THREE DIMENSIONAL GEOMETRY

Chapter 11

Three Dimensional Geometry

Big idea: to describe a straight line in space you only need two things, a point on it and a direction. Angles and distances between lines then come from calculations with those directions.

A drone flies in a straight line from one point to another. A laser beam crosses a room. Two pipes run past each other in a building, one above the other. All of these are straight lines in space. In a flat drawing, a line needs only two coordinates. In space, every point needs three: $(x, y, z)$.

In Class XI you learned the distance formula in space. The distance between $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is

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

For example, take $A = (1, 1, 1)$ and $B = (3, 3, 2)$. The steps are $2$, $2$ and $1$ along the three axes, so $AB = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

In Chapter 10 you met vectors. Here we use vectors to study straight lines. In this chapter you will learn:

• direction cosines and direction ratios: numbers that describe which way a line points;
• the equation of a line in space, in vector form and in Cartesian form;
• the angle between two lines;
• the shortest distance between two lines, whether they are skew or parallel.

Draw a line through the origin and pick a direction along it. This directed line makes three angles with the positive $x$-, $y$- and $z$-axes. Call them $\alpha$, $\beta$ and $\gamma$. The three numbers

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

are the direction cosines of the line. They say how much the line leans towards each axis. If a unit vector points along the line, then $(l, m, n)$ are exactly its three components.

A unit vector has length $1$, so its components satisfy

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

Every set of direction cosines passes this test. If three numbers do not, they cannot be direction cosines.

Direction cosines are awkward because of the square roots. So we also use any three numbers $a, b, c$ proportional to $l, m, n$. They are called direction ratios. If $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ points along the line, then $a, b, c$ are direction ratios, and

$$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$$

and in the same way $m = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}}$ and $n = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}}$.

• Direction cosines are unique for a directed line. Direction ratios are not: if $(a, b, c)$ works, so does $(2a, 2b, 2c)$ or $(-a, -b, -c)$.
• A directed line has a direction. Reversing the direction changes the sign of every direction cosine.

Find the direction cosines of the line along $2\hat{i} - \hat{j} + 2\hat{k}$. Here $a^2 + b^2 + c^2 = 9$, so the length is $\sqrt{9} = 3$. The direction cosines are $(\frac{2}{3},\; -\frac{1}{3},\; \frac{2}{3})$.

Check: $l^2 + m^2 + n^2 = \frac{4 + 1 + 4}{9} = 1$. The angles with the axes are $\alpha = \cos^{-1}\frac{2}{3} \approx 48.19°$, $\beta = \cos^{-1}\left(-\frac{1}{3}\right) \approx 109.47°$ and $\gamma \approx 48.19°$.

A line makes equal angles with the three axes. Then $l = m = n$, and $l^2 + m^2 + n^2 = 1$ gives $3l^2 = 1$, so $l = \pm\dfrac{1}{\sqrt{3}}$. Each angle is $\cos^{-1}\dfrac{1}{\sqrt{3}} \approx 54.74°$ (or $125.26°$ for the opposite direction). A direction ratio set is $(1, 1, 1)$.

Each row was computed exactly. In the last two rows the square root does not simplify, so the cosines keep it.

Reverse the vector in Example 1 to $-2\hat{i} + \hat{j} - 2\hat{k}$. The cosines become $(-\frac{2}{3},\; \frac{1}{3},\; -\frac{2}{3})$, and the angles become $131.81°$, $70.53°$ and $131.81°$. Each new angle is $180°$ minus the old one, because $\cos(180° - \alpha) = -\cos\alpha$.

1. Find the direction cosines of the line with direction ratios $(1, 2, 2)$.
2. Find the direction cosines of the line with direction ratios $(3, -4, 0)$. What angle does it make with the $z$-axis?

Answers

1. Length $\sqrt{9} = 3$. Direction cosines $(\frac{1}{3},\; \frac{2}{3},\; \frac{2}{3})$.
2. Length $\sqrt{25} = 5$. Direction cosines $(\frac{3}{5},\; -\frac{4}{5},\; 0)$. The $z$-cosine is $0$, so the angle with the $z$-axis is $90°$.

Often a line is given by two points on it, $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. The vector $\overrightarrow{PQ}$ lies along the line, and its components are the steps from $P$ to $Q$. So the direction ratios are

$$x_2 - x_1, \qquad y_2 - y_1, \qquad z_2 - z_1.$$

Divide each by the length $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ to get the direction cosines (for the direction from $P$ to $Q$).

Find the direction cosines of the line through $P = (2, 1, -1)$ and $Q = (4, -1, 0)$. The steps are $(2, -2, 1)$ and the length is $\sqrt{9} = 3$. So the direction cosines are $(\frac{2}{3},\; -\frac{2}{3},\; \frac{1}{3})$.

For $P = (0, 1, 2)$ and $Q = (1, 3, 5)$ the steps are $(1, 2, 3)$ and $a^2 + b^2 + c^2 = 14$. The direction cosines are $(\frac{1}{\sqrt{14}},\; \frac{2}{\sqrt{14}},\; \frac{3}{\sqrt{14}})$. It is fine to leave $\sqrt{14}$ in the answer.

Three points $A$, $B$, $C$ are collinear (on one line) exactly when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are in the same direction or opposite directions. That means their direction ratios are proportional.

Test $A = (1, 1, 1)$, $B = (3, 2, 0)$, $C = (7, 4, -2)$. Here $\overrightarrow{AB} = (2, 1, -1)$ and $\overrightarrow{AC} = (6, 3, -3)$. The second is exactly $3$ times the first, so the three points lie on one line.

Now test $D = (1, 1, 1)$, $E = (2, 3, 4)$, $F = (4, 6, 9)$. Here $\overrightarrow{DE} = (1, 2, 3)$ and $\overrightarrow{DF} = (3, 5, 8)$. They are not proportional, because the ratios $\frac{3}{1}$, $\frac{5}{2}$, $\frac{8}{3}$ are different. (Their cross product is $(1, 1, -1) \ne \vec{0}$, which confirms it.) So $D$, $E$, $F$ are not on one line.

1. Write the direction cosines of the $x$-axis, the $y$-axis and the $z$-axis.
2. Can $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}\right)$ be the direction cosines of a line? What about $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$? If the first can, find the three angles.
3. Find the direction cosines of the line along the vector $4\hat{i} - 3\hat{j} + 12\hat{k}$.
4. Find the direction cosines of the line through $(-2, 4, -5)$ and $(1, 2, 3)$.
5. A line makes an angle of $60°$ with the $x$-axis and $45°$ with the $y$-axis. Find the possible angles with the $z$-axis and the direction cosines.
6. A line has direction ratios $2, -3, 6$. Find its direction cosines and the angle it makes with each axis.
7. Show that the points $A(2, 3, -4)$, $B(1, -2, 3)$ and $C(3, 8, -11)$ are collinear.
8. Lines have direction ratios $(1, 2, 3)$ and $(-2, -4, -6)$. Are they parallel? Are their direction cosines equal?

Answers

1. For the $x$-axis, direction ratios $(1, 0, 0)$ give $(1,\; 0,\; 0)$. For the $y$-axis, $(0,\; 1,\; 0)$. For the $z$-axis, $(0,\; 0,\; 1)$. (Each makes an angle $0°$ with itself and $90°$ with the other two axes.)
2. The first: $\frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1$, so yes. The angles are $\cos^{-1}\frac{1}{2} = 60°$, $60°$ and $\cos^{-1}\frac{1}{\sqrt{2}} = 45°$. The second: $\frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 0.75$, which is not $1$, so no.
3. $a^2 + b^2 + c^2 = 169$ and the length is $13$. The direction cosines are $(\frac{4}{13},\; -\frac{3}{13},\; \frac{12}{13})$.
4. The steps are $(3, -2, 8)$, and $a^2 + b^2 + c^2 = 77$. The direction cosines are $(\frac{3}{\sqrt{77}},\; -\frac{2}{\sqrt{77}},\; \frac{8}{\sqrt{77}})$.
5. $l = \frac{1}{2}$ and $m = \frac{1}{\sqrt{2}}$, so $n^2 = 1 - \frac{1}{4} - \frac{1}{2} = 0.25$ and $n = \pm\frac{1}{2}$. The angle with the $z$-axis is $60°$ (for $n = \frac{1}{2}$) or $120°$ (for $n = -\frac{1}{2}$).
6. $a^2 + b^2 + c^2 = 49$, so the length is $7$. The direction cosines are $(\frac{2}{7},\; -\frac{3}{7},\; \frac{6}{7})$. The angles are about $73.4°$, $115.38°$ and $31°$.
7. $\overrightarrow{AB} = (-1, -5, 7)$ and $\overrightarrow{AC} = (1, 5, -7)$. Since $\overrightarrow{AC} = -\overrightarrow{AB}$, they are along the same line, so $A$, $B$, $C$ are collinear. ($C$ is on the other side of $A$ from $B$.)
8. The second is $-2$ times the first, so the lines are parallel. The direction cosines are $(\frac{1}{\sqrt{14}},\; \frac{2}{\sqrt{14}},\; \frac{3}{\sqrt{14}})$ and $(-\frac{1}{\sqrt{14}},\; -\frac{2}{\sqrt{14}},\; -\frac{3}{\sqrt{14}})$, which are opposite in sign. The lines have the same direction line but opposite directions.

On a flat page, a line has the equation $y = mx + c$. In space this does not work: one equation in $x$, $y$, $z$ describes a surface, not a line. We need another idea.

The idea is to imagine a moving point. Start at a known point $A$ and walk in a fixed direction $\vec{b}$. After "time" $\lambda$, you are at $A + \lambda\vec{b}$. As $\lambda$ takes every real value, forwards and backwards, you trace out the whole line. The letter $\lambda$ is called the parameter.

The picture shows the line through $A = (1, -1, 2)$ with direction $\vec{b} = (2, 1, 1)$. The marked points are $\lambda = -1, 0, 1, 2$.

There are two common ways to fix a line in space:

• a point on the line and a direction it is parallel to (Section 11.3.1);
• two points on the line (also in Section 11.3.1: the direction is the vector joining them).

Let a line pass through the point $A$ with position vector $\vec{a}$, and be parallel to the vector $\vec{b}$. A general point $P$ on the line has position vector $\vec{r}$. The vector $\overrightarrow{AP}$ lies along the line, so it is a multiple of $\vec{b}$: $\overrightarrow{AP} = \lambda\vec{b}$. Since $\vec{r} = \vec{a} + \overrightarrow{AP}$,

$$\vec{r} = \vec{a} + \lambda\vec{b}, \qquad \lambda \in \mathbb{R}.$$

This is the vector equation of the line. Different values of $\lambda$ give different points. $\lambda = 0$ gives $A$ itself.

Write the vector equation of the line through $A = (2, -1, 3)$ parallel to $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.

$$\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 2\hat{k})$$

For $\lambda = 2$ the point is $(4, 3, -1)$. Does $(5, 5, -3)$ lie on the line? We need a single $\lambda$ that gives all three coordinates. The computer found $\lambda = 3$ works for $(5, 5, -3)$, so yes. For $(5, 5, -2)$ no single $\lambda$ works, so it is not on the line.

If the line passes through $A$ (position vector $\vec{a}$) and $B$ (position vector $\vec{b}$), its direction is $\overrightarrow{AB} = \vec{b} - \vec{a}$. So

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

Here $\lambda = 0$ gives $A$ and $\lambda = 1$ gives $B$. Values between $0$ and $1$ give points between $A$ and $B$. For example $\lambda = \frac{1}{2}$ gives the midpoint.

Find the vector equation of the line through $T_1 = (1, 0, 2)$ and $T_2 = (3, 4, -2)$. The direction is $T_2 - T_1 = (2, 4, -4)$. So $\vec{r} = (\hat{i} + 2\hat{k}) + \lambda(2\hat{i} + 4\hat{j} - 4\hat{k})$. At $\lambda = 1$ we get $T_2$, and at $\lambda = \frac{1}{2}$ we get the midpoint $(2, 2, 0)$ (the computer confirmed both).

Any multiple of a direction is also a direction. The vector $(2, 4, -4)$ is $2$ times $(1, 2, -2)$. So the same line is also $\vec{r} = (\hat{i} + 2\hat{k}) + \mu(\hat{i} + 2\hat{j} - 2\hat{k})$. The parameter $\mu$ is half as large: $T_2$ is now at $\mu = 2$.

Let the line pass through $A(x_1, y_1, z_1)$ and have direction ratios $a, b, c$. Write the general point as $P(x, y, z)$. The vector equation $\vec{r} = \vec{a} + \lambda\vec{b}$ says, component by component:

$$x = x_1 + \lambda a, \qquad y = y_1 + \lambda b, \qquad z = z_1 + \lambda c.$$

Solve each for $\lambda$: $\lambda = \frac{x - x_1}{a}$, $\lambda = \frac{y - y_1}{b}$, $\lambda = \frac{z - z_1}{c}$. All three are equal to the same $\lambda$, so

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}.$$

This is the Cartesian form (or symmetric form). It describes the same line with no parameter. If you write direction cosines $l, m, n$ in the denominators, the line is the same, because $(l, m, n)$ is proportional to $(a, b, c)$.

Write the line $\vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 5\hat{j} - \hat{k})$ in Cartesian form. The point is $(3, -1, 2)$ and the direction ratios are $(2, 5, -1)$. So

$$\dfrac{x - 3}{2} = \dfrac{y + 1}{5} = \dfrac{z - 2}{-1}.$$

Write $\dfrac{x + 1}{3} = \dfrac{y - 2}{-2} = \dfrac{z}{4}$ as a vector equation. Read the point from the numerators (watch the signs): $x - (-1)$ gives $-1$, $y - 2$ gives $2$, and $z$ gives $0$. The direction ratios are the denominators $(3, -2, 4)$. So $\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - 2\hat{j} + 4\hat{k})$. Check: for $\lambda = 1$ the point is $(2, 0, 4)$, and the computer confirmed it is on the line.

For the line through $(1, 0, 2)$ and $(3, 4, -2)$ the direction ratios are $(2, 4, -4)$, or after dividing by $2$, $(1, 2, -2)$. The Cartesian form is

$$\dfrac{x - 1}{1} = \dfrac{y}{2} = \dfrac{z - 2}{-2}.$$

In general, the line through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\dfrac{x - x_1}{x_2 - x_1} = \dfrac{y - y_1}{y_2 - y_1} = \dfrac{z - z_1}{z_2 - z_1}$.

Division by $0$ is not allowed, so a zero in the denominator needs care. Take the line through $(1, 2, 3)$ with direction ratios $(0, 2, 1)$. The $x$-component never changes: $x = 1$ for every point. We still write the form in a compact way:

$$\frac{x - 1}{0} = \frac{y - 2}{2} = \frac{z - 3}{1}.$$

This is a notation that means: $x - 1 = 0$, and $\frac{y - 2}{2} = \frac{z - 3}{1}$. For example, $(1, 4, 4)$ is on this line (it is the point with $\lambda = 1$).

1. Write $\vec{r} = (\hat{i} - 2\hat{k}) + \lambda(3\hat{i} - \hat{j} + 4\hat{k})$ in Cartesian form.
2. Write $\dfrac{x - 2}{1} = \dfrac{y + 3}{4} = \dfrac{z - 1}{-2}$ in vector form.

Answers

1. $\dfrac{x - 1}{3} = \dfrac{y}{-1} = \dfrac{z + 2}{4}$
2. $\vec{r} = (2\hat{i} - 3\hat{j} + \hat{k}) + \lambda(\hat{i} + 4\hat{j} - 2\hat{k})$

Two lines in space may meet, or may pass each other without meeting (we call these skew lines). In both cases we can still speak of the angle between them. It is the angle between their directions. For skew lines, imagine sliding one line, without turning it, until it meets the other.

Suppose the lines are parallel to $\vec{b}_1$ and $\vec{b}_2$. From the dot product, $\vec{b}_1 \cdot \vec{b}_2 = |\vec{b}_1||\vec{b}_2|\cos\theta$. Two lines always make two angles, $\theta$ and $180° - \theta$. We take the acute one (at most $90°$), so we use the absolute value:

$$\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1|\,|\vec{b}_2|}.$$

In terms of direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:

$$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}.$$

With direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ the denominator is $1$, so $\cos\theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.

• Perpendicular lines: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ (the dot product is zero).
• Parallel lines: the direction ratios are proportional: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

Find the angle between the lines with directions $(1, 2, 2)$ and $(2, 1, -2)$. The dot product is $1\cdot 2 + 2\cdot 1 + 2\cdot(-2) = 0$. So the lines are perpendicular: $\theta = 90°$.

Directions $(1, 1, 0)$ and $(0, 1, 1)$. The dot product is $1$, and both lengths are $\sqrt{2}$. So $\cos\theta = \dfrac{1}{\sqrt{2}\sqrt{2}} = \dfrac{1}{2}$ and $\theta = 60°$.

Find the angle between the lines $\dfrac{x - 1}{2} = \dfrac{y + 3}{-1} = \dfrac{z - 4}{2}$ and $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z + 2}{2}$. Read the direction ratios from the denominators: $(2, -1, 2)$ and $(1, 2, 2)$. Then $\vec{b}_1\cdot\vec{b}_2 = 4$ and both lengths are $\sqrt{9} = 3$. So $\cos\theta = \dfrac{4}{3 \cdot 3} = \dfrac{4}{9}$ and $\theta \approx 63.61°$.

Take $\vec{b}_1 = (1, 0, 0)$ and $\vec{b}_2 = (-1, 1, 0)$. The dot product is $-1$, which is negative. Without the absolute value, $\cos\theta = -\dfrac{1}{\sqrt{2}}$ would give $135°$. That is the angle between the vectors, but the angle between the lines is the acute one: $\cos\theta = \dfrac{\frac{\sqrt{2}}{2}}{1}$, so $\theta = 45°$.

Parallel lines have angle $0°$: for directions $(2, -4, 6)$ and $(-1, 2, -3)$ the angle is $0°$, and the first is $-2$ times the second.

Two pipes run past each other in a building. You want to join them with the shortest possible rod. Where should the rod go, and how long is it?

Take any point on the first line and any point on the second line. The distance between them depends on which points you choose. Among all such pairs there is a smallest distance. That is the shortest distance between the lines. It is measured along the segment that is perpendicular to both lines (the common perpendicular).

There are three situations. The method depends on which one you have.

In Section 11.5.1 you will find the distance between skew lines. In Section 11.5.2 you will find the distance between parallel lines.

Let the skew lines be $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$. We look for the common perpendicular. A direction that is perpendicular to both $\vec{b}_1$ and $\vec{b}_2$ is their cross product $\vec{n} = \vec{b}_1 \times \vec{b}_2$. (It is not zero, because skew lines are not parallel.)

Join a point $A_1$ on the first line to a point $A_2$ on the second line. The vector is $\overrightarrow{A_1A_2} = \vec{a}_2 - \vec{a}_1$. The shortest distance is how much of this vector points along $\vec{n}$, because the common perpendicular is parallel to $\vec{n}$. That is the length of the projection of $\vec{a}_2 - \vec{a}_1$ on $\vec{n}$:

$$d = \frac{\left|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}.$$

Find the shortest distance between $\vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j})$.

1. The directions are not proportional, so the lines are not parallel.
2. $\vec{b}_1 \times \vec{b}_2 = (1, 1, -2)$, and $|\vec{b}_1 \times \vec{b}_2|^2 = 6$.
3. $\vec{a}_2 - \vec{a}_1 = (1, 3, -1)$, so $(\vec{a}_2 - \vec{a}_1)\cdot(\vec{b}_1 \times \vec{b}_2) = 6$.
4. $d = \dfrac{6}{\sqrt{6}} = \sqrt{6} \approx 2.4495$.

The computer also found the two feet of the common perpendicular, by solving for the closest points: they are $P = (2, 1, 3)$ on the first line ($\lambda = 1$) and $Q = (3, 2, 1)$ on the second line ($\mu = 1$). The segment $PQ = (1, 1, -2)$ is perpendicular to both directions (both dot products are $0$), and its length is $\sqrt{6}$, the same as the formula gave.

When $(\vec{a}_2 - \vec{a}_1)\cdot(\vec{b}_1 \times \vec{b}_2) = 0$, the shortest distance is $0$. The two lines and the connecting vector lie in one plane (they are coplanar), so the lines intersect (or are parallel). Example: $\vec{r} = \lambda(\hat{i} + \hat{j})$ and $\vec{r} = (2\hat{i}) + \mu(\hat{j})$. The triple product is $0$, so $d = 0$. They meet at $(2, 2, 0)$ (found for $\lambda = 2$, $\mu = 2$).

Suppose the lines are given as $\dfrac{x - x_1}{a_1} = \dfrac{y - y_1}{b_1} = \dfrac{z - z_1}{c_1}$ and $\dfrac{x - x_2}{a_2} = \dfrac{y - y_2}{b_2} = \dfrac{z - z_2}{c_2}$. Then $\vec{a}_2 - \vec{a}_1 = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$, and the cross product is $\vec{b}_1 \times \vec{b}_2 = (b_1 c_2 - b_2 c_1,\; c_1 a_2 - c_2 a_1,\; a_1 b_2 - a_2 b_1)$. The top of the fraction is a determinant:

$$d = \frac{\left|\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}\right|}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}$$

Find the shortest distance between $\dfrac{x + 1}{3} = \dfrac{y - 1}{1} = \dfrac{z - 2}{2}$ and $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z}{1}$.

1. Points: $(-1, 1, 2)$ and $(2, -1, 0)$, so $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (3, -2, -2)$. Directions: $(3, 1, 2)$ and $(1, 2, 1)$.
2. The determinant is $\begin{vmatrix} 3 & -2 & -2 \\ 3 & 1 & 2 \\ 1 & 2 & 1 \end{vmatrix} = -17$.
3. The cross product is $(-3, -1, 5)$, and the sum of squares of its components is $35$.
4. $d = \dfrac{|-17|}{\sqrt{35}} = \dfrac{17}{\sqrt{35}} \approx 2.8735$.

The determinant is not zero, and the directions are not proportional, so the lines are skew and this distance is correct.

Let the lines be $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$. They have the same direction $\vec{b}$. The distance between two parallel lines is the same everywhere. It is the length of the perpendicular from any point of one line to the other line.

Take $A_1$ on the first line and $A_2$ on the second, and let $\vec{w} = \vec{a}_2 - \vec{a}_1$. The vectors $\vec{b}$ and $\vec{w}$ make a parallelogram. Its area is $|\vec{b} \times \vec{w}|$ (base times height). The base is $|\vec{b}|$, and the height is exactly the distance $d$ between the lines. So $|\vec{b}|\cdot d = |\vec{b} \times \vec{w}|$, that is

$$d = \frac{\left|\vec{b} \times (\vec{a}_2 - \vec{a}_1)\right|}{\left|\vec{b}\right|}.$$

Find the distance between $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k})$ and $\vec{r} = (3\hat{i} + \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$.

1. $\vec{a}_2 - \vec{a}_1 = (2, -2, 2)$.
2. $\vec{b} \times (\vec{a}_2 - \vec{a}_1) = (6, 0, -6)$, with squared length $72$. Also $|\vec{b}|^2 = 9$.
3. $d = \sqrt{\dfrac{72}{9}} = \sqrt{8} = 2\sqrt{2} \approx 2.8284$.

The foot of the perpendicular from $A_2 = (3, 0, 1)$ to the first line is at $\lambda = 2/3$, that is $(\frac{7}{3}, \frac{8}{3}, \frac{1}{3})$. Its distance from $A_2$ is $\sqrt{8}$, as found.

A point is a "line of zero length", so the same idea gives the distance from a point to a line. Find the distance from $P = (3, 1, -2)$ to the line through $A = (0, 1, 1)$ with direction $(2, 2, 1)$. Here $\overrightarrow{AP} = (3, 0, -3)$ and $\overrightarrow{AP} \times \vec{d} = (6, -9, 6)$. The distance is $\sqrt{\dfrac{153}{9}} = \sqrt{17} \approx 4.1231$. The foot of the perpendicular is at $t = 1/3$, that is $(\frac{2}{3}, \frac{5}{3}, \frac{4}{3})$.

1. Find the angle between the lines with direction ratios $(1, -1, 2)$ and $(2, 1, -1)$.
2. Show that the lines $\dfrac{x - 3}{2} = \dfrac{y + 1}{2} = \dfrac{z}{1}$ and $\dfrac{x}{1} = \dfrac{y - 2}{-2} = \dfrac{z + 1}{2}$ are perpendicular.
3. Find $k$ so that the lines with direction ratios $(k, 3, 2)$ and $(3, k, -3)$ are perpendicular.
4. Write $\vec{r} = (-3\hat{i} + 5\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k})$ in Cartesian form. Then write $\dfrac{x - 4}{3} = \dfrac{y + 2}{5} = \dfrac{z - 1}{-2}$ in vector form.
5. Find the shortest distance between $\vec{r} = (\hat{k}) + \lambda(\hat{i} + \hat{j})$ and $\vec{r} = (\hat{i}) + \mu(\hat{j} + \hat{k})$. Do the lines meet?
6. Find the shortest distance between $\vec{r} = (\hat{j}) + \lambda(\hat{i} + \hat{k})$ and $\vec{r} = (2\hat{i} + \hat{k}) + \mu(\hat{j} - \hat{k})$.
7. Find the distance between the parallel lines $\vec{r} = (\hat{i}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{j}) + \mu(\hat{i} + \hat{j} + \hat{k})$.
8. Find the distance of the point $(1, 1, 1)$ from the line through the origin along the $z$-axis.
9. Are the lines with direction ratios $(2, -4, 6)$ and $(-1, 2, -3)$ parallel? What are their direction cosines?
10. Show that the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu(\hat{i} + 2\hat{j} + 3\hat{k})$ meet, and find the point.