Direction Cosines and Direction Ratios

If a line makes angles $\alpha, \beta, \gamma$ with the positive $x$-, $y$-, $z$-axes, its direction cosines are $$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma, \qquad l^2 + m^2 + n^2 = 1$$ Any triple $a, b, c$ proportional to $l, m, n$ gives direction ratios. To convert DRs to DCs, divide by $\sqrt{a^2 + b^2 + c^2}$: $$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$ DCs are unique up to an overall sign (the two senses of the line); DRs are infinitely many. For the line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, one set of DRs is $x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1$.

Equations of a Line

Through a point with position vector $\vec{a}$, parallel to $\vec{b}$: $$\vec{r} = \vec{a} + \lambda\vec{b}$$ In cartesian form, through $(x_1, y_1, z_1)$ with DRs $a, b, c$: $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$

Through two points $\vec{a}$ and $\vec{b}$: $\vec{r} = \vec{a} + \lambda\left(\vec{b} - \vec{a}\right)$, i.e. use the joining vector as the direction.

Reading a cartesian equation: the standard form needs coefficient $1$ on each of $x, y, z$ — rewrite $\frac{1 - x}{2}$ as $\frac{x - 1}{-2}$ before reading off the DRs. Rushing this step is the chapter's most common error.

Angle Between Two Lines

For lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$: $$\cos\theta = \left\vert\frac{\vec{b_1}\cdot\vec{b_2}}{\vert\vec{b_1}\vert\vert\vec{b_2}\vert}\right\vert$$ The modulus keeps $\theta$ acute — the convention for the angle between lines. Consequences: perpendicular $\iff a_1a_2 + b_1b_2 + c_1c_2 = 0$; parallel $\iff \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Shortest Distance Between Two Lines

Skew lines — neither parallel nor intersecting — exist only in three dimensions, and the shortest distance is along the common perpendicular.

Skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$: $$d = \frac{\left\vert\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2} - \vec{a_1}\right)\right\vert}{\vert\vec{b_1}\times\vec{b_2}\vert}$$

Parallel lines ($\vec{b_1} = \vec{b_2} = \vec{b}$) make that formula collapse ($\vec{b}\times\vec{b} = \vec{0}$), so use $$d = \frac{\left\vert\vec{b}\times\left(\vec{a_2} - \vec{a_1}\right)\right\vert}{\vert\vec{b}\vert}$$

Test $d = 0$: the lines intersect. The five-step routine for any shortest-distance question: extract $\vec{a_1}, \vec{b_1}, \vec{a_2}, \vec{b_2}$; check parallel or not; compute the cross product; dot with $\vec{a_2} - \vec{a_1}$ (skew case); divide by the right magnitude.

Worked Examples — Direction Cosines and Line Equations

Example 1 — DCs from the angles

A line makes angles $90^\circ$, $60^\circ$, $30^\circ$ with the $x$-, $y$-, $z$-axes. Find its direction cosines.

Step 1 — take cosines: $l = \cos 90^\circ = 0$, $m = \cos 60^\circ = \dfrac{1}{2}$, $n = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$.

Step 2 — verify: $0 + \dfrac{1}{4} + \dfrac{3}{4} = 1$.

Answer: $\left(0, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$ — the identity check is a free mark, always show it.

Example 2 — DCs from DRs

Find the direction cosines of a line with direction ratios $2, -1, 2$.

Step 1 — the normaliser: $\sqrt{4 + 1 + 4} = 3$.

Answer: $\left(\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}\right)$ — or the same triple with all signs flipped; both describe the line.

Example 3 — DCs of a line through two points

Find the direction cosines of the line through $A(-2, 4, -5)$ and $B(1, 2, 3)$.

Step 1 — DRs by subtraction: $3, -2, 8$.

Step 2 — normalise: $\sqrt{9 + 4 + 64} = \sqrt{77}$.

Answer: $\left(\dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}}\right)$.

Example 4 — Line through a point, parallel to a vector

Find the vector and cartesian equations of the line through $(5, 2, -4)$ parallel to $3\hat{i} + 2\hat{j} - 8\hat{k}$.

Step 1 — vector form: $\vec{r} = 5\hat{i} + 2\hat{j} - 4\hat{k} + \lambda\left(3\hat{i} + 2\hat{j} - 8\hat{k}\right)$.

Step 2 — cartesian form:

Answer: $\dfrac{x - 5}{3} = \dfrac{y - 2}{2} = \dfrac{z + 4}{-8}$ — note $z - (-4)$ becomes $z + 4$; sign slips here cost a full part.

Example 5 — Line through two points

Find the equations of the line through $A(-1, 0, 2)$ and $B(3, 4, 6)$.

Step 1 — direction: $\overrightarrow{AB} = 4\hat{i} + 4\hat{j} + 4\hat{k}$, and any proportional triple works — take $1, 1, 1$.

Step 2 — write both forms: $\vec{r} = -\hat{i} + 2\hat{k} + \lambda\left(\hat{i} + \hat{j} + \hat{k}\right)$.

Answer: $\dfrac{x + 1}{1} = \dfrac{y}{1} = \dfrac{z - 2}{1}$ — simplifying the DRs before writing the equation keeps everything clean.

Example 6 — Reading a cartesian equation correctly

Write the direction ratios and a point on the line $\dfrac{1 - x}{2} = \dfrac{y + 3}{5} = \dfrac{2z - 6}{4}$.

Step 1 — force standard form: $\dfrac{1-x}{2} = \dfrac{x - 1}{-2}$ and $\dfrac{2z-6}{4} = \dfrac{z - 3}{2}$.

Step 2 — now read off: $$\frac{x - 1}{-2} = \frac{y + 3}{5} = \frac{z - 3}{2}$$

Answer: point $(1, -3, 3)$, DRs $-2, 5, 2$ — reading DRs before standardising would give the wrong triple $2, 5, 4$.

Example 7 — A line equally inclined to the axes

Find the direction cosines of a line equally inclined to the three coordinate axes.

Step 1 — set $l = m = n$: then $3l^2 = 1$.

Answer: $l = m = n = \pm\dfrac{1}{\sqrt{3}}$ — the DRs are simply $1, 1, 1$.

Worked Examples — Angles and Shortest Distances

Example 8 — Angle between two lines (DRs given)

Find the angle between the lines with direction ratios $3, 5, 4$ and $1, 1, 2$.

Step 1 — the three ingredients: $\vec{b_1}\cdot\vec{b_2} = 3 + 5 + 8 = 16$, $\vert\vec{b_1}\vert = \sqrt{50} = 5\sqrt{2}$, $\vert\vec{b_2}\vert = \sqrt{6}$.

Step 2 — the formula: $\cos\theta = \dfrac{16}{5\sqrt{2}\cdot\sqrt{6}} = \dfrac{16}{10\sqrt{3}} = \dfrac{8\sqrt{3}}{15}$.

Answer: $\theta = \cos^{-1}\dfrac{8\sqrt{3}}{15}$ — rationalise before boxing the answer.

Example 9 — Angle between two vector-form lines

Find the angle between $\vec{r} = \vec{a_1} + \lambda\left(3\hat{i} + 2\hat{j} + 6\hat{k}\right)$ and $\vec{r} = \vec{a_2} + \mu\left(\hat{i} + 2\hat{j} + 2\hat{k}\right)$.

Step 1 — only the direction vectors matter: $\vec{b_1}\cdot\vec{b_2} = 3 + 4 + 12 = 19$, $\vert\vec{b_1}\vert = 7$, $\vert\vec{b_2}\vert = 3$.

Answer: $\theta = \cos^{-1}\dfrac{19}{21}$ — the points $\vec{a_1}, \vec{a_2}$ play no role in the angle.

Example 10 — Perpendicularity check

Show that the lines with direction ratios $2, 1, -2$ and $1, 2, 2$ are perpendicular.

Step 1 — the dot-product test: $2(1) + 1(2) + (-2)(2) = 2 + 2 - 4 = 0$.

Answer: the sum vanishes, so the lines are perpendicular — one line of arithmetic, full marks.

Example 11 — Shortest distance between skew lines

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

Step 1 — extract the four vectors: $\vec{a_1} = (1,2,1)$, $\vec{b_1} = (1,-1,1)$, $\vec{a_2} = (2,-1,-1)$, $\vec{b_2} = (2,1,2)$.

Step 2 — cross product: $$\vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}, \qquad \vert\vec{b_1}\times\vec{b_2}\vert = 3\sqrt{2}$$

Step 3 — dot with $\vec{a_2} - \vec{a_1} = (1, -3, -2)$: $(-3)(1) + 0(-3) + 3(-2) = -9$.

Step 4 — divide: $d = \dfrac{\vert -9\vert}{3\sqrt{2}} = \dfrac{3}{\sqrt{2}}$.

Answer: $\dfrac{3\sqrt{2}}{2}$ units.

Example 12 — Shortest distance, cartesian data

Find the shortest distance between $\dfrac{x + 1}{7} = \dfrac{y + 1}{-6} = \dfrac{z + 1}{1}$ and $\dfrac{x - 3}{1} = \dfrac{y - 5}{-2} = \dfrac{z - 7}{1}$.

Step 1 — extract: $\vec{a_1} = (-1,-1,-1)$, $\vec{b_1} = (7,-6,1)$; $\vec{a_2} = (3,5,7)$, $\vec{b_2} = (1,-2,1)$.

Step 2 — cross product: $\vec{b_1}\times\vec{b_2} = -4\hat{i} - 6\hat{j} - 8\hat{k}$, with magnitude $\sqrt{116} = 2\sqrt{29}$.

Step 3 — dot with $\vec{a_2} - \vec{a_1} = (4, 6, 8)$: $-16 - 36 - 64 = -116$.

Step 4 — divide: $d = \dfrac{116}{2\sqrt{29}} = \dfrac{58}{\sqrt{29}}$.

Answer: $2\sqrt{29}$ units — simplify the surd fully: $\frac{58}{\sqrt{29}} = \frac{58\sqrt{29}}{29} = 2\sqrt{29}$.

Example 13 — Distance between parallel lines

Find the distance between $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)$ and $\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)$.

Step 1 — same $\vec{b}$, so use the parallel formula: $\vec{a_2} - \vec{a_1} = (2, 1, -1)$.

Step 2 — cross product: $$\vec{b}\times\left(\vec{a_2} - \vec{a_1}\right) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} = -9\hat{i} + 14\hat{j} - 4\hat{k}, \qquad \vert\cdot\vert = \sqrt{293}$$

Step 3 — divide by $\vert\vec{b}\vert = 7$:

Answer: $\dfrac{\sqrt{293}}{7}$ units — spotting that the direction vectors match (and switching formulas) is the first mark of the question.