{a} = 2\hat{i}+\hat{j}$ and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$.
practice Find the area of the triangle with adjacent sides $\vec{a} = 3\hat{i}+4\hat{j}$ and $\vec{b} = \hat{i}+2\hat{j}$.
practice Find the area of the parallelogram whose adjacent sides are $\vec{a} = 2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}-2\hat{j}-\hat{k}$.
practice Find the area of the triangle with vertices $A(1,2,0)$, $B(3,3,1)$ and $C(2,4,3)$.
practice Find a unit vector perpendicular to both $\vec{a} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}+3\hat{k}$.
practice For $\vec{a} = 2\hat{i}+\hat{j}$ and $\vec{b} = \hat{i}+3\hat{j}$, verify that $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.
practice For $\vec{a} = \hat{i}+2\hat{j}$ and $\vec{b} = 2\hat{i}+\hat{j}$, find $\vec{a} \cdot \vec{b}$ and $\vec{b} \cdot \vec{a}$.
practice For the same $\vec{a} = \hat{i}+2\hat{j}$ and $\vec{b} = 2\hat{i}+\hat{j}$, find $\vec{a} \times \vec{b}$, and state how $\vec{b} \times \vec{a}$ compares.
practice A force $\vec{F} = 2\hat{i}-3\hat{j}+\hat{k}$ (in newtons) acts at a point with position vector $\vec{r} = \hat{i}+\hat{j}+\hat{k}$ (in metres) relative to the origin. Find the moment of the force about the origin.
practice A wrench of length $0.5$ m is used to turn a bolt, with a force of $40$ N applied perpendicular to the wrench. Find the magnitude of the torque.
practice A unit vector makes an angle $\pi/3$ with $\hat{i}$ and $\pi/4$ with $\hat{j}$. If the angle it makes with $\hat{k}$ is acute, with direction cosine $n$, find $n$.
practice Two nonzero vectors $\vec{a}$ and $\vec{b}$ satisfy $\left|\vec{a} \times \vec{b}\right| = \vec{a} \cdot \vec{b}$. What can be concluded about the angle $\theta$ between them?
$\theta = \pi/6$
$\theta = \pi/4$
$\theta = \pi/3$
$\theta = \pi/2$
practice If $\left|\vec{a}\right| = 10$, $\left|\vec{b}\right| = 2$ and $\vec{a} \cdot \vec{b} = 12$, find $\left|\vec{a} \times \vec{b}\right|$.
practice The diagonals of a parallelogram are $\vec{d_1} = 3\hat{i}+\hat{j}-2\hat{k}$ and $\vec{d_2} = \hat{i}-3\hat{j}+4\hat{k}$. Find its area. *(Hint, write the diagonals as $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$.)*
practice Find the projection vector of $\vec{a} = 2\hat{i}+3\hat{j}+2\hat{k}$ on $\vec{b} = \hat{i}+2\hat{j}+\hat{k}$.
practice For $\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b} = -\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c} = 3\hat{i}+\hat{j}$, find $\lambda$ such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$.
Answers
$-8$
$17$
$\cos^{-1}(-1/3)$
$\cos^{-1}(24/25)$
$1/3$
$0$
$24$
$0$: the force is perpendicular to the displacement, so no work is done.
$3$
$0$: the dot product vanishes, so the two vectors are perpendicular.
$\sqrt{37}$
$4$
$1$
No. Take $\vec{a}=\hat{k}$, $\vec{b}=\hat{i}$, $\vec{c}=\hat{j}$: both dot products equal $0$, but $\vec{b}$ and $\vec{c}$ are different vectors.
$1$
$3$
$\left|\vec{a}+\vec{b}\right| = \sqrt{6} \approx 2.449$ and $\left|\vec{a}\right|+\left|\vec{b}\right| = 2\sqrt{2} \approx 2.828$, so the inequality holds.
(D) $a = 1/\left|\lambda\right|$.
$-4\hat{i}+10\hat{j}+8\hat{k}$
$\hat{i}-2\hat{j}+\hat{k}$
$1$
$3\sqrt{3}$
$\sqrt{35}/2$
$(2\hat{i}-\hat{j}-\hat{k})/\sqrt{6}$
$-5\hat{k}$: the exact negative of $\vec{a} \times \vec{b}$, confirming anti-commutativity.
$4$: the same value either way, since the dot product is commutative.
The Miscellaneous Exercise mixes everything you have met. The questions do not say which tool to use. So first ask what each one wants: a length, an angle, a point, the work done or an area.
Miscellaneous: Vector algebra
practice Write a unit vector in the $X Y$-plane making an angle of $60^\circ$ with the positive $x$-axis.
practice Find the scalar components and magnitude of the vector joining $P(3,2,-1)$ and $Q(-2,4,3)$, directed from $P$ to $Q$.
practice A cyclist rides $5$ km north, then $12$ km east. Find the magnitude of the cyclist’s net displacement from the starting point.
practice If $\left|\vec{a}\right| = \left|\vec{b}\right|$, is it true that $\vec{a} = \vec{b}$? Justify your answer.
practice Find the value of $x$ for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.
practice Find a vector of magnitude $10$ units, parallel to the resultant of $\vec{a} = 2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b} = \hat{i}-2\hat{j}+\hat{k}$.
practice If $\vec{a} = \hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b} = 3\hat{i}-\hat{j}+2\hat{k}$, find a unit vector parallel to $\vec{a}+\vec{b}$.
practice Show that the points $A(1,-2,-3)$, $B(5,2,1)$ and $C(13,10,9)$ are collinear, and find the ratio in which $B$ divides $A C$.
practice Find the position vector of a point $R$ which divides the join of $P$ (position vector $\vec{p} = 2\hat{i}-\hat{j}+\hat{k}$) and $Q$ (position vector $\vec{q} = \hat{i}+\hat{j}-\hat{k}$) externally in the ratio $1:2$.
practice The two adjacent sides of a parallelogram are $\vec{a} = 2\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{b} = \hat{i}-2\hat{j}-3\hat{k}$. Find the unit vector parallel to the diagonal $\vec{a}+\vec{b}$.
practice Find the area of that same parallelogram, with adjacent sides $\vec{a} = 2\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{b} = \hat{i}-2\hat{j}-3\hat{k}$.
practice If $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c} = \vec{0}$, find $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.
practice With respect to the right-handed system of mutually perpendicular unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$, let $\vec{\alpha} = \hat{i}+2\hat{j}$ and $\vec{\beta} = 3\hat{i}-\hat{j}+2\hat{k}$. Express $\vec{\beta} = \vec{\beta_1}+\vec{\beta_2}$, where $\vec{\beta_1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta_2}$ is perpendicular to $\vec{\alpha}$. Find $\vec{\beta_1}$ and $\vec{\beta_2}$.
practice A delivery boy cycles $3$ km east and then $4$ km north to reach a shop. Find the magnitude of his displacement from the start.
practice Find the area of the triangle with two sides given by $\vec{a} = \hat{i}+\hat{j}-2\hat{k}$ and $\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$.
practice A force $\vec{F} = \hat{i}-2\hat{j}+3\hat{k}$ (newtons) moves a body through a displacement $\vec{d} = 3\hat{i}+\hat{j}+2\hat{k}$ (metres). Find the work done.
practice The same force $\vec{F} = \hat{i}-2\hat{j}+3\hat{k}$ acts at the point with position vector $\vec{r} = 2\hat{i}+\hat{j}-\hat{k}$ relative to the origin. Find the moment of the force about the origin.
practice $A$ and $B$ have position vectors $\vec{a} = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b} = 4\hat{i}+\hat{j}-\hat{k}$. $R$ divides $A B$ internally in the ratio $1:2$. Find $\vec{O R} \cdot \vec{a}$, where $O$ is the origin.
practice If $\vec{a}$ and $\vec{b}$ are unit vectors with $\left|\vec{a}-\vec{b}\right| = 1$, find the angle between them.
practice If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \geq 0$ only when:
$0 < \theta < \pi/2$
$0 \leq \theta \leq \pi/2$
$0 < \theta < \pi$
$0 \leq \theta \leq \pi$
practice Let $\hat{a}$ and $\hat{b}$ be unit vectors and $\theta$ the angle between them. Then $\hat{a}+\hat{b}$ is a unit vector if $\theta$ equals:
practice The direction cosines of the vector $2 \hat{i} - 2 \hat{j} + \hat{k}$ are ______. (1 mark)
$2$, $-2$, $1$
$2/3$, $-2/3$, $1/3$
$2/3$, $2/3$, $1/3$
$2/9$, $-2/9$, $1/9$
practice Assertion (A). For any two vectors $\vec{a}$ and $\vec{b}$, $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. Reason (R). For any two vectors $\vec{a}$ and $\vec{b}$, $\left|\vec{a} \times \vec{b}\right| = \left|\vec{b} \times \vec{a}\right|$. (1 mark)
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
practice Find the projection of $\vec{a} = 4 \hat{i} + 2 \hat{j} - \hat{k}$ on $\vec{b} = 2 \hat{i} - 2 \hat{j} + \hat{k}$. (2 marks)
practice Find the area of the triangle with vertices $A(1, 0, 1)$, $B(3, 1, 2)$ and $C(2, 3, 0)$. (3 marks)
practice Let $\vec{a} = \hat{i} + 4 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{k}$. Find a vector of magnitude 9 that is perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$. (5 marks)
practice Meera walks from her home $H$ 6 km east to a shop, and then 8 km north to her school. Take $H$ as the origin, with $\hat{i}$ one km east and $\hat{j}$ one km north. (i) Write the two legs as vectors $\vec{a}$ and $\vec{b}$, and her displacement $\vec{d}$ from home. (1 mark) (ii) Find how far the school is from her home, and the unit vector in the direction of $\vec{d}$. (1 mark) (iii) Find the angle between the first leg and $\vec{d}$, using the dot product. (2 marks)
Answers
(B) $2/3$, $-2/3$, $1/3$.
(B) Both A and R are true but R is not the correct explanation of A.
