What is sign errors in the cross-product determinant?

The middle (j) component carries a minus sign; dropping it flips that component.

Find a·b for a = i + 2j + 3k and b = 4i - j + k. [2 marks]

State a unit vector perpendicular to both i and j. [1 mark]

OCR OCR 2019-style practice: The vectors are a = 2i - j + 2k and b = i + 2j + 2k. Find the angle between them, giving your answer to the nearest degree.

Scalar product (M1): a·b = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4 (A1). Magnitudes (M1): |a| = sqrt(4 + 1 + 4) = 3 and |b| = sqrt(1 + 4 + 4) = 3. Use cosθ = a·b|a||b| (M1): cosθ = 49, so θ = 49 ≈ 63.6^ ≈ 64^ (A1). Markers reward the scalar product, both magnitudes, the cosine formula, and the angle to the nearest degree.

OCR OCR 2022-style practice: Find a vector perpendicular to both a = i + j - k and b = 2i - j + 3k, and hence find the area of the triangle with two sides a and b.

Use the vector product (M1): a×b = vmatrix i & j & k \\ 1 & 1 & -1 \\ 2 & -1 & 3 vmatrix. Expand (M1, A1): i((1)(3) - (-1)(-1)) - j((1)(3) - (-1)(2)) + k((1)(-1) - (1)(2)) = i(3 - 1) - j(3 + 2) + k(-1 - 2) = 2i - 5j - 3k (A1). This vector is perpendicular to both. Its magnitude is |a×b| = sqrt(4 + 25 + 9) = sqrt(38) (M1). The triangle area is half the parallelogram area (A1): 12sqrt(38). Markers reward the determinant set-up, the correct cross product, its magnitude, and halving for the triangle.

EnglandFurther MathsSyllabus dot point

How do the scalar and vector products work, and what geometric information do they give?

The scalar (dot) product and its use for angles and perpendicularity, the vector (cross) product and its use for a perpendicular direction and areas, and the modulus of the vector product as an area.

A focused answer to the OCR A-Level Further Mathematics A content on the scalar and vector products, covering the dot product and its use for the angle between vectors and for testing perpendicularity, the cross product and its use to find a vector perpendicular to two given vectors, and the modulus of the cross product as the area of a parallelogram or triangle.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The scalar product is $\mathbf{a}\cdot\mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$ , a number; it gives the angle via $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ , and two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b} = 0$ . The vector product $\mathbf{a}\times\mathbf{b}$ is a vector, computed as the determinant $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ ; it is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ , and its modulus $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ is the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ (half that for the triangle).

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What this dot point is asking
The scalar product
The angle between two vectors
The vector product
The modulus as an area
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What this dot point is asking

OCR wants you to compute and use the scalar (dot) product to find the angle between two vectors and to test for perpendicularity, and to compute and use the vector (cross) product to find a vector perpendicular to two given vectors and to find the area of a parallelogram or triangle, using the fact that the modulus of the cross product equals that area.

The scalar product

The scalar product multiplies corresponding components and adds, giving a single number. Its power is the link to the angle between the vectors and the instant test for perpendicularity.

The angle between two vectors

The scalar product gives the angle directly. Compute $\mathbf{a}\cdot\mathbf{b}$ and the two magnitudes, then take the inverse cosine of their ratio. The result lies between $0$ and $\pi$ . A positive scalar product means the angle is acute, a negative value means it is obtuse, and zero means the vectors are perpendicular, so the sign alone tells you the rough geometry before you compute the angle. If a question asks for the acute angle between two lines, take the modulus of the dot product so the cosine is non-negative.

The vector product

The vector product produces a vector perpendicular to both inputs, with direction given by the right-hand rule. It is computed as a symbolic determinant with $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ across the top row.

The modulus as an area

The length of the vector product is $|\mathbf{a}||\mathbf{b}|\sin\theta$ , which is exactly the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$ . Halving gives the area of the triangle, a standard application.

The scalar and vector products are the engine of the rest of the vectors strand: angles between lines and planes use the dot product, and the cross product supplies the normal to a plane.

Try this

Q1. Find $\mathbf{a}\cdot\mathbf{b}$ for $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}$ . [2 marks]

Cue. $(1)(4) + (2)(-1) + (3)(1) = 4 - 2 + 3 = 5$ .

Q2. State a unit vector perpendicular to both $\mathbf{i}$ and $\mathbf{j}$ . [1 mark]

Cue. $\mathbf{i}\times\mathbf{j} = \mathbf{k}$ , which already has magnitude $1$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksThe vectors are

\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}

and

\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}

. Find the angle between them, giving your answer to the nearest degree.

Show worked answer →

Scalar product (M1): $\mathbf{a}\cdot\mathbf{b} = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4$ (A1).

Magnitudes (M1): $|\mathbf{a}| = \sqrt{4 + 1 + 4} = 3$ and $|\mathbf{b}| = \sqrt{1 + 4 + 4} = 3$ .

Use $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ (M1): $\cos\theta = \dfrac{4}{9}$ , so $\theta = \arccos\dfrac{4}{9} \approx 63.6^\circ \approx 64^\circ$ (A1).

Markers reward the scalar product, both magnitudes, the cosine formula, and the angle to the nearest degree.

OCR 20226 marksFind a vector perpendicular to both

\mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k}

and

\mathbf{b} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}

, and hence find the area of the triangle with two sides

\mathbf{a}

and

\mathbf{b}

Show worked answer →

Use the vector product (M1): $\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & -1 & 3 \end{vmatrix}$ .

Expand (M1, A1): $\mathbf{i}((1)(3) - (-1)(-1)) - \mathbf{j}((1)(3) - (-1)(2)) + \mathbf{k}((1)(-1) - (1)(2)) = \mathbf{i}(3 - 1) - \mathbf{j}(3 + 2) + \mathbf{k}(-1 - 2) = 2\mathbf{i} - 5\mathbf{j} - 3\mathbf{k}$ (A1).

This vector is perpendicular to both. Its magnitude is $|\mathbf{a}\times\mathbf{b}| = \sqrt{4 + 25 + 9} = \sqrt{38}$ (M1).

The triangle area is half the parallelogram area (A1): $\tfrac{1}{2}\sqrt{38}$ .

Markers reward the determinant set-up, the correct cross product, its magnitude, and halving for the triangle.

Related dot points

Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)