Find the magnitude of pmatrix 2 \\ -3 \\ 6 pmatrix. [2 marks]

Show that pmatrix 1 \\ 2 \\ -1 pmatrix and pmatrix 3 \\ -1 \\ 1 pmatrix are perpendicular. [2 marks]

SQA SQA Higher 2022-style practice: Vectors are given by u = pmatrix 3 \\ -1 \\ 2 pmatrix and v = pmatrix 1 \\ 4 \\ -2 pmatrix. Calculate the size of the angle between u and v, giving your answer to the nearest degree.

Scalar product: u · v = 3(1) + (-1)(4) + 2(-2) = 3 - 4 - 4 = -5 (1 mark). Magnitudes: |u| = sqrt(9 + 1 + 4) = sqrt(14) and |v| = sqrt(1 + 16 + 4) = sqrt(21) (2 marks). Then cosθ = -5sqrt(14)sqrt(21) = -5sqrt(294) ≈ -0.2916 (1 mark), so θ = cos^-1(-0.2916) ≈ 107^ (1 mark). Markers reward the scalar product, both magnitudes, and the angle from the cosine, noting the negative value gives an obtuse angle.

ScotlandMathsSyllabus dot point

How do you describe position and direction in three dimensions, and how does the scalar product reveal the angle between two vectors?

Vectors in three dimensions, the magnitude and unit vector, addition and scalar multiplication, the section formula and collinearity, and the scalar product including its use to find the angle between two vectors and to test for perpendicularity.

A focused answer to the SQA Higher Mathematics vectors content, covering three-dimensional vectors, magnitude and unit vectors, addition and scalar multiplication, collinearity and the section formula, and the scalar product used to find the angle between vectors and test perpendicularity.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Magnitude and unit vectors
Collinearity and the section formula
The scalar product
Examples in context
Try this

What this dot point is asking

The SQA wants you to handle vectors in three dimensions, find a magnitude and a unit vector, add and scale vectors, test for collinearity and use the section formula, and apply the scalar product to find the angle between two vectors and to test whether they are perpendicular.

Magnitude and unit vectors

A vector in three dimensions records a displacement in the $x$ , $y$ and $z$ directions. Its magnitude (length) comes from a three-dimensional Pythagoras, and a unit vector is a vector of length $1$ pointing the same way, obtained by dividing by the magnitude.

Find a unit vector in the direction of

\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}

A unit vector points in the same direction as $\mathbf{a}$ but has length $1$ . Find it.

Step 1: Calculate the magnitude of $\mathbf{a}$

The magnitude uses the three-dimensional version of Pythagoras, squaring and summing all three components:

|\mathbf{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

Step 2: Divide by the magnitude to scale to length $1$

Dividing each component by $|\mathbf{a}| = 3$ shrinks the vector to length $1$ without changing its direction:

\hat{\mathbf{a}} = \dfrac{1}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} \tfrac{2}{3} \\[4pt] -\tfrac{1}{3} \\[4pt] \tfrac{2}{3} \end{pmatrix}

Step 3: Verify the length is $1$

Check by computing the magnitude of $\hat{\mathbf{a}}$ :

\left(\tfrac{2}{3}\right)^2 + \left(\tfrac{1}{3}\right)^2 + \left(\tfrac{2}{3}\right)^2 = \tfrac{4 + 1 + 4}{9} = 1 \checkmark

Final answer: The unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \begin{pmatrix} \tfrac{2}{3} \\[2pt] -\tfrac{1}{3} \\[2pt] \tfrac{2}{3} \end{pmatrix}$ .

Collinearity and the section formula

Three points are collinear when one connecting vector is a scalar multiple of another and they share a common point. So to test collinearity of $A$ , $B$ , $C$ , form $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (or $\overrightarrow{AC}$ ) and check whether one is a scalar multiple of the other. The section formula gives the point dividing $AB$ in a given ratio by combining the position vectors in proportion: the point dividing $AB$ in ratio $m : n$ has position vector $\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$ .

The scalar product

The scalar (dot) product turns two vectors into a single number, and crucially that number is also $|\mathbf{a}||\mathbf{b}|\cos\theta$ . Equating the two forms lets you extract the angle between the vectors, and a zero scalar product signals perpendicularity.

Find the angle between

\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}

Find the angle between the two vectors by equating the component and geometric forms of the scalar product.

Step 1: Compute the scalar product using components

Multiply corresponding components and add:

\mathbf{a} \cdot \mathbf{b} = 1(2) + 2(0) + 2(1) = 2 + 0 + 2 = 4

Step 2: Find the magnitude of each vector

Apply the three-dimensional Pythagoras formula to each:

|\mathbf{a}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \qquad |\mathbf{b}| = \sqrt{4 + 0 + 1} = \sqrt{5}

Step 3: Use the geometric form to find the angle

The geometric form of the scalar product gives $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ . Rearranging for $\cos\theta$ and substituting:

\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \dfrac{4}{3\sqrt{5}} \approx 0.5963

\theta = \cos^{-1}(0.5963) \approx 53.4^\circ

Final answer: The angle between $\mathbf{a}$ and $\mathbf{b}$ is approximately $53.4^\circ$ .

Examples in context

In three-dimensional design and physics the scalar product tests whether two struts meet at a right angle. If two beams point along $\mathbf{p} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$ , then $\mathbf{p} \cdot \mathbf{q} = 2 - 3 + 1 = 0$ , so the beams are perpendicular and the joint is square, a conclusion an engineer reaches with one quick calculation rather than measuring an angle.

Try this

Q1. Find the magnitude of $\begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$ . [2 marks]

Cue. $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .

Q2. Show that $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$ are perpendicular. [2 marks]

Cue. Scalar product $= 3 - 2 - 1 = 0$ , so they are perpendicular.

Exam-style practice questions

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

SQA Higher 20205 marksThe points

A(1, -1, 2)

B(3, 0, 4)

and

C(7, 2, 8)

are given. Show that

A

B

and

C

are collinear, and state the ratio in which

B

divides

AC

Show worked answer →

Find direction vectors. $\overrightarrow{AB} = B - A = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ and $\overrightarrow{AC} = C - A = \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix}$ (2 marks).

Observe $\overrightarrow{AC} = 3\overrightarrow{AB}$ , so the vectors are parallel; since they share the common point $A$ , the three points are collinear (2 marks).

Because $\overrightarrow{AC} = 3\overrightarrow{AB}$ , $B$ is one third of the way from $A$ to $C$ , so $\overrightarrow{AB} : \overrightarrow{BC} = 1 : 2$ (1 mark). Markers reward both direction vectors, the scalar-multiple link plus the common-point statement, and the ratio.

SQA Higher 20225 marksVectors are given by

\mathbf{u} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}

and

\mathbf{v} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}

. Calculate the size of the angle between

\mathbf{u}

and

\mathbf{v}

, giving your answer to the nearest degree.

Show worked answer →

Scalar product: $\mathbf{u} \cdot \mathbf{v} = 3(1) + (-1)(4) + 2(-2) = 3 - 4 - 4 = -5$ (1 mark).

Magnitudes: $|\mathbf{u}| = \sqrt{9 + 1 + 4} = \sqrt{14}$ and $|\mathbf{v}| = \sqrt{1 + 16 + 4} = \sqrt{21}$ (2 marks).

Then $\cos\theta = \dfrac{-5}{\sqrt{14}\sqrt{21}} = \dfrac{-5}{\sqrt{294}} \approx -0.2916$ (1 mark), so $\theta = \cos^{-1}(-0.2916) \approx 107^\circ$ (1 mark). Markers reward the scalar product, both magnitudes, and the angle from the cosine, noting the negative value gives an obtuse angle.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)