Lines in Space and the Angle Between Them

Why this mat­ters#

To pin down a line in space, you need just two things: one point on it and the direc­tion it runs in. So we start by learn­ing to describe a direc­tion, using direc­tion cosines and direc­tion ratios. Then we write the equa­tion of a line in vec­tor and Carte­sian form, and finally we work out the angle between two lines.

Direc­tion cosines and ratios#

If a line makes angles $\alpha, \beta, \gamma$ with the axes, its direc­tion cosines are $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$, with $l^2 + m^2 + n^2 = 1$. Any num­bers $a, b, c$ in the same pro­por­tion are called direc­tion ratios, and you can get back to the cosines using $\displaystyle l = \pm\tfrac{a}{\sqrt{a^2 + b^2 + c^2}}$, etc.

The line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ has direc­tion ratios $x_2 - x_1, y_2 - y_1, z_2 - z_1$. Just sub­tract the coor­di­nates.

Equa­tion of a line#

The line through the point with posi­tion vec­tor $\vec a$, run­ning par­al­lel to $\vec b$, is

$$\vec r = \vec a + \lambda\vec b.$$

In Carte­sian form, the line through $(x_1, y_1, z_1)$ with direc­tion ratios $a, b, c$ is

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

If you are given two points instead, sim­ply take $\vec b = \vec a_2 - \vec a_1$.

Angle between two lines#

If the lines have direc­tion vec­tors $\vec b_1$ and $\vec b_2$, with ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$, then

$$\displaystyle \cos\theta = \frac{\lvert a_1a_2 + b_1b_2 + c_1c_2 \rvert}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}.$$

Two spe­cial cases are worth remem­ber­ing. The lines are per­pen­dic­u­lar when $a_1a_2 + b_1b_2 + c_1c_2 = 0$, and par­al­lel when their ratios are pro­por­tional.

Prac­tice#

1. Find the direc­tion cosines of the line with ratios $6, -2, 3$.
2. Find the direc­tion cosines of the line through $(-1, 2, 5)$ and $(3, 0, 1)$.
3. Show that $(1, -1, 2)$, $(3, 3, 0)$, $(-2, -7, 5)$ are collinear.
4. A line makes equal angles with all three axes. Find its direc­tion cosines.
5. Find the vec­tor and Carte­sian equa­tions of the line through $(2, 0, -3)$ par­al­lel to $\hat i - 4\hat j + 5\hat k$.
6. Find the Carte­sian equa­tion of the line through $(1, 2, 3)$ and $(4, 0, -1)$.
7. Write $\displaystyle \tfrac{x + 3}{2} = \tfrac{4 - y}{5} = \tfrac{3z + 1}{3}$ in stan­dard form and give a point on it and its direc­tion.
8. Find the angle between $\vec r = \hat i + \lambda(2\hat i + \hat j + 2\hat k)$ and $\vec r = 3\hat j + \mu(\hat i + 2\hat j - 2\hat k)$.
9. Find $p$ so that the lines $\displaystyle \tfrac{x - 1}{3} = \tfrac{y - 2}{2p} = \tfrac{z - 3}{2}$ and $\displaystyle \tfrac{x - 1}{3p} = \tfrac{y - 1}{1} = \tfrac{z - 6}{-5}$ are per­pen­dic­u­lar.
10. Show that the line through $(2, 3, 4)$ and $(4, 7, 8)$ is par­al­lel to the line through $(0, 1, 2)$ and $(1, 3, 4)$.

Answers#