Three Dimensional Geometry

From arrows to geometry

You have the tools from Chapter 10: vectors, with the dot product and the cross product. This chapter puts them to work on straight lines in space, starting in the room in the picture. This first section shows what fixes a line in place. The next section turns that into the line’s equation.

Stand in a corner of your classroom. Two edges of the floor meet there, and so does the upright edge of the wall. In Class 11 you took such a corner as the origin. The three edges were the x-axis, the y-axis and the z-axis. Then every point in the room had three coordinates.

Now stretch a string from that corner to the far top corner of the room. The string is straight. Think of it extended through the walls, both ways, with no end. That is a line in space. It has no ends, so no length you can measure, and no single arrow is the whole of it.

Chapter 10 wrote an arrow as numbers, like the walk of 3 km north and 4 km east. A line needs a little more. It needs one point to pass through, and one direction to run in.

The figure fixes a line this way. It passes through the point $A(1, 2, 3)$. It runs along $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$. One step along $\vec{b}$ from $A$ reaches $(3, 4, 4)$. Step back, or step on, and you stay on the line.

Here are the questions to carry to the last page. How do you write a whole line as an equation? How do you say which way it points? What is the angle between two lines? And how close do two lines come when they never meet?

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Equation of a line in space

The first section fixed a line by one point and one direction. This section turns those two facts into an equation, which answers the first of the four questions. Every later section starts from that equation: it hands you the line’s direction, and the direction gives you angles and distances.

Think of a straight road on a city map. You stand at a bus stop on it, and someone tells you which way the road runs. That is enough to find every point of the road. Go some distance one way, or the other way, and you are still on it.

So, as in the room, a line needs two things: one point on it, and one direction along it. Every other point is the start plus some number of steps in that direction.

The whole line

Now name the parts. Let $\vec{a}$ be the position vector of a point $A$ on the line. Let $\vec{b}$ be a vector along the line, with $\vec{b} \neq \vec{0}$. Take any point $R$, with position vector $\vec{r}$. It is on the line exactly when the vector from $A$ to $R$ is some number of copies of $\vec{b}$:

$\vec{r} - \vec{a} = \lambda \vec{b}$.

Then add $\vec{a}$ to both sides. A point is on the line exactly when

$\vec{r} = \vec{a} + \lambda \vec{b}$

for some real number $\lambda$. This is the VECTOR EQUATION of the line. The number $\lambda$ is a PARAMETER. Each value of $\lambda$ gives one point. Every point of the line comes from exactly one value.

Let us check it on the figures. There $A$ is $(1, 2, 1)$, and $\vec{b} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$. Each value of $\lambda$ gives one point:

The sign says which way you go. With $\lambda > 0$ you move one way from $A$, and with $\lambda < 0$ you move the other way.

The three numbers in $\vec{b}$, here $2, 3, 2$, are called the DIRECTION RATIOS of the line. Any three numbers in the same ratio, such as $4, 6, 4$, point the same way and do just as well. The section after this one says more about them, and how they relate to angles.

Go back to the road on the map. As you move along it, your east and north positions both change. Each one changes at its own steady rate, set by the road’s direction. In space a third number, your height, changes the same way.

Eliminating the parameter

So write the vector equation one coordinate at a time, then remove $\lambda$. Take the point $A(x_1, y_1, z_1)$ and the direction $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$, so $a$, $b$, $c$ are its direction ratios. Here it is, step by step:

The last step gives the CARTESIAN EQUATION of the line. The point sits in the numerators, each part with a minus sign. The direction ratios sit in the denominators.

Look at the figure. One multiple of the direction splits into three direction ratios: 2 along $x$, 3 along $y$ and 2 along $z$. They go in the denominators. The point $A(1, 2, 1)$ goes in the numerators, so the line is $(x - 1)/2 = (y - 2)/3 = (z - 1)/2$.

Check it at the point $(5, 8, 5)$. The three parts are $(5 - 1)/2$, $(8 - 2)/3$ and $(5 - 1)/2$, and each is 2. They agree, so the point is on the line, and 2 is its $\lambda$.

Sometimes a denominator is 0, as in $(x - 1)/2 = (y - 2)/3 = (z - 1)/0$. Read that 0 as a direction ratio of 0, not as a division. It means $z$ never changes along the line. Here $z = 1$ all the way.

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Direction cosines and direction ratios of a line

You can now write any line from a point and a direction $\vec{b}$. But one line has many such arrows: $2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ and $4 \hat{i} + 6 \hat{j} + 4 \hat{k}$ run along the same line. This section answers the second question: how do you say which way a line points, with one fixed set of three numbers? The angle section after it is built on that set.

Go back to the string in the corner of the room. It leans away from all three edges at once, so it makes an angle with each edge. Here is why those three angles matter. Together they fix which way the string points, however long it is. Their cosines come straight from the three numbers in $\vec{b}$ that the equation already uses. And once each line has its three numbers, two lines can be compared, which is how the angle section finds the angle between them.

Turn the string, and at least one of the three angles changes. Slide it without turning, and none of them do.

Give the angles names. A line makes the angle $\alpha$ with the $x$-axis, $\beta$ with the $y$-axis and $\gamma$ with the $z$-axis. Their cosines are the DIRECTION COSINES of the line:

$l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$.

Facing the other way

Chapter 10 found these for a vector. A line has one catch. It runs both ways, so you can face along it either way. Face the other way, and each angle becomes $\pi$ minus itself. Then all three cosines change sign. So a line has two sets of direction cosines, $l, m, n$ and $-l, -m, -n$. They are the same up to one sign for all three.

The squares add to 1

Why cosines? Take the point at distance 1 along the line, and drop it onto the $x$-axis. The right triangle has hypotenuse 1 and the angle $\alpha$, so the foot is at $\cos \alpha$, which is $l$. In the same way the feet on the $y$ and $z$ axes are $m$ and $n$.

So the cosines are the components of a vector of length 1 along the line, $l \hat{i} + m \hat{j} + n \hat{k}$. Its length $\sqrt{l^2 + m^2 + n^2}$ is 1. Square both sides, as in Chapter 10:

$l^2 + m^2 + n^2 = 1$.

Let us check it on the figure on the previous page, with direction ratios $2, 3, 6$ and its three angles marked.

You met direction ratios in the last section: the three numbers in $\vec{b}$. Any three numbers in the same ratio as $l$, $m$, $n$ are direction ratios of the line. Here $2, 3, 6$ is one set, and $4, 6, 12$ is another. To get the cosines from ratios $a$, $b$, $c$, divide each by their length. Here is why that works:

Use the same sign for all three. Dividing by $-\sqrt{a^2 + b^2 + c^2}$ also gives an arrow of length 1, but facing the other way along the line. That is the other set, $-l, -m, -n$. Change only one sign, as in the figure, and you have a different line.

Read any direction in four steps. First write its three components: they are direction ratios. If the line is given by two points, subtract the first point’s coordinates from the second’s: those differences are the ratios. Then find their length, the square root of the sum of their squares. Divide each ratio by the length to get the direction cosines. Last, check that the squares of the cosines add to 1.

Check each line of this table against the rule. Then cover the last two columns and work them out yourself.

The figure above shows one direction with three sets of ratios: $1, 2, 2$ and $2, 4, 4$ and $3, 6, 6$. Their lengths are 3, 6 and 9. Divide each set by its own length, and every one gives $1/3$, $2/3$, $2/3$.

Three angles to test

Your turn: a line makes angles of 45, 60 and 45 degrees with the three axes. Can these be its direction angles? Square the cosines and add. Do you get $5/4$ rather than 1?

You do not need to measure any angle. Two points on a line already fix its direction.

Take the string again. It has two ends. To go from one end to the other, you move some way along each edge of the room. Those three moves say which way the string runs.

Now write the ends as $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. From $P$ to $Q$ you move $x_2 - x_1$ along the $x$-axis, $y_2 - y_1$ along the $y$-axis and $z_2 - z_1$ along the $z$-axis. These three moves are direction ratios of the line $P Q$.

The figure draws them as the three edges of a box. The line $P Q$ is the box’s long diagonal. Its length is the distance from Class 11:

$P Q = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

From direction ratios to cosines

Divide each direction ratio by $P Q$, and you have the direction cosines:

$l = (x_2 - x_1)/(P Q)$, $m = (y_2 - y_1)/(P Q)$, $n = (z_2 - z_1)/(P Q)$.

Let us try it on the room. Say it is 6 m long, 6 m wide and 3 m high. The string runs from the corner $(0, 0, 0)$ to the far top corner $(6, 6, 3)$.

Take the ends the other way round, from $(6, 6, 3)$ back to $(0, 0, 0)$, and all three moves change sign. That is the same line facing the other way, so nothing is wrong.

Two points to try

Your turn: a line passes through $(1, 2, 3)$ and $(3, 5, 9)$. Find its direction cosines, and check that their squares add to 1. Do you get $2/7$, $3/7$, $6/7$?

Direction ratios and direction cosines are not the same thing. Direction cosines satisfy $l^2 + m^2 + n^2 = 1$. Direction ratios need not.

Step 1

The direction ratios are 2, 4 and 4.

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Step 2

Direction ratios point the same way as the line.

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Step 3

So the direction cosines are 2, 4 and 4.

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Here is a working for the direction cosines of a line with direction ratios 2, 4, 4. Which step is wrong?

• The direction ratios are 2, 4 and 4.
• Direction ratios point the same way as the line.
• So the direction cosines are 2, 4 and 4.

Keep your answer, then test it. Square and add: $4 + 16 + 16 = 36$, not 1. So 2, 4, 4 cannot be direction cosines.

The third step is wrong. The ratios do point the same way, but their length is 6, not 1. Divide each one by 6:

$l = 2/6 = 1/3$, $m = 4/6 = 2/3$, $n = 4/6 = 2/3$.

Now $1/9 + 4/9 + 4/9 = 1$. The figure draws both. The ratios make an arrow 6 long. The cosines make an arrow 1 long, in the same direction.

So before you write down direction cosines, square them and add. You must get 1.

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Angle between two lines

Each line now has a direction you can write down, as direction ratios or as direction cosines. This section takes two lines at once and answers the third question: what is the angle between them? In direction cosines, $\cos \theta = \left|l_1 l_2 + m_1 m_2 + n_1 n_2\right|$. In direction ratios, which this chapter works with throughout, the same formula divides through by the two lengths. It also tells you when two lines are perpendicular and when they are parallel.

Stretch two strings from the same corner of the room to two other corners. Where they meet, there is an angle between them. Now move one string to another place without turning it. The two strings may no longer meet. But they still point the same ways, so the angle between their directions has not changed.

So the angle between two lines depends only on their directions. For lines that do not meet, slide one without turning it until it meets the other. The angle there is the ANGLE BETWEEN THE TWO LINES.

A line can face either way. So two lines make two angles, $\theta$ and $\pi - \theta$. The angle between the lines is taken as the one from 0 to $\pi/2$. That is why the formula has an absolute value.

The cosine formula

Take two directions, $\vec{b_1} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$ and $\vec{b_2} = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}$. Their direction ratios are $a_1, b_1, c_1$ and $a_2, b_2, c_2$. Here is how the dot product of Chapter 10 gives the angle, step by step:

Let us check the formula on the first panel of the figure, where the directions are $2, 2, 1$ and $-1, 2, 2$:

Perpendicular and parallel

Two cases come straight out of the formula.

• The lines are PERPENDICULAR exactly when $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, since then $\cos \theta = 0$. In the second panel, $2, 2, 1$ and $1, -2, 2$ give $2 - 4 + 2 = 0$.
• The lines are PARALLEL exactly when their ratios are proportional, since then $\theta = 0$. When none of $a_2$, $b_2$, $c_2$ is 0, that reads $a_1/a_2 = b_1/b_2 = c_1/c_2$. In the third panel, $3, 3, 1.5$ is 1.5 times $2, 2, 1$.

Your turn: two lines have direction ratios $1, -2, 3$ and $-2, 4, -6$. Find $\cos \theta$. What does it tell you about the lines? Do you get $\cos \theta = 1$, so that the lines are parallel?

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Shortest distance between two lines

Two lines that meet make an angle, and the last section found it with the dot product. In space, two lines can also pass each other without meeting, and without being parallel. This section answers the last question: how close do two lines come? It uses the cross product, and with it the chapter’s four questions are answered.

Stand under a flyover in a city. The road on the flyover runs one way. The road below runs another way. They are not parallel, and they never meet. One passes over the other.

Two such lines in space are SKEW LINES. On flat paper this cannot happen. There, two lines either meet or are parallel. In space there is a third case.

How close do the two roads come? The shortest gap is the one at right angles to both roads. Under the flyover it is the vertical, straight up from the road below to the road above.

Now write it in vectors. Let the lines be $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$. The direction at right angles to both is $\vec{b_1} \times \vec{b_2}$, the cross product of Chapter 10. The formula below needs $\vec{b_1} \times \vec{b_2} \neq \vec{0}$. A cross product is $\vec{0}$ exactly when the two vectors are parallel, so this holds exactly when the lines are not parallel.

The vector joining a point of one line to a point of the other is $\vec{a_2} - \vec{a_1}$. The SHORTEST DISTANCE between the lines is the length of $\vec{a_2} - \vec{a_1}$'s projection on $\vec{b_1} \times \vec{b_2}$.

Why the projection? Let $\hat{n}$ be the unit vector along $\vec{b_1} \times \vec{b_2}$. A general point of line 1 is $\vec{a_1} + \lambda \vec{b_1}$, and of line 2 is $\vec{a_2} + \mu \vec{b_2}$. The vector joining any such pair is

$(\vec{a_2} + \mu \vec{b_2}) - (\vec{a_1} + \lambda \vec{b_1}) = (\vec{a_2} - \vec{a_1}) + \mu \vec{b_2} - \lambda \vec{b_1}$.

Dot this with $\hat{n}$: the last two terms are multiples of $\vec{b_1}$ and $\vec{b_2}$, both at right angles to $\hat{n}$, so they vanish, and every such joining vector has the same part along $\hat{n}$, equal to $(\vec{a_2} - \vec{a_1}) \cdot \hat{n}$. None is shorter than that part, and the joining vector at right angles to both lines runs along $\hat{n}$, so it is exactly that long.

Now turn it into a formula, step by step:

The formula holds for any two lines that are not parallel. Then $d = 0$ exactly when they meet, as on the left of the figure, and $d > 0$ when they are skew, as on the right.

Let us check it on the figure. One line goes through $(0, 0, 1)$ along $1, 3, 0$. The other goes through $(4, 1, 1)$ along $1, 0, 3$. On the page they seem to cross, but in space they pass apart.

Step 1

The directions 2î − ĵ + k̂ and 3î − 5ĵ + 2k̂ are not proportional. So the lines are not parallel.

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Step 2

Two lines that are not parallel always meet.

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Step 3

So solving the two equations together must give a meeting point.

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Here is a working for the two lines of the worked example. Which step is wrong?

• The directions $2 \hat{i} - \hat{j} + \hat{k}$ and $3 \hat{i} - 5 \hat{j} + 2 \hat{k}$ are not proportional. So the lines are not parallel.
• Two lines that are not parallel always meet.
• So solving the two equations together must give a meeting point.

Keep your answer, then try it. Set the coordinates of the two lines equal, and solve, step by step:

The second step is wrong. Two lines that are not parallel always meet on flat paper, but not in space. These lines are skew.

No meeting point does not mean nothing to find. Skew lines still have an angle between them, from their directions. They also have a shortest distance, here $10/\sqrt{59}$.

Look along a railway track. The two rails run side by side and never meet. They point the same way, and the gap between them is the same everywhere.

Two PARALLEL LINES are like the rails. They share one direction $\vec{b}$ and pass through different points. Write them as $\vec{r} = \vec{a_1} + \lambda \vec{b}$ and $\vec{r} = \vec{a_2} + \mu \vec{b}$.

The skew-line formula does not work here. Both lines run along the same $\vec{b}$, so the angle between their directions is 0. Chapter 10 gave the length of a cross product as $\left|\vec{b}\right| \left|\vec{b}\right| \sin \theta$, and $\sin 0 = 0$:

$\vec{b} \times \vec{b} = \vec{0}$.

For parallel lines the cross product of the two directions is $\vec{0}$, and you cannot divide by 0. Use the one direction instead. The DISTANCE BETWEEN PARALLEL LINES is

$d = \left|\vec{b} \times (\vec{a_2} - \vec{a_1})\right|/\left|\vec{b}\right|$.

Here is why. The vectors $\vec{b}$ and $\vec{a_2} - \vec{a_1}$ are two sides of a parallelogram, as in the figure. Its height is the gap between the rails:

Check it on the figure. One line passes through $A_1(1, 1, 1)$, and the other through $A_2(3, 1, -1)$. Both run along $1, 4, 1$.

The same gap, by formula

Your turn: use the formula on the same two lines. Do you get $2 \sqrt{2}$ again?

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One toolkit, every question in this chapter

Each section answered one question and handed its answer on. The equation gave a direction, the direction gave the cosines, two directions gave the angle, and the cross product gave the gap. This last section puts the four answers in one place, ready for the exercises.

The chapter opened with four questions. How do you write a whole line as an equation? How do you say which way it points? What is the angle between two lines? And how close do two lines come when they never meet?

• A LINE through $\vec{a}$ along $\vec{b} \neq \vec{0}$: $\vec{r} = \vec{a} + \lambda \vec{b}$. With $A(x_1, y_1, z_1)$ and $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$, it is $(x - x_1)/a = (y - y_1)/b = (z - z_1)/c$.
• DIRECTION RATIOS: the three numbers in $\vec{b}$, or any three in the same ratio. DIRECTION COSINES: the ratios divided by their length, $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$, with $l^2 + m^2 + n^2 = 1$. A line has two sets, $l, m, n$ and $-l, -m, -n$. Ratios $2, 3, 6$ give $2/7$, $3/7$, $6/7$.
• ANGLE between two lines: $\cos \theta = \left|\vec{b_1} \cdot \vec{b_2}\right|/(\left|\vec{b_1}\right| \left|\vec{b_2}\right|)$, with $0 \leq \theta \leq \pi/2$. Perpendicular when the dot product is 0. Parallel when the directions are proportional.
• NON-PARALLEL LINES, where $\vec{b_1} \times \vec{b_2} \neq \vec{0}$: $d = \left|(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})\right|/\left|\vec{b_1} \times \vec{b_2}\right|$. Then $d = 0$ exactly when they meet; otherwise they are skew and $d > 0$.
• PARALLEL LINES along $\vec{b}$: $d = \left|\vec{b} \times (\vec{a_2} - \vec{a_1})\right|/\left|\vec{b}\right|$.

The figure sets two cases side by side. Two lines that meet at $(1, 1, 1)$ have $\cos \theta = 4/9$. Two skew lines are $\sqrt{11}$ apart.

Back to the string in the room, 6 m by 6 m by 3 m. It runs from $(0, 0, 0)$ along $6, 6, 3$. So its equation is $\vec{r} = \lambda (6 \hat{i} + 6 \hat{j} + 3 \hat{k})$, and its direction cosines are $2/3$, $2/3$ and $1/3$.

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Practice set: Exercise 11.1

Exercise 11.1 practises direction cosines and direction ratios. You find them from angles, from ratios and from two points on a line. Check every answer: the squares of the cosines add to 1.

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Practice set: Exercise 11.2

Exercise 11.2 practises the equation of a line, in vector form and in Cartesian form. You also find the angle between two lines. Some questions ask for the distance between skew lines or parallel lines.

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Practice set: miscellaneous

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 direction, an equation, an angle or a distance.