Vectors are quantities with both magnitude and direction, in contrast to scalars which have magnitude only.
A vector in 2D: $\mathbf{a} = a_1\,\mathbf{i} + a_2\,\mathbf{j} = \begin{pmatrix}a_1\a_2\end{pmatrix}$.
A vector in 3D: $\mathbf{a} = a_1\,\mathbf{i} + a_2\,\mathbf{j} + a_3\,\mathbf{k} = \begin{pmatrix}a_1\a_2\a_3\end{pmatrix}$.
Where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the standard unit vectors along the $x$-, $y$-, and $z$-axes.
| Property | 2D formula | 3D formula |
|---|---|---|
| Magnitude | $ | \mathbf{a} |
| Unit vector | $\hat{\mathbf{a}} = \mathbf{a}/ | \mathbf{a} |
| Zero vector | $\mathbf{0} = \begin{pmatrix}0\0\end{pmatrix}$ | $\begin{pmatrix}0\0\0\end{pmatrix}$ |
Direction in 2D: The vector $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$ makes angle $\theta = \arctan(a_2/a_1)$ with the positive $x$-axis (quadrant-adjusted).
The position vector of point $P(x,y,z)$ relative to origin $O$ is $\overrightarrow{OP} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.
The vector from $A$ to $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ where $\mathbf{a}, \mathbf{b}$ are position vectors of $A, B$.
Midpoint of $AB$: position vector $\dfrac{\mathbf{a}+\mathbf{b}}{2}$.
Dividing in ratio $m:n$ from $A$ to $B$: position vector $\dfrac{n\mathbf{a} + m\mathbf{b}}{m+n}$.
$\mathbf{a} \parallel \mathbf{b}$ if and only if $\mathbf{a} = \lambda\mathbf{b}$ for some scalar $\lambda \neq 0$.
Example: Are $\mathbf{a} = 2\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}$ and $\mathbf{b} = -\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$ parallel?
$\mathbf{a} = -2\mathbf{b}$, so yes, they are parallel (and anti-parallel, i.e., opposite direction).
$$\mathbf{a} + \mathbf{b} = \begin{pmatrix}a_1+b_1\a_2+b_2\a_3+b_3\end{pmatrix}, \qquad \lambda\mathbf{a} = \begin{pmatrix}\lambda a_1\ \lambda a_2\ \lambda a_3\end{pmatrix}$$
Example: $\mathbf{a} = \begin{pmatrix}1\-2\3\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}4\1\-1\end{pmatrix}$. Find $2\mathbf{a} - \mathbf{b}$.
$$2\mathbf{a} - \mathbf{b} = \begin{pmatrix}2\-4\6\end{pmatrix} - \begin{pmatrix}4\1\-1\end{pmatrix} = \begin{pmatrix}-2\-5\7\end{pmatrix}$$
KEY TAKEAWAY: Vectors in 3D extend 2D concepts by adding a $z$-component and the unit vector $\mathbf{k}$. Magnitude uses the 3D Pythagorean theorem. All algebraic operations extend componentwise.
EXAM TIP: Always write vectors in full component form before performing operations. Component errors are a common source of lost marks.
VCAA FOCUS: Position vectors, the vector from $A$ to $B$, and dividing a line in a given ratio are foundational for nearly every vector question.
