Unit 4 extends vector applications to 3D geometry (lines, planes) and to mechanics (forces, equilibrium, relative velocity).
Line through $A$ in direction $\mathbf{d}$:
$$\mathbf{r} = \mathbf{a} + t\mathbf{d}, \quad t \in \mathbb{R}$$
Plane through $A$ with normal $\mathbf{n}$:
$$\mathbf{n}\cdot(\mathbf{r}-\mathbf{a}) = 0 \quad \Leftrightarrow \quad \mathbf{n}\cdot\mathbf{r} = \mathbf{n}\cdot\mathbf{a} = d$$
Plane through three points $A, B, C$:
Normal: $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$.
Substitute $\mathbf{r} = \mathbf{a}+t\mathbf{d}$ into the plane equation $\mathbf{n}\cdot\mathbf{r} = d$:
$$t = \frac{d - \mathbf{n}\cdot\mathbf{a}}{\mathbf{n}\cdot\mathbf{d}}$$
If $\mathbf{n}\cdot\mathbf{d} = 0$: line is parallel to the plane (no intersection, or lies in it).
Forces as vectors: A force $\mathbf{F}$ has both magnitude and direction.
Equilibrium: A particle is in equilibrium when $\sum\mathbf{F} = \mathbf{0}$.
In 3D: resolve into $x$, $y$, $z$ components and set each sum to zero.
Example 1: Three forces act on a particle:
$\mathbf{F}_1 = 3\mathbf{i}+\mathbf{j}-2\mathbf{k}$, $\mathbf{F}_2 = -\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\mathbf{F}_3 = a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$.
For equilibrium: $\mathbf{F}_3 = -(\mathbf{F}_1+\mathbf{F}_2) = -(2\mathbf{i}+3\mathbf{j}-\mathbf{k}) = -2\mathbf{i}-3\mathbf{j}+\mathbf{k}$.
Velocity of $B$ relative to $A$: $\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$.
To find when two particles are closest:
1. Express position of each as $\mathbf{r}(t)$.
2. Form $\mathbf{s}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$.
3. Minimise $|\mathbf{s}(t)|^2$ by differentiating and setting to zero.
Example 2: At $t=0$: $A$ is at $(0,0)$ with velocity $(2,1)$; $B$ is at $(6,2)$ with velocity $(-1,3)$ (m, m/s).
$\mathbf{r}_A = 2t\mathbf{i}+t\mathbf{j}$; $\mathbf{r}_B = (6-t)\mathbf{i}+(2+3t)\mathbf{j}$.
$\mathbf{s}(t) = (6-3t)\mathbf{i}+(2+2t)\mathbf{j}$.
$|\mathbf{s}|^2 = (6-3t)^2+(2+2t)^2 = 9t^2-36t+36+4t^2+8t+4 = 13t^2-28t+40$.
$\dfrac{d}{dt}|\mathbf{s}|^2 = 26t-28 = 0 \Rightarrow t = 14/13 \approx 1.08$ s.
Minimum distance$^2 = 13(14/13)^2-28(14/13)+40 = \ldots$; compute numerically.
KEY TAKEAWAY: In 3D mechanics, forces and velocities are vectors. Equilibrium means the vector sum of all forces is zero. Relative velocity problems use $\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$.
EXAM TIP: For equilibrium in 2D or 3D, resolve forces into components and set up a system of equations. Check that the number of equations matches the number of unknowns.
VCAA FOCUS: Line-plane intersections, particle equilibrium with multiple forces, and closest-approach problems are all high-frequency vector topics in Unit 4.
