IGCSE · Practice questions · Vector notation, magnitude and unit vectors

Component by component, every time.

Six original Cambridge-style questions on two-dimensional vectors: column and \(\mathbf{i}\), \(\mathbf{j}\) forms, resultants, magnitudes, unit vectors, vectors of a given magnitude, \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), parallel vectors and equating components, all without a calculator.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Split into components, then rebuild.

Convert: \(x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}\); work with whichever form the question uses and answer in the same form.
Combine: add, subtract and scale the \(\mathbf{i}\) and \(\mathbf{j}\) parts separately; a resultant is the sum of all the vectors.
Measure: \(|x\mathbf{i} + y\mathbf{j}| = \sqrt{x^{2} + y^{2}}\); unit vector \(\dfrac{\mathbf{a}}{|\mathbf{a}|}\); magnitude \(k\) along \(\mathbf{a}\) is \(\dfrac{k}{|\mathbf{a}|}\mathbf{a}\).
Between points: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). Parallel: one vector is \(k\) times the other, with the same \(k\) for both components.
Unknowns: collect the \(\mathbf{i}\) and \(\mathbf{j}\) terms, equate each component, and solve the pair of equations.

Recall

[2 marks]

The vector \(\mathbf{p} = 7\mathbf{i} - 24\mathbf{j}\).

(a) Write \(\mathbf{p}\) as a column vector.

(b) Find \(|\mathbf{p}|\).

(a) \(\mathbf{p} = \begin{pmatrix} 7 \\ -24 \end{pmatrix}\). ✓
(b) \(|\mathbf{p}| = \sqrt{49 + 576} = \sqrt{625} = 25\). ✓

Calculation

[3 marks]

\(\mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\) and \(\mathbf{c} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}\).

Find \(2\mathbf{a} - \mathbf{b} + \mathbf{c}\) and its magnitude, giving the magnitude in the form \(k\sqrt{41}\).

\(2\mathbf{a} - \mathbf{b} + \mathbf{c} = \begin{pmatrix} 4 + 1 + 5 \\ -6 - 4 + 2 \end{pmatrix} = \begin{pmatrix} 10 \\ -8 \end{pmatrix}\). ✓
Magnitude: \(\sqrt{100 + 64} = \sqrt{164}\). ✓
\(\sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41}\). ✓

Calculation

[3 marks]

(a) Find the unit vector in the direction of \(8\mathbf{i} - 6\mathbf{j}\).

(b) Hence find the vector of magnitude 25 in the direction of \(8\mathbf{i} - 6\mathbf{j}\).

(a) \(|8\mathbf{i} - 6\mathbf{j}| = \sqrt{64 + 36} = 10\), so the unit vector is \(\dfrac{1}{10}(8\mathbf{i} - 6\mathbf{j}) = 0.8\mathbf{i} - 0.6\mathbf{j}\). ✓
(b) \(25(0.8\mathbf{i} - 0.6\mathbf{j}) = 20\mathbf{i} - 15\mathbf{j}\). ✓
Check: \(\sqrt{400 + 225} = \sqrt{625} = 25\). ✓

Calculation

[4 marks]

Relative to an origin \(O\), the points \(A\), \(B\) and \(C\) have position vectors \(\overrightarrow{OA} = 3\mathbf{i} + 2\mathbf{j}\), \(\overrightarrow{OB} = 7\mathbf{i} - \mathbf{j}\) and \(\overrightarrow{OC} = -5\mathbf{i} + 8\mathbf{j}\).

(a) Find \(\overrightarrow{AB}\) and \(|\overrightarrow{AB}|\).

(b) Find \(\overrightarrow{AC}\) and show that it is parallel to \(\overrightarrow{AB}\).

(a) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 4\mathbf{i} - 3\mathbf{j}\). ✓
\(|\overrightarrow{AB}| = \sqrt{16 + 9} = 5\). ✓
(b) \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = -8\mathbf{i} + 6\mathbf{j}\). ✓
\(-8\mathbf{i} + 6\mathbf{j} = -2(4\mathbf{i} - 3\mathbf{j}) = -2\overrightarrow{AB}\), one multiplier for both components, so \(\overrightarrow{AC}\) is parallel to \(\overrightarrow{AB}\) (opposite direction, twice the length). ✓

Analysis

[5 marks]

(a) Find the constants \(\lambda\) and \(\mu\) such that \(\lambda(3\mathbf{i} + \mathbf{j}) + \mu(-\mathbf{i} + 2\mathbf{j}) = 4\mathbf{i} + 13\mathbf{j}\).

(b) Find the unit vector in the direction of \(4\mathbf{i} + 13\mathbf{j}\).

(a) Collect terms: \((3\lambda - \mu)\mathbf{i} + (\lambda + 2\mu)\mathbf{j} = 4\mathbf{i} + 13\mathbf{j}\). ✓
Equate components: \(3\lambda - \mu = 4\) and \(\lambda + 2\mu = 13\). ✓
From the first, \(\mu = 3\lambda - 4\). Substitute: \(\lambda + 6\lambda - 8 = 13\), so \(\lambda = 3\) and \(\mu = 5\). ✓
Check: \(3(3\mathbf{i} + \mathbf{j}) + 5(-\mathbf{i} + 2\mathbf{j}) = 9\mathbf{i} + 3\mathbf{j} - 5\mathbf{i} + 10\mathbf{j} = 4\mathbf{i} + 13\mathbf{j}\). ✓
(b) \(|4\mathbf{i} + 13\mathbf{j}| = \sqrt{16 + 169} = \sqrt{185}\), so the unit vector is \(\dfrac{1}{\sqrt{185}}(4\mathbf{i} + 13\mathbf{j})\). ✓

Analysis

[6 marks]

The vector \(\mathbf{v} = p\mathbf{i} + (p + 7)\mathbf{j}\), where \(p\) is a constant, has magnitude 13.

(a) Find the two possible values of \(p\).

(b) Given that \(p > 0\), find the unit vector in the direction of \(\mathbf{v}\), and the vector of magnitude 6.5 in the direction opposite to \(\mathbf{v}\).

(a) \(p^{2} + (p + 7)^{2} = 169\), so \(2p^{2} + 14p + 49 = 169\). ✓
\(2p^{2} + 14p - 120 = 0\), so \(p^{2} + 7p - 60 = 0\) and \((p + 12)(p - 5) = 0\). ✓
\(p = 5\) or \(p = -12\). Check: \(5\mathbf{i} + 12\mathbf{j}\) and \(-12\mathbf{i} - 5\mathbf{j}\) both have length \(\sqrt{169} = 13\). ✓
(b) With \(p = 5\), \(\mathbf{v} = 5\mathbf{i} + 12\mathbf{j}\) and the unit vector is \(\dfrac{1}{13}(5\mathbf{i} + 12\mathbf{j})\). ✓
Opposite direction, magnitude 6.5: \(-\dfrac{6.5}{13}(5\mathbf{i} + 12\mathbf{j}) = -\dfrac{1}{2}(5\mathbf{i} + 12\mathbf{j}) = -2.5\mathbf{i} - 6\mathbf{j}\). ✓

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.