Use vectors in three dimensions, compute scalar (dot) and vector (cross) products, find angles between vectors, scalar and vector projections, and apply these to geometric problems and the equations of lines and planes

What this dot point is asking

QCAA wants you to work with vectors in three dimensions: write them in component form, find magnitudes and unit vectors, compute dot and cross products, find angles and projections, and use vectors to describe lines and to solve geometric problems. This is heavily assessed in IA2 and the external assessment, where setting up the right product for the question is the key decision.

The answer

Component form and magnitude

A vector in three dimensions is written $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ or as a column. Its magnitude is

A unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$. The vector from point $A$ to point $B$ is $\vec{AB} = \mathbf{b} - \mathbf{a}$, the position vector of $B$ minus that of $A$.

The dot (scalar) product

The dot product produces a scalar:

where $\theta$ is the angle between the vectors. Rearranging gives the angle:

If $\mathbf{a}\cdot\mathbf{b} = 0$ and neither vector is zero, the vectors are perpendicular. The dot product is the tool for angles and perpendicularity.

Scalar and vector projections

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is the signed length of the shadow of $\mathbf{a}$ along $\mathbf{b}$:

The vector projection multiplies this by the unit vector $\hat{\mathbf{b}}$:

The scalar projection is negative when the angle is obtuse, which the sign of $\mathbf{a}\cdot\mathbf{b}$ captures automatically.

The cross (vector) product

The cross product produces a vector perpendicular to both inputs:

Its magnitude is $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$, which equals the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. The direction follows the right-hand rule. The cross product is the tool for finding a normal vector and for areas.

Equations of lines

A line through point $A$ with position vector $\mathbf{a}$ and direction $\mathbf{d}$ has vector equation

Splitting into components gives the parametric equations $x = a_1 + t d_1$, $y = a_2 + t d_2$, $z = a_3 + t d_3$. Two lines are parallel when their direction vectors are scalar multiples, and they intersect when a single pair $(t, s)$ satisfies all three component equations.

Properties of the dot product

The dot product is commutative, $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$, and distributes over addition, $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$. A vector dotted with itself returns the square of its length, $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$, which is the algebraic identity behind every magnitude calculation. The base vectors satisfy $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1$ and $\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0$, because they are mutually perpendicular unit vectors. These identities let you expand a product such as $(\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 + \mathbf{a} \cdot \mathbf{b} - 2|\mathbf{b}|^2$ without resorting to components, a move QCAA rewards in proof-style items.

Properties of the cross product

The cross product is anti-commutative, $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$, so the order matters and reversing it flips the direction of the normal. A vector crossed with itself or with a parallel vector gives the zero vector, $\mathbf{a} \times \mathbf{a} = \mathbf{0}$, which is the algebraic test for parallel vectors. For the base vectors the cyclic rule holds: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$, with a sign change if you reverse any pair. Because $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ equals the parallelogram area, half of it is the area of the triangle spanned by the two vectors, which is the standard route to a triangle area from three position vectors.

Choosing the right product

The single most important decision in a vectors question is which product to use. Reach for the dot product whenever the question mentions an angle, perpendicularity, work done, or a projection, because the dot product converts directly into $\cos\theta$. Reach for the cross product whenever the question mentions a normal direction, an area, or a vector perpendicular to two others, because the cross product converts into $\sin\theta$ and produces a vector. Stating which product you will use and why is itself worth communication credit in extended-response items.