Vectors in two and three dimensions in $\mathbf{i}, \mathbf{j}, \mathbf{k}$ form, magnitude and unit vectors, the scalar (dot) product and the angle between vectors, vector projection, and the use of the scalar product to test for perpendicular and parallel vectors

What this dot point is asking

VCAA wants you to operate on vectors in two and three dimensions using $\mathbf{i}, \mathbf{j}, \mathbf{k}$ component form: adding, scaling, finding magnitudes and unit vectors, computing the scalar (dot) product, finding the angle between two vectors, resolving one vector along another (projection), and using the dot product to test whether vectors are parallel or perpendicular. Vectors recur in the kinematics and vector-calculus parts of the course, so this foundation matters.

Component form, magnitude and unit vectors

In three dimensions a vector is written $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$, where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the unit vectors along the coordinate axes. Addition and scalar multiplication act component by component:

The magnitude (length) is

The unit vector in the direction of $\mathbf{a}$ (for $\mathbf{a} \neq \mathbf{0}$) is

which has magnitude $1$ and points the same way as $\mathbf{a}$. The two-dimensional case is the same with the $\mathbf{k}$ component set to zero.

The scalar (dot) product

The scalar product of $\mathbf{a}$ and $\mathbf{b}$ is defined by either of two equivalent expressions:

where $\theta$ is the angle between the vectors with $0 \leq \theta \leq \pi$. The dot product is commutative and distributes over addition, and $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$. Rearranging the geometric form gives the angle between vectors:

Parallel and perpendicular tests

Two key tests follow from the dot product.

• Perpendicular: $\mathbf{a} \perp \mathbf{b}$ if and only if $\mathbf{a} \cdot \mathbf{b} = 0$ (with both nonzero), since $\cos 90^\circ = 0$.
• Parallel: $\mathbf{a} \parallel \mathbf{b}$ if and only if $\mathbf{a} = \lambda\mathbf{b}$ for some scalar $\lambda$; equivalently the dot product reaches its extreme $\pm|\mathbf{a}||\mathbf{b}|$ because $\cos\theta = \pm 1$.

Scalar and vector resolutes (projection)

Resolving $\mathbf{a}$ in the direction of $\mathbf{b}$ splits $\mathbf{a}$ into a component along $\mathbf{b}$ and a component perpendicular to it. The scalar resolute of $\mathbf{a}$ in the direction of $\mathbf{b}$ is

a signed length. The vector resolute (the projection of $\mathbf{a}$ onto $\mathbf{b}$) is

The component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ is then $\mathbf{a} - \operatorname{proj}_{\mathbf{b}}\mathbf{a}$, which lets you decompose $\mathbf{a}$ into parallel and perpendicular parts.

Why the dot product is central

The dot product converts geometry into arithmetic. Lengths come from $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$, angles from the cosine formula, perpendicularity from a zero product, and projections from resolving along a unit vector. Almost every vector exam question reduces to one of these four uses, so recognising which one a problem needs is the key skill.

Examples in context

Example 1. Perpendicularity. $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = 4\mathbf{i} - 2\mathbf{j}$ satisfy $\mathbf{a} \cdot \mathbf{b} = 4 - 4 = 0$, so they are perpendicular.

Example 2. Unit vector. For $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$, $|\mathbf{a}| = 5$, so $\hat{\mathbf{a}} = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}$.

Try this

Q1. Find the angle between $\mathbf{a} = \mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{i}$. [2 marks]

• Cue. $\cos\theta = \frac{1}{\sqrt{2}\cdot 1} = \frac{1}{\sqrt 2}$, so $\theta = 45^\circ$.

Q2. Find the scalar resolute of $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ in the direction of $\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$. [2 marks]

• Cue. $\mathbf{a}\cdot\mathbf{b} = 2 - 2 + 4 = 4$, $|\mathbf{b}| = 3$, so scalar resolute $= \frac{4}{3}$.

Q3. Determine the value of $t$ for which $\mathbf{a} = t\mathbf{i} + 3\mathbf{j}$ is perpendicular to $\mathbf{b} = 2\mathbf{i} - \mathbf{j}$. [2 marks]

• Cue. $2t - 3 = 0$, so $t = \frac{3}{2}$.