AS & A Level · Practice questions · Vectors in two and three dimensions

Three components, no fear.

Six original Cambridge-style questions on vectors in two and three dimensions: notation, magnitude, unit vectors, parallel and collinear points, and position vectors in context.

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

End minus start, then Pythagoras.

Displacement: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
Magnitude: \(\sqrt{x^{2} + y^{2} + z^{2}}\); unit vector = vector ÷ magnitude.
Parallel: one vector is a single multiple \(\lambda\) of the other.
Collinear: \(\overrightarrow{AB}\) parallel to \(\overrightarrow{BC}\) with \(B\) shared.
New points: known point + (fraction of a displacement) or + (distance × unit vector).

Recall

[2 marks]

The points \(A\) and \(B\) have position vectors \(3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\) and \(-\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\). Find \(\overrightarrow{AB}\) and the exact length of \(AB\).

\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = -4\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\). ✓
\(AB = \sqrt{16 + 36 + 4} = \sqrt{56} = 2\sqrt{14}\). ✓

Recall

[2 marks]

Find the unit vector in the direction of \(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\).

Magnitude \(\sqrt{36 + 4 + 9} = \sqrt{49} = 7\). ✓
Unit vector \(\tfrac{1}{7}(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})\). ✓

Calculation

[3 marks]

The vectors \(p\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) and \(2\mathbf{i} + q\mathbf{j} + 4\mathbf{k}\) are parallel. Find \(p\) and \(q\).

\((p, 3, -6) = \lambda(2, q, 4)\). ✓
\(\mathbf{k}\): \(-6 = 4\lambda\), so \(\lambda = -\tfrac{3}{2}\). ✓
\(\mathbf{i}\): \(p = 2\lambda = -3\). \(\mathbf{j}\): \(3 = -\tfrac{3}{2}q\), so \(q = -2\). ✓

Proof

[3 marks]

The points \(P\), \(Q\) and \(R\) have coordinates \((2, -1, 0)\), \((5, 1, -4)\) and \((11, 5, -12)\). Show that \(P\), \(Q\) and \(R\) are collinear and find the ratio \(PQ : QR\).

\(\overrightarrow{PQ} = (3, 2, -4)\). ✓
\(\overrightarrow{QR} = (6, 4, -8) = 2\overrightarrow{PQ}\), so \(\overrightarrow{PQ}\) and \(\overrightarrow{QR}\) are parallel. ✓
They share the point \(Q\), so \(P\), \(Q\), \(R\) are collinear, and \(PQ : QR = 1 : 2\). ✓

Analysis

[4 marks]

\(OABC\) is a parallelogram, where \(O\) is the origin, \(\overrightarrow{OA} = 4\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) and \(\overrightarrow{OB} = 7\mathbf{i} + 5\mathbf{j}\).

(a) Find \(\overrightarrow{OC}\). (b) Find the exact lengths of the diagonals \(OB\) and \(AC\).

(a) \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\). ✓
(b) \(OB = \sqrt{49 + 25 + 0} = \sqrt{74}\). ✓
\(\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (3-4,\ 4-1,\ 2-(-2)) = (-1, 3, 4)\), so \(AC = \sqrt{1 + 9 + 16} = \sqrt{26}\). ✓

Analysis

[5 marks]

Relative to a fixed origin \(O\), with distances in metres, a drone starts at the point \(A\) with position vector \(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\). It flies in a straight line in the direction of \(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) at a constant speed of \(12\ \text{m s}^{-1}\) for 4 seconds, arriving at \(P\).

(a) Find the position vector of \(P\). (b) Find the distance \(OP\), giving your answer to 3 significant figures.

Distance flown \(= 12 \times 4 = 48\) m. ✓
\(|2\mathbf{i} + 2\mathbf{j} + \mathbf{k}| = 3\), so the unit direction is \(\tfrac{1}{3}(2, 2, 1)\). ✓
Displacement \(= 48 \times \tfrac{1}{3}(2, 2, 1) = (32, 32, 16)\). ✓
(a) \(\overrightarrow{OP} = (2 + 32,\ -1 + 32,\ 3 + 16) = 34\mathbf{i} + 31\mathbf{j} + 19\mathbf{k}\). ✓
(b) \(OP = \sqrt{34^{2} + 31^{2} + 19^{2}} = \sqrt{2478} = 49.8\) m (3 s.f.). ✓

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.