Vectors in three dimensions (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants confident vector algebra in three dimensions: adding and scaling vectors, computing magnitude and unit vectors, using the scalar product to find angles and scalar/vector projections, and using the vector product to find perpendiculars and the area of a triangle or parallelogram.

Components, magnitude and unit vectors

Write $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$. Then $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$. A unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \tfrac{1}{|\mathbf{a}|}\mathbf{a}$. The vector from point $A$ to point $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (position vectors).

The scalar (dot) product

This gives the angle between vectors via $\cos\theta = \tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. Two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b} = 0$.

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is $\tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$, and the vector projection is $\tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$.

The vector (cross) product

The result is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ (right-hand rule), with $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$. It is anticommutative: $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$, and $\mathbf{a}\times\mathbf{a} = \mathbf{0}$.

Choosing the right tool

The two products do different jobs, and SCSA questions reward picking the right one. Use the scalar product when the question asks for an angle, a test of perpendicularity, or a projection of one vector along another, because it returns a number tied to $\cos\theta$. Use the vector product when the question asks for a vector perpendicular to two given vectors, a normal to a plane, a test of whether vectors are parallel, or an area, because it returns a vector with magnitude tied to $\sin\theta$. A quick way to remember it: dot for direction comparison, cross for area and normals.

Position vectors and points in space

Geometry questions are usually phrased in terms of points, so the first move is almost always to convert points into vectors. The position vector of a point $A$ is the vector from the origin to $A$, and the displacement from $A$ to $B$ is $\vec{AB} = \mathbf{b} - \mathbf{a}$. To find an angle at a vertex, take the two edges leaving that vertex and dot them; to find an area, cross them. The midpoint of $AB$ has position vector $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$, and a point dividing $AB$ in the ratio $m:n$ has position vector $\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$, both of which appear in collinearity and ratio problems.