A line through point $A$ (position vector $\mathbf{a}$) in direction $\mathbf{d}$:
$$\mathbf{r} = \mathbf{a} + t\mathbf{d}, \quad t \in \mathbb{R}$$
Or in component form:
$$\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3} = t$$
Example 1: Write the vector equation of the line through $A(1,2,-1)$ and $B(3,0,2)$.
$\mathbf{a} = \begin{pmatrix}1\2\-1\end{pmatrix}$, $\mathbf{d} = \overrightarrow{AB} = \begin{pmatrix}2\-2\3\end{pmatrix}$.
$$\mathbf{r} = \begin{pmatrix}1\2\-1\end{pmatrix} + t\begin{pmatrix}2\-2\3\end{pmatrix}$$
Two lines are:
- Parallel if their direction vectors are parallel ($\mathbf{d}_1 = \lambda\mathbf{d}_2$)
- Intersecting if they share a common point (solve the system $\mathbf{a}_1 + s\mathbf{d}_1 = \mathbf{a}_2 + t\mathbf{d}_2$)
- Skew if not parallel and not intersecting (in 3D)
Normal form: $\mathbf{n} \cdot \mathbf{r} = d$ where $\mathbf{n}$ is the normal vector.
Expanding: $n_1 x + n_2 y + n_3 z = d$.
Point-normal form: $\mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0$ where $A$ is a point on the plane.
Parametric form: $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$ where $\mathbf{b}, \mathbf{c}$ lie in the plane.
Example 2: Find the equation of the plane through $A(1,0,2)$ with normal $\mathbf{n} = 2\mathbf{i}-\mathbf{j}+3\mathbf{k}$.
$2(x-1) - (y-0) + 3(z-2) = 0 \Rightarrow 2x - y + 3z = 8$.
Example 3: Find the normal to the plane through $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$.
$\overrightarrow{AB} = (-1,1,0)$, $\overrightarrow{AC} = (-1,0,1)$.
$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = (1,1,1)$.
Plane equation: $x + y + z = 1$.
Angle between two lines: $\cos\theta = \dfrac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ (take acute angle).
Angle between two planes: Same as angle between their normals.
Angle between line and plane: $\sin\phi = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ (complement of angle between line and normal).
Distance from point $P$ to line $\mathbf{r} = \mathbf{a}+t\mathbf{d}$:
$$\text{dist} = \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|}$$
Distance from point $P(x_0,y_0,z_0)$ to plane $ax+by+cz=d$:
$$\text{dist} = \frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$$
Example 4: Distance from $P(1,1,1)$ to plane $2x-y+3z=8$.
$$\text{dist} = \frac{|2-1+3-8|}{\sqrt{4+1+9}} = \frac{4}{\sqrt{14}}$$
KEY TAKEAWAY: Vector methods reduce geometric problems to algebraic manipulation. Lines use a point plus direction; planes use a point plus normal or two direction vectors.
EXAM TIP: When finding where a line intersects a plane, substitute the parametric form of the line into the plane equation and solve for $t$, then back-substitute.
VCAA FOCUS: Finding angles between lines/planes and distances from points to lines/planes are high-frequency exam tasks. Know all four distance/angle formulas.
