IGCSE · Practice questions · Magnitude of a vector

Square, add, root.

Extended. Six original Cambridge-style questions on the magnitude of a vector, ending with a two-leg drone flight.

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

Find the column vector, then Pythagoras.

Column vector first: for \(\overrightarrow{AB}\) or \(2\mathbf{a} - \mathbf{b}\), work out the single column vector.
Magnitude: \(\sqrt{x^2 + y^2}\), squaring negatives with brackets.
Exact or rounded: a surd such as \(\sqrt{50} = 5\sqrt{2}\), or 3 significant figures.
Unknowns: set \(x^2 + y^2\) equal to the magnitude squared; remember \(\pm\).
In context: the magnitude of a velocity is a speed.

Recall

[2 marks]

Find the magnitude of (a) \(\begin{pmatrix}-7\\24\end{pmatrix}\) (b) \(\begin{pmatrix}5\\-5\end{pmatrix}\), giving (b) correct to 3 significant figures.

(a) \(\sqrt{49 + 576} = \sqrt{625} = 25\) ✓
(b) \(\sqrt{25 + 25} = \sqrt{50} = 7.07\) (3 s.f.) ✓

Calculation

[2 marks]

\(A\) is the point \((1, -2)\) and \(B\) is the point \((7, 6)\). Find \(|\overrightarrow{AB}|\).

\(\overrightarrow{AB} = \begin{pmatrix}6\\8\end{pmatrix}\). ✓
\(|\overrightarrow{AB}| = \sqrt{36 + 64} = 10\). ✓

Calculation

[3 marks]

\(\mathbf{a} = \begin{pmatrix}2\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-5\\1\end{pmatrix}\). Find \(|2\mathbf{a} + \mathbf{b}|\), giving your answer as an exact surd in its simplest form.

\(2\mathbf{a} + \mathbf{b} = \begin{pmatrix}4 - 5\\6 + 1\end{pmatrix} = \begin{pmatrix}-1\\7\end{pmatrix}\). ✓
\(|2\mathbf{a} + \mathbf{b}| = \sqrt{1 + 49} = \sqrt{50}\). ✓
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\). ✓

Analysis

[3 marks]

\(\left|\begin{pmatrix}k\\2k\end{pmatrix}\right| = 10\) and \(k > 0\). Find \(k\), giving your answer as an exact surd.

\(k^2 + (2k)^2 = 100\), so \(5k^2 = 100\). ✓
\(k^2 = 20\), so \(k = \sqrt{20} = 2\sqrt{5}\) (taking \(k > 0\)). ✓

Proof

[3 marks]

\(\mathbf{a} = \begin{pmatrix}4\\-3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}0\\5\end{pmatrix}\).

(a) Show that \(|\mathbf{a}| = |\mathbf{b}|\) but \(\mathbf{a} \ne \mathbf{b}\).

(b) Find \(|\mathbf{a} - \mathbf{b}|\) as an exact surd.

(a) \(|\mathbf{a}| = \sqrt{16 + 9} = 5\) and \(|\mathbf{b}| = \sqrt{0 + 25} = 5\). Equal lengths, but the components differ, so the directions differ and \(\mathbf{a} \ne \mathbf{b}\). ✓
(b) \(\mathbf{a} - \mathbf{b} = \begin{pmatrix}4\\-8\end{pmatrix}\), so \(|\mathbf{a} - \mathbf{b}| = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\). ✓

Analysis

[4 marks]

A drone flies from its base by the vector \(\begin{pmatrix}30\\40\end{pmatrix}\) metres, then by \(\begin{pmatrix}-50\\80\end{pmatrix}\) metres.

(a) Find the total distance the drone flies.

(b) Find how far the drone finishes from its base.

Give your answers correct to 3 significant figures.

(a) First leg: \(\sqrt{900 + 1600} = 50\) m. Second leg: \(\sqrt{2500 + 6400} = \sqrt{8900} = 94.33\ldots\) m. ✓
Total \(= 50 + 94.33\ldots = 144\) m (3 s.f.). ✓
(b) Overall vector \(\begin{pmatrix}-20\\120\end{pmatrix}\), so the distance is \(\sqrt{400 + 14\,400} = \sqrt{14\,800} = 122\) 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.