AS & A Level · Practice questions · The vector product

Cross it, then check it.

Six original Cambridge-style questions on the vector product: computing it in components, using it for the sine of an angle, normals to planes, areas of triangles, converting a plane to scalar product form and the zero product of parallel vectors.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Set out, cross, then dot to check.

Write both vectors as columns in the order the question gives; \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\).
Cross with the determinant layout: cover each column in turn, and put a minus on the \(\mathbf{j}\) term.
Check by dotting the answer with both vectors: each must give 0.
Plane: \(\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}\) or \(\mathbf{b} \times \mathbf{c}\), scaled if you like, then \(p = \mathbf{a} \cdot \mathbf{n}\).
Area: parallelogram \(|\mathbf{a} \times \mathbf{b}|\), triangle \(\dfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|\), using the unscaled product.

Recall

[2 marks]

Find \((3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} - \mathbf{k})\), and verify that your answer is perpendicular to both vectors.

\(\mathbf{i}\): \((-1)(-1) - (2)(2) = -3\); \(\mathbf{j}\): \((2)(1) - (3)(-1) = 5\); \(\mathbf{k}\): \((3)(2) - (-1)(1) = 7\). ✓
So the product is \(-3\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}\). ✓
Check: \((-3)(3) + (5)(-1) + (7)(2) = -9 - 5 + 14 = 0\) and \((-3)(1) + (5)(2) + (7)(-1) = -3 + 10 - 7 = 0\). ✓

Analysis

[5 marks]

\(\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{b} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\).

(a) Find \(\mathbf{a} \times \mathbf{b}\).

(b) Hence find the exact value of \(\sin\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

(a) \(\begin{pmatrix} (-2)(-2) - (2)(1) \\ (2)(2) - (1)(-2) \\ (1)(1) - (-2)(2) \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 5 \end{pmatrix}\). ✓
(b) \(|\mathbf{a} \times \mathbf{b}| = \sqrt{4 + 36 + 25} = \sqrt{65}\), and \(|\mathbf{a}| = |\mathbf{b}| = 3\), so \(\sin\theta = \dfrac{\sqrt{65}}{9}\). ✓
(c) Both an acute angle and its supplement have this sine, so you need the sign of \(\cos\theta\). ✓
\(\mathbf{a} \cdot \mathbf{b} = 2 - 2 - 4 = -4 < 0\), so \(\theta\) is obtuse: \(\cos\theta = -\dfrac{4}{9}\). ✓
\(\theta = 180^{\circ} - 63.6^{\circ} = 116.4^{\circ}\) (1 d.p.). ✓

Calculation

[4 marks]

The points \(A(2, 1, 0)\), \(B(0, 3, 1)\) and \(C(1, 0, 2)\) lie in the plane \(\Pi\). Find a Cartesian equation of \(\Pi\).

\(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix}\). ✓
\(\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (2)(2) - (1)(-1) \\ (1)(-1) - (-2)(2) \\ (-2)(-1) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix}\). ✓
\(p = \overrightarrow{OA} \cdot \mathbf{n} = 10 + 3 + 0 = 13\), so \(5x + 3y + 4z = 13\). ✓
Check: \(B\) gives \(0 + 9 + 4 = 13\) and \(C\) gives \(5 + 0 + 8 = 13\). ✓

Calculation

[3 marks]

Find the exact area of the triangle with vertices \(P(1, 1, 1)\), \(Q(3, 2, 1)\) and \(R(1, 3, 4)\).

\(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\), \(\overrightarrow{PR} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}\). ✓
\(\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{pmatrix} (1)(3) - (0)(2) \\ (0)(0) - (2)(3) \\ (2)(2) - (1)(0) \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 4 \end{pmatrix}\). ✓
Area \(= \dfrac{1}{2}\sqrt{9 + 36 + 16} = \dfrac{\sqrt{61}}{2} \approx 3.91\). ✓

Calculation

[4 marks]

The plane \(\Pi\) has equation \(\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).

Find an equation of \(\Pi\) in the form \(\mathbf{r} \cdot \mathbf{n} = p\), and hence its Cartesian equation.

\(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} (1)(-1) - (0)(0) \\ (0)(1) - (2)(-1) \\ (2)(0) - (1)(1) \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix}\); take \(\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). ✓
Check: \(\mathbf{n} \cdot \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = 2 - 2 = 0\) and \(\mathbf{n} \cdot \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = 1 - 1 = 0\). ✓
\(p = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 1 + 2 + 3 = 6\). ✓
\(\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 6\), that is \(x - 2y + z = 6\). ✓

Reasoning

[4 marks]

\(\mathbf{a} = \mathbf{i} + k\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{b} = 2\mathbf{i} + 4\mathbf{j} + m\mathbf{k}\), where \(k\) and \(m\) are constants. Given that \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\), find \(k\) and \(m\), and state the geometrical relationship between \(\mathbf{a}\) and \(\mathbf{b}\).

\(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} km - 8 \\ 4 - m \\ 4 - 2k \end{pmatrix}\). ✓
Setting each component to 0: \(4 - m = 0 \Rightarrow m = 4\) and \(4 - 2k = 0 \Rightarrow k = 2\). ✓
Check the first component: \(km - 8 = 8 - 8 = 0\). ✓
A zero vector product of non-zero vectors means \(\sin\theta = 0\): \(\mathbf{a}\) and \(\mathbf{b}\) are parallel (indeed \(\mathbf{b} = 2\mathbf{a}\)). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.