AS & A Level · Practice questions · The scalar product

Dot, divide, inverse cosine.

Six original Cambridge-style questions on the scalar product: angles between vectors and lines, perpendicularity, the foot of the perpendicular from a point to a line, and a pyramid.

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

Components for the number, cosine for the angle.

Calculate: \(\mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\).
Angle: \(\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\), with both vectors pointing away from the vertex.
Lines: use the direction vectors and give the acute angle.
Perpendicular: set the scalar product equal to zero.
Foot of perpendicular: general point \(N\), then \(\overrightarrow{PN} \cdot \mathbf{b} = 0\).

Recall

[2 marks]

Show that the vectors \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) and \(4\mathbf{i} + \mathbf{j} - 5\mathbf{k}\) are perpendicular.

\(2(4) + (-3)(1) + 1(-5) = 8 - 3 - 5 = 0\). ✓
The scalar product of two non-zero vectors is zero, so they are perpendicular. ✓

Calculation

[2 marks]

Find the value of \(p\) for which \(\mathbf{i} + p\mathbf{j} + 2p\mathbf{k}\) is perpendicular to \(4\mathbf{i} + 3\mathbf{j} - \mathbf{k}\).

\(4 + 3p - 2p = 0\). ✓
\(p = -4\). ✓

Calculation

[3 marks]

The points \(A\) and \(B\) have position vectors \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) and \(3\mathbf{i} + 4\mathbf{k}\). Find angle \(AOB\).

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 3 + 0 + 8 = 11\). ✓
\(|\overrightarrow{OA}| = 3\), \(|\overrightarrow{OB}| = 5\), so \(\cos\theta = \tfrac{11}{15}\). ✓
Angle \(AOB = 42.8^\circ\) (1 d.p.). ✓

Analysis

[5 marks]

The points \(A(1, 0, 2)\), \(B(3, 1, 0)\) and \(C(2, 4, 3)\) form a triangle. (a) Find angle \(BAC\). (b) Hence find the area of triangle \(ABC\).

(a) \(\overrightarrow{AB} = (2, 1, -2)\), \(\overrightarrow{AC} = (1, 4, 1)\); product \(2 + 4 - 2 = 4\). ✓
\(|\overrightarrow{AB}| = 3\), \(|\overrightarrow{AC}| = \sqrt{18}\), \(\cos A = \dfrac{4}{3\sqrt{18}}\), so angle \(BAC = 71.7^\circ\). ✓
(b) Area \(= \tfrac{1}{2} \times 3 \times \sqrt{18} \times \sin 71.7^\circ = 6.04\) (3 s.f.). ✓

Analysis

[5 marks]

The line \(l\) has equation \(\mathbf{r} = (1, 0, -1) + t(2, 1, 2)\) and \(P\) is the point \((6, 6, 0)\). Find the position vector of the foot of the perpendicular from \(P\) to \(l\), and hence the perpendicular distance from \(P\) to \(l\).

General point \(N = (1 + 2t, t, -1 + 2t)\), so \(\overrightarrow{PN} = (2t - 5,\ t - 6,\ 2t - 1)\). ✓
\(\overrightarrow{PN} \cdot (2, 1, 2) = 4t - 10 + t - 6 + 4t - 2 = 9t - 18 = 0\), so \(t = 2\). ✓
Foot: \(5\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\). ✓
\(\overrightarrow{PN} = (-1, -4, 3)\), distance \(\sqrt{26} = 5.10\) (3 s.f.). ✓

Analysis

[6 marks]

A pyramid \(OABCV\) has a square base \(OABC\) in the horizontal plane, with \(\overrightarrow{OA} = 6\mathbf{i}\), \(\overrightarrow{OC} = 6\mathbf{j}\), and apex \(V\) at \((3, 3, 4)\), vertically above the centre of the base.

(a) Find \(\overrightarrow{VA}\) and \(\overrightarrow{VB}\). (b) Find angle \(AVB\). (c) Show that the edge \(VA\) is perpendicular to the diagonal \(OB\).

(a) \(B = (6, 6, 0)\). \(\overrightarrow{VA} = (6, 0, 0) - (3, 3, 4) = (3, -3, -4)\), \(\overrightarrow{VB} = (3, 3, -4)\). ✓
(b) \(\overrightarrow{VA} \cdot \overrightarrow{VB} = 9 - 9 + 16 = 16\); both magnitudes are \(\sqrt{34}\). ✓
\(\cos AVB = \tfrac{16}{34}\), so angle \(AVB = 61.9^\circ\) (1 d.p.). ✓
(c) \(\overrightarrow{OB} = (6, 6, 0)\); \(\overrightarrow{VA} \cdot \overrightarrow{OB} = 18 - 18 + 0 = 0\), so \(VA\) is perpendicular to \(OB\). ✓

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.