IGCSE · Practice questions · Vectors in two dimensions

Across, then up.

Extended. Six original Cambridge-style questions on column vectors: writing, combining and scaling them, ending with a boat crossing a current.

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

Top with top, bottom with bottom.

Between points: \(\overrightarrow{AB}\) = end minus start.
Add or subtract: combine the top numbers, then the bottom numbers.
Scalar: \(k\mathbf{a}\) multiplies both numbers by \(k\).
Parallel: one vector is a number times the other.
Unknowns: make one equation from the tops and one from the bottoms.

Recall

[2 marks]

\(A\) is the point \((-3, 2)\) and \(B\) is the point \((1, -4)\). Write as column vectors (a) \(\overrightarrow{AB}\) (b) \(\overrightarrow{BA}\).

(a) \(\begin{pmatrix}1 - (-3)\\-4 - 2\end{pmatrix} = \begin{pmatrix}4\\-6\end{pmatrix}\) ✓
(b) \(\begin{pmatrix}-4\\6\end{pmatrix}\) ✓

Calculation

[2 marks]

\(\mathbf{a} = \begin{pmatrix}5\\-2\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-1\\3\end{pmatrix}\). Find (a) \(\mathbf{a} + \mathbf{b}\) (b) \(\mathbf{a} - \mathbf{b}\).

(a) \(\begin{pmatrix}4\\1\end{pmatrix}\) ✓
(b) \(\begin{pmatrix}5 - (-1)\\-2 - 3\end{pmatrix} = \begin{pmatrix}6\\-5\end{pmatrix}\) ✓

Calculation

[2 marks]

With \(\mathbf{a}\) and \(\mathbf{b}\) as in the previous question, find \(3\mathbf{a} - 2\mathbf{b}\).

\(3\mathbf{a} = \begin{pmatrix}15\\-6\end{pmatrix}\), \(2\mathbf{b} = \begin{pmatrix}-2\\6\end{pmatrix}\). ✓
\(3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix}17\\-12\end{pmatrix}\). ✓

Analysis

[3 marks]

Find \(p\) and \(q\) when \(p\begin{pmatrix}3\\-1\end{pmatrix} + q\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}7\\7\end{pmatrix}\).

Tops: \(3p + q = 7\). Bottoms: \(-p + 2q = 7\). ✓
From the first, \(q = 7 - 3p\). Substitute: \(-p + 14 - 6p = 7\), so \(7p = 7\) and \(p = 1\). ✓
\(q = 7 - 3 = 4\). Check: \(\begin{pmatrix}3 + 4\\-1 + 8\end{pmatrix} = \begin{pmatrix}7\\7\end{pmatrix}\). ✓

Proof

[2 marks]

Show that the vectors \(\begin{pmatrix}10\\-4\end{pmatrix}\) and \(\begin{pmatrix}-15\\6\end{pmatrix}\) are parallel, and state whether they point the same way.

\(-15 = -1.5 \times 10\) and \(6 = -1.5 \times (-4)\), so \(\begin{pmatrix}-15\\6\end{pmatrix} = -1.5\begin{pmatrix}10\\-4\end{pmatrix}\). ✓
They are parallel; the multiplier is negative, so they point in opposite directions. ✓

Analysis

[4 marks]

A boat would move by \(\begin{pmatrix}5\\2\end{pmatrix}\) km each hour in still water. A current moves it by \(\begin{pmatrix}-1\\2\end{pmatrix}\) km each hour. The boat starts at the origin.

(a) Find the boat's actual movement each hour as a column vector.

(b) Find its position after 3 hours.

(a) \(\begin{pmatrix}5\\2\end{pmatrix} + \begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}4\\4\end{pmatrix}\). ✓
(b) \(3\begin{pmatrix}4\\4\end{pmatrix} = \begin{pmatrix}12\\12\end{pmatrix}\), so it is at \((12, 12)\). ✓
(c) \(\begin{pmatrix}20\\20\end{pmatrix} = 5\begin{pmatrix}4\\4\end{pmatrix}\), so 5 hours. ✓

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.