Home / A-Level Further Maths 9231 / Vectors / The vector product
Cambridge International AS & A Level · Topic 1.6
A-Level Further Maths 9231 / Topic 1.6 / AS & A Level

A vector at right angles to both.

The scalar product of two vectors is a number. The vector product \(\mathbf{a} \times \mathbf{b}\) is a vector: it points at right angles to both \(\mathbf{a}\) and \(\mathbf{b}\), and its length is the area of the parallelogram they make. That one idea gives you normals to planes, areas of triangles and a quick route from the parametric form of a plane to \(\mathbf{r} \cdot \mathbf{n} = p\).

Topic 1.6All vectors lessons →
The key idea

\(\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta \, \hat{\mathbf{n}}\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\) and \(\hat{\mathbf{n}}\) is the unit vector perpendicular to both, chosen by the right-hand rule. In components, \(\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2)\mathbf{i} + (a_3 b_1 - a_1 b_3)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}\). Swapping the order reverses it, \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\), and parallel vectors give \(\mathbf{0}\).

h = |b| sin θ area = |a × b| a b θ a × b b × a O The vector product a × b is perpendicular to both a and b |a × b| = |a||b| sin θ = area of parallelogram b × a = −(a × b) Right-hand rule fingers turn from a to b, thumb points along a × b
Fig. 1: a and b drawn from O lie in the shaded plane. a × b is perpendicular to both (right-hand rule: curl from a to b, thumb up), and b × a points the opposite way. The parallelogram on a and b has base |a| and height |b| sin θ, so its area is |a × b|
Section 01

Perpendicular to both.

For two vectors \(\mathbf{a}\) and \(\mathbf{b}\) with angle \(\theta\) between them (\(0 \le \theta \le \pi\)), the vector product (or cross product) is \[\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta \, \hat{\mathbf{n}},\] where \(\hat{\mathbf{n}}\) is a unit vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). There are two such unit vectors, pointing opposite ways; the right-hand rule picks one. Curl the fingers of your right hand from \(\mathbf{a}\) round to \(\mathbf{b}\) and your thumb points along \(\mathbf{a} \times \mathbf{b}\) (Fig. 1). The length \(|\mathbf{a}||\mathbf{b}|\sin\theta\) is base times height of the parallelogram on \(\mathbf{a}\) and \(\mathbf{b}\), so it is that parallelogram's area.

Worked example Products of i, j and k
Find (a) \(\mathbf{j} \times \mathbf{i}\) (b) \(\mathbf{i} \times \mathbf{k}\) (c) \((2\mathbf{j}) \times (5\mathbf{j})\) (d) \((3\mathbf{k}) \times (4\mathbf{i})\).
  1. (a) \(\mathbf{i} \times \mathbf{j} = \mathbf{k}\), and reversing the order changes the sign: \(\mathbf{j} \times \mathbf{i} = -\mathbf{k}\).
  2. (b) The cyclic product is \(\mathbf{k} \times \mathbf{i} = \mathbf{j}\), so \(\mathbf{i} \times \mathbf{k} = -\mathbf{j}\).
  3. (c) \(2\mathbf{j}\) and \(5\mathbf{j}\) are parallel, so the product is \(\mathbf{0}\).
  4. (d) \((3\mathbf{k}) \times (4\mathbf{i}) = 12(\mathbf{k} \times \mathbf{i}) = 12\mathbf{j}\).
Answer: (a) \(-\mathbf{k}\)   (b) \(-\mathbf{j}\)   (c) \(\mathbf{0}\)   (d) \(12\mathbf{j}\)
Worked example Length from the angle
\(|\mathbf{a}| = 6\), \(|\mathbf{b}| = 5\) and the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(\dfrac{\pi}{6}\). Find \(|\mathbf{a} \times \mathbf{b}|\), the area of the parallelogram with sides \(\mathbf{a}\) and \(\mathbf{b}\), and the area of the triangle with sides \(\mathbf{a}\) and \(\mathbf{b}\).
  1. \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = 6 \times 5 \times \sin\dfrac{\pi}{6} = 30 \times \dfrac{1}{2} = 15\).
  2. The parallelogram's area is \(|\mathbf{a} \times \mathbf{b}| = 15\).
  3. The triangle is half of the parallelogram: \(\dfrac{1}{2} \times 15 = 7.5\).
Answer: \(|\mathbf{a} \times \mathbf{b}| = 15\); parallelogram \(15\); triangle \(7.5\)
Your turn
\(|\mathbf{a}| = 3\), \(|\mathbf{b}| = 8\) and the angle between them is \(150^{\circ}\). Find \(|\mathbf{a} \times \mathbf{b}|\).
Examiner trap

\(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{b} \times \mathbf{a}\) are NOT equal: they have the same length but opposite directions. The scalar product commutes, the vector product does not, so keep the order the question gives.

Section 02

The component formula.

In components, with \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), \[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix}.\] A reliable way to remember it is the determinant layout: put \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) on the top row, then \(\mathbf{a}\), then \(\mathbf{b}\), and expand along the top row, remembering the MINUS on the middle term.

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\]

\[= \mathbf{i}\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}\]

\[- \mathbf{j}\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}\]

\[+ \mathbf{k}\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}\]

Cover the column of the unit vector you are working on and take the 2 by 2 determinant of what is left. Always check the answer: dot it with \(\mathbf{a}\) and with \(\mathbf{b}\); both results must be 0.

Worked example Compute and check
Find \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}\).
  1. \(\mathbf{i}\): cover the first column, \((-1)(-2) - (3)(4) = 2 - 12 = -10\).
  2. \(\mathbf{j}\): cover the second column, \((2)(-2) - (3)(1) = -4 - 3 = -7\), then the minus sign gives \(+7\).
  3. \(\mathbf{k}\): cover the third column, \((2)(4) - (-1)(1) = 8 + 1 = 9\).
  4. Check: \((-10)(2) + (7)(-1) + (9)(3) = -20 - 7 + 27 = 0\) and \((-10)(1) + (7)(4) + (9)(-2) = -10 + 28 - 18 = 0\).
Answer: \(\begin{pmatrix} -10 \\ 7 \\ 9 \end{pmatrix}\)
Worked example Sine of the angle
Find the exact value of \(\sin\theta\), where \(\theta\) is the angle between \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{b} = 3\mathbf{i} + 4\mathbf{k}\).
  1. \(\mathbf{a} \times \mathbf{b} = (2 \times 4 - 2 \times 0)\mathbf{i} + (2 \times 3 - 1 \times 4)\mathbf{j} + (1 \times 0 - 2 \times 3)\mathbf{k} = 8\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}\).
  2. \(|\mathbf{a} \times \mathbf{b}| = \sqrt{64 + 4 + 36} = \sqrt{104} = 2\sqrt{26}\).
  3. \(|\mathbf{a}| = \sqrt{1 + 4 + 4} = 3\) and \(|\mathbf{b}| = \sqrt{9 + 0 + 16} = 5\).
  4. \(\sin\theta = \dfrac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|} = \dfrac{2\sqrt{26}}{15}\). (Check with the scalar product: \(\cos\theta = \dfrac{11}{15}\) and \(\dfrac{121}{225} + \dfrac{104}{225} = 1\).)
Answer: \(\sin\theta = \dfrac{2\sqrt{26}}{15}\)
Your turn
Find \((\mathbf{i} + \mathbf{j}) \times (\mathbf{j} + \mathbf{k})\).
Examiner trap

The middle component is \(a_3 b_1 - a_1 b_3\), the NEGATIVE of the 2 by 2 determinant \(a_1 b_3 - a_3 b_1\). Forgetting the minus on \(\mathbf{j}\) is the most common slip, and the dot-product check catches it every time.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which is equal to \(\mathbf{b} \times \mathbf{a}\)?

Quick check+10 XP

Find \(\mathbf{i} \times \mathbf{k}\).

Quick check+10 XP

Find \(\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\).

Quick check+10 XP

\(|\mathbf{a}| = 2\), \(|\mathbf{b}| = 7\) and \(\mathbf{a}\) is perpendicular to \(\mathbf{b}\). Find \(|\mathbf{a} \times \mathbf{b}|\).

Section 03

Normals, areas and planes.

Because \(\mathbf{a} \times \mathbf{b}\) is perpendicular to both vectors, it is a normal to any plane that contains directions \(\mathbf{a}\) and \(\mathbf{b}\). For a plane through three points \(A\), \(B\), \(C\), take \(\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}\), then \(p = \overrightarrow{OA} \cdot \mathbf{n}\). The same product gives areas.

Worked example Plane and area from three points
The points \(A(1, 0, 2)\), \(B(3, 1, 0)\) and \(C(2, -1, 4)\) lie in a plane \(\Pi\). Find a Cartesian equation of \(\Pi\) and the area of triangle \(ABC\).
  1. \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\).
  2. \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (1)(2) - (-2)(-1) \\ (-2)(1) - (2)(2) \\ (2)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ -6 \\ -3 \end{pmatrix}\).
  3. Take \(\mathbf{n} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\) (divide by \(-3\)). Then \(p = \overrightarrow{OA} \cdot \mathbf{n} = 0 + 0 + 2 = 2\), so \(2y + z = 2\).
  4. Check with \(B\): \(2 + 0 = 2\); with \(C\): \(-2 + 4 = 2\).
  5. Area \(= \dfrac{1}{2}\left|\begin{pmatrix} 0 \\ -6 \\ -3 \end{pmatrix}\right| = \dfrac{1}{2}\sqrt{0 + 36 + 9} = \dfrac{\sqrt{45}}{2} = \dfrac{3\sqrt{5}}{2}\). Use the UNSCALED product for areas.
Answer: \(2y + z = 2\); area \(\dfrac{3\sqrt{5}}{2} \approx 3.35\)
Worked example Parametric to scalar product form
Write the plane \(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}\) in the form \(\mathbf{r} \cdot \mathbf{n} = p\).
  1. \(\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (0)(-1) - (2)(1) \\ (2)(0) - (1)(-1) \\ (1)(1) - (0)(0) \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}\).
  2. Check: \(\mathbf{n} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = -2 + 0 + 2 = 0\) and \(\mathbf{n} \cdot \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = 0 + 1 - 1 = 0\).
  3. \(p = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = -2 + 2 - 1 = -1\).
Answer: \(\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = -1\), or in Cartesian form \(2x - y - z = 1\)
Your turn
Find the area of the parallelogram with sides \(\mathbf{i} + 2\mathbf{j}\) and \(\mathbf{j} + 3\mathbf{k}\).
Examiner trap

You may scale a normal freely (\(\begin{pmatrix} 0 \\ -6 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\) give the same plane), but NOT before finding an area: the area needs the length of the actual product, and a triangle needs the \(\dfrac{1}{2}\).

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Further Mathematics Paper 1 to 4 standard.

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Exam style+20 XP

Find a Cartesian equation of the plane through \(A(1, 1, 0)\), \(B(2, 0, 1)\) and \(C(0, 2, 3)\).

Exam style+20 XP

Find the area of the triangle with vertices \(O\), \(P(2, 1, 2)\) and \(Q(1, 0, 3)\).

Exam style+20 XP

The plane \(\Pi\) has equation \(\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\). Find a Cartesian equation of \(\Pi\).

Exam style+20 XP

\(\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + \mathbf{k}\). Use the vector product to find \(\sin\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

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Six original Cambridge-style questions
Component products with a check, the sine of an obtuse angle, a plane through three points, a triangle's area, a parametric plane in \(\mathbf{r} \cdot \mathbf{n} = p\) form and parallel vectors.