Solved Examples

How to Use This Section

Twenty fully worked problems covering the whole chapter, in four graded batches: direction cosines and ratios, then line equations, then angles, then shortest distances and mixed problems. Attempt each on paper before reading the solution.

The chapter's entire machinery is four moves: normalise ratios into cosines, read a line's point and direction from either form of its equation, dot directions for angles, and cross for perpendiculars and distances. Every problem below is a composition of those four.

Key Point: in this chapter more than any other, marks die at the reading step — an unstandardised Cartesian equation ($3 - x$ in a numerator, a $2z$, cosines confused with ratios) poisons everything downstream. Standardise first, always.

Batch 1 — Direction Cosines and Ratios (Easy)

Example 1: The third direction angle

A line makes $135^\circ$ with the $x$-axis and $60^\circ$ with the $y$-axis. Find the acute angle it makes with the $z$-axis.

Solution:

1. Identity: $\cos^2 135^\circ + \cos^2 60^\circ + \cos^2\gamma = 1$, i.e. $\dfrac{1}{2} + \dfrac{1}{4} + \cos^2\gamma = 1$.
2. Solve: $\cos^2\gamma = \dfrac{1}{4}$, acute choice $\cos\gamma = \dfrac{1}{2}$.

Answer: $\gamma = 60^\circ$. The obtuse $135^\circ$ contributes $\cos^2 = \dfrac{1}{2}$ just like $45^\circ$ would — squares forget signs.

Example 2: Ratios to cosines

Find the direction cosines of a line with direction ratios $6, -2, 3$.

Solution:

1. Normalise: $\sqrt{36 + 4 + 9} = 7$.
2. Divide: $\left(\dfrac{6}{7}, -\dfrac{2}{7}, \dfrac{3}{7}\right)$.

Answer: $\left(\dfrac{6}{7}, -\dfrac{2}{7}, \dfrac{3}{7}\right)$ — a 6-2-3-7 quadruple worth remembering alongside 1-2-2-3 and 2-3-6-7.

Example 3: Cosines from two points

Find the direction cosines of the line from $P(4, 3, -5)$ to $Q(-2, 1, -8)$.

Solution:

1. Ratios (terminal minus initial): $(-6, -2, -3)$.
2. Distance: $\sqrt{36 + 4 + 9} = 7$.
3. Divide: $\left(-\dfrac{6}{7}, -\dfrac{2}{7}, -\dfrac{3}{7}\right)$.

Answer: $\left(-\dfrac{6}{7}, -\dfrac{2}{7}, -\dfrac{3}{7}\right)$ — the all-negated set describes the same line traversed from $Q$ to $P$.

Example 4: Three points on one line

Show that $(2, 3, 4)$, $(3, 4, 5)$ and $(4, 5, 6)$ are collinear.

Solution:

1. Consecutive ratios: first to second: $(1, 1, 1)$; second to third: $(1, 1, 1)$.
2. Identical (hence proportional) ratios through a shared point — one straight line.

Answer: collinear; indeed all three lie on $\dfrac{x - 2}{1} = \dfrac{y - 3}{1} = \dfrac{z - 4}{1}$.

Example 5: Completing a cosine triple

If $\left(\dfrac{1}{2}, \dfrac{1}{2}, n\right)$ are direction cosines of a line with $n > 0$, find $n$.

Solution:

1. Identity: $\dfrac{1}{4} + \dfrac{1}{4} + n^2 = 1$, so $n^2 = \dfrac{1}{2}$.
2. Positive root: $n = \dfrac{1}{\sqrt{2}}$.

Answer: $n = \dfrac{1}{\sqrt{2}}$ — the line makes $60^\circ, 60^\circ, 45^\circ$ with the axes.

Batch 2 — Line Equations (Easy-Medium)

Example 6: Point and direction, both forms

Find the equation of the line through $(1, 2, 3)$ parallel to $3\hat{i} + 2\hat{j} - 2\hat{k}$, in vector and Cartesian form.

Solution:

1. Vector form: $\vec{r} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(3\hat{i} + 2\hat{j} - 2\hat{k}\right)$.
2. Cartesian form: $\dfrac{x - 1}{3} = \dfrac{y - 2}{2} = \dfrac{z - 3}{-2}$.

Answer: as above.

Example 7: From a position vector

Find both forms of the line through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ in the direction $\hat{i} + 2\hat{j} - \hat{k}$.

Solution:

1. Vector form: $\vec{r} = \left(2\hat{i} - \hat{j} + 4\hat{k}\right) + \lambda\left(\hat{i} + 2\hat{j} - \hat{k}\right)$.
2. Cartesian form: $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z - 4}{-1}$.

Answer: as above — the position vector supplies the point $(2, -1, 4)$, the direction supplies the denominators.

Example 8: Parallel to a given line

Find the Cartesian equation of the line through $(-2, 4, -5)$ parallel to $\dfrac{x + 3}{3} = \dfrac{y - 4}{5} = \dfrac{z + 8}{6}$.

Solution:

1. Borrow the direction: ratios $(3, 5, 6)$ — parallel lines share direction ratios.
2. Anchor at the new point: $\dfrac{x + 2}{3} = \dfrac{y - 4}{5} = \dfrac{z + 5}{6}$.

Answer: $\dfrac{x + 2}{3} = \dfrac{y - 4}{5} = \dfrac{z + 5}{6}$.

Example 9: Cartesian to vector

The Cartesian equation of a line is $\dfrac{x - 5}{3} = \dfrac{y + 4}{7} = \dfrac{z - 6}{2}$. Write its vector form.

Solution:

1. Point from numerators: $(5, -4, 6)$.
2. Direction from denominators: $(3, 7, 2)$.

Answer: $\vec{r} = \left(5\hat{i} - 4\hat{j} + 6\hat{k}\right) + \lambda\left(3\hat{i} + 7\hat{j} + 2\hat{k}\right)$.

Example 10: A line with two zero ratios

Find the equation of the line through $(3, -2, -5)$ and $(3, -2, 6)$.

Solution:

1. Direction: $(0, 0, 11)$ — only $z$ changes; the line is parallel to the $z$-axis.
2. Vector form: $\vec{r} = \left(3\hat{i} - 2\hat{j} - 5\hat{k}\right) + \lambda\hat{k}$.
3. Cartesian description: $x = 3$, $y = -2$, $z$ arbitrary (formally $\dfrac{x-3}{0} = \dfrac{y+2}{0} = \dfrac{z+5}{11}$, with zero denominators read as "numerator identically zero").

Answer: the vertical line $x = 3, y = -2$ — zero direction ratios are statements, not divisions.

Batch 3 — Angles between Lines (Medium)

Example 11: A routine angle

Find the angle between the lines with direction ratios $(2, 5, -3)$ and $(-1, 8, 4)$.

Solution:

1. Dot: $-2 + 40 - 12 = 26$.
2. Norms: $\sqrt{4 + 25 + 9} = \sqrt{38}$ and $\sqrt{1 + 64 + 16} = 9$.
3. Angle: $\cos\theta = \dfrac{26}{9\sqrt{38}}$.

Answer: $\theta = \cos^{-1}\dfrac{26}{9\sqrt{38}}$.

Example 12: A surd-heavy pair that collapses

Find the angle between lines with direction ratios $(1, 1, 2)$ and $\left(\sqrt{3} - 1, \ -\sqrt{3} - 1, \ 4\right)$.

Solution:

1. Dot: $(\sqrt{3} - 1) + (-\sqrt{3} - 1) + 8 = 6$.
2. Norms: $\sqrt{6}$ and $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2 + 16} = \sqrt{4 - 2\sqrt{3} + 4 + 2\sqrt{3} + 16} = \sqrt{24} = 2\sqrt{6}$ — the cross terms cancel.
3. Angle: $\cos\theta = \dfrac{6}{\sqrt{6}\cdot 2\sqrt{6}} = \dfrac{6}{12} = \dfrac{1}{2}$.

Answer: $60^\circ$ — designed surds; trust the algebra and the mess evaporates.

Example 13: Angle between the diagonals of a cube

Find the angle between two (space) diagonals of a cube.

Solution:

1. Model the cube on unit axes: two diagonals have direction ratios $(1, 1, 1)$ and $(1, 1, -1)$.
2. Dot: $1 + 1 - 1 = 1$; norms $\sqrt{3}$ each.
3. Angle: $\cos\theta = \dfrac{1}{3}$.

Answer: $\cos^{-1}\dfrac{1}{3} \approx 70.5^\circ$ — a classic result exams quote directly.

Example 14: Cube diagonal and edge

Find the angle between a space diagonal of a cube and one of its edges.

Solution:

1. Directions: diagonal $(1, 1, 1)$, edge $(1, 0, 0)$.
2. Compute: $\cos\theta = \dfrac{1}{\sqrt{3}}$.

Answer: $\cos^{-1}\dfrac{1}{\sqrt{3}} \approx 54.7^\circ$ — the equal-inclination angle from Section 1, met again from the cube's corner.

Example 15: Perpendicular to two given directions

Find the direction cosines of the line perpendicular to both lines with direction ratios $(1, 2, 2)$ and $(0, 2, 1)$.

Solution:

1. Cross: $(1, 2, 2) \times (0, 2, 1) = (2 - 4, \ 0 - 1, \ 2 - 0) = (-2, -1, 2)$.
2. Normalise: magnitude $3$.

Answer: $\left(-\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3}\right)$ (or negated) — "perpendicular to two directions" always means the cross product.

Batch 4 — Distances and Mixed Problems (Medium-Hard)

Example 16: A clean skew distance

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

Solution:

1. Cross: $\left(\hat{i} - \hat{j}\right) \times \left(\hat{i} + \hat{j}\right) = 2\hat{k}$, magnitude $2$.
2. Joining vector: $(1, -1, -2)$.
3. Triple product: $2\hat{k} \cdot (1, -1, -2) = -4$; modulus $4$.

Answer: $d = \dfrac{4}{2} = 2$.

Example 17: A line perpendicular to two lines

Find the vector equation of the line through $(1, 2, -4)$ perpendicular to both $\dfrac{x - 8}{3} = \dfrac{y + 19}{-16} = \dfrac{z - 10}{7}$ and $\dfrac{x - 15}{3} = \dfrac{y - 29}{8} = \dfrac{z - 5}{-5}$.

Solution:

1. Direction = cross of the two directions: $(3, -16, 7) \times (3, 8, -5) = (80 - 56, \ 21 + 15, \ 24 + 48) = (24, 36, 72) = 12(2, 3, 6)$.
2. Anchor at the given point: $\vec{r} = \left(\hat{i} + 2\hat{j} - 4\hat{k}\right) + \lambda\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)$.

Answer: $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)$ — always reduce the cross by its common factor before writing the answer.

Example 18: Do the lines meet?

Show that the lines $\vec{r} = \left(\hat{i} + \hat{j}\right) + \lambda\hat{i}$ and $\vec{r} = \mu\hat{j}$ intersect, and find the point of intersection.

Solution:

1. Coplanarity test: $\left(\hat{i} \times \hat{j}\right)\cdot\left(\vec{a}_2 - \vec{a}_1\right) = \hat{k}\cdot(-1, -1, 0) = 0$ — coplanar; being non-parallel, they intersect.
2. Find the point: equate $(1 + \lambda, 1, 0) = (0, \mu, 0)$: $\lambda = -1$, $\mu = 1$.

Answer: they meet at $(0, 1, 0)$. The zero triple product predicts the meeting; the parameter equations locate it.

Example 19: Distance from a point to a line

Find the distance from the point $P(1, 0, 0)$ to the line $x = y = z$.

Solution:

1. Set up: the line passes through $A(0,0,0)$ with direction $\vec{b} = (1, 1, 1)$; $\overrightarrow{AP} = (1, 0, 0)$.
2. Cross: $\overrightarrow{AP} \times \vec{b} = (0\cdot1 - 0\cdot1, \ 0\cdot1 - 1\cdot1, \ 1\cdot1 - 0\cdot1) = (0, -1, 1)$, magnitude $\sqrt{2}$.
3. Divide by $|\vec{b}|$: $d = \dfrac{\sqrt{2}}{\sqrt{3}} = \dfrac{\sqrt{6}}{3}$.

Answer: $\dfrac{\sqrt{6}}{3}$ — the point-to-line distance is the parallel-lines formula with one "line" shrunk to a point.

Example 20: The nearest point on a line

Find the point on the line $\dfrac{x - 1}{1} = \dfrac{y - 2}{1} = \dfrac{z - 3}{1}$ closest to the origin.

Solution:

1. Parametrise: $(1 + t, \ 2 + t, \ 3 + t)$.
2. Perpendicularity condition: the position vector of the nearest point is perpendicular to the direction $(1,1,1)$: $(1 + t) + (2 + t) + (3 + t) = 0$, so $t = -2$.
3. Substitute: the point is $(-1, 0, 1)$, at distance $\sqrt{2}$.

Answer: $(-1, 0, 1)$, distance $\sqrt{2}$ — parametrise, impose one perpendicularity dot, substitute: the universal "foot of perpendicular" recipe.