Three Dimensional Geometry — chapter notes

If a directed line makes angles $\alpha$, $\beta$, $\gamma$ with the positive $x$-, $y$- and $z$-axes, its direction cosines are $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$. They are the components of a unit vector along the line, so $l^2 + m^2 + n^2 = 1$. Reversing the direction changes the signs of all three.

Example: a line making $120^\circ$ with the $x$-axis and $45^\circ$ with the $y$-axis has $\cos^2\gamma = 1 - \dfrac14 - \dfrac12 = \dfrac14$, so $\gamma = 60^\circ$ or $120^\circ$.

Numbers $a, b, c$ proportional to $l, m, n$ are direction ratios. Then $l = \pm\dfrac{a}{\sqrt{a^2 + b^2 + c^2}}$, and similarly for $m$, $n$ (same sign throughout).

For the line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, direction ratios are $x_2 - x_1$, $y_2 - y_1$, $z_2 - z_1$; dividing by $PQ$ gives the direction cosines. Example: $A(2, -1, 3)$, $B(4, 5, 0)$: ratios $2, 6, -3$, $AB = 7$, cosines $\dfrac27, \dfrac67, -\dfrac37$.

Collinearity: $A$, $B$, $C$ are collinear if the ratios of $AB$ and $BC$ are proportional (they share $B$).

Vector form. Through the point $\vec{a}$ parallel to $\vec{b}$: $\vec{r} = \vec{a} + \lambda\vec{b}$. Through two points $\vec{a}$, $\vec{c}$: $\vec{r} = \vec{a} + \lambda(\vec{c} - \vec{a})$.

Cartesian form. Through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$: $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$ ($= \lambda$, giving the general point $(x_1 + \lambda a, y_1 + \lambda b, z_1 + \lambda c)$).

Standard form first: in $\dfrac{2x - 1}{3} = \dfrac{y + 2}{2} = \dfrac{4 - z}{5}$, rewrite as $\dfrac{x - \frac12}{\frac32} = \dfrac{y + 2}{2} = \dfrac{z - 4}{-5}$: point $\left(\dfrac12, -2, 4\right)$, ratios $3, 4, -10$.

The angle between two lines is the angle between their direction vectors (taken acute):

$\cos\theta = \dfrac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}$.

• Perpendicular: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
• Parallel: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ (then check a point: they may coincide).

Example: directions $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ give $\cos\theta = \dfrac12$, $\theta = 60^\circ$.

Two distinct lines in space are parallel, intersecting or skew (neither parallel nor intersecting; not coplanar).

Do they meet? Write general points of both lines, equate two coordinates, solve for $\lambda$ and $\mu$, and test the third coordinate. Example: $\vec{r} = \hat{i} - \hat{j} + 2\hat{k} + \lambda(2\hat{i} + 3\hat{j} + \hat{k})$ and $\vec{r} = -\hat{i} + 4\hat{j} + \hat{k} + \mu(2\hat{i} - \hat{j} + \hat{k})$ give $\lambda = 1$, $\mu = 2$ and the common point $(3, 2, 3)$.

Quick test: for non-parallel lines, $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 0$ exactly when they intersect.

The shortest distance is the length of the segment perpendicular to both lines (the common perpendicular). For $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$:

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

Example: $\vec{a}_1 = \hat{i} + 2\hat{j}$, $\vec{b}_1 = \hat{i} + \hat{k}$, $\vec{a}_2 = \hat{j} + 3\hat{k}$, $\vec{b}_2 = \hat{j} - \hat{k}$: $\vec{b}_1 \times \vec{b}_2 = -\hat{i} + \hat{j} + \hat{k}$, dot product $3$, $d = \sqrt3$.

In Cartesian form, read $\vec{a}$ and $\vec{b}$ from the standard form first. The scalar triple product notation and planes are not in the 2026–27 syllabus; write the dot and cross products out.

For parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$: $d = \dfrac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$. Example: $\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{a}_2 - \vec{a}_1 = 2\hat{i} - 2\hat{j} + 2\hat{k}$ give $d = \dfrac{2\sqrt2}{3}$.

Foot of the perpendicular from a point $P$ to a line: take the general point $F$ of the line, set $\vec{PF} \cdot \vec{b} = 0$, solve for $\lambda$. Example: $P(5, 3, -2)$ and $\vec{r} = \hat{i} + 2\hat{k} + \lambda(\hat{i} + 2\hat{j} - 2\hat{k})$ give $\lambda = 2$, $F(3, 4, -2)$, distance $\sqrt5$.