Exam questions · Maths · Vectors, Constructions and Loci
Column Vectors and Vector Arithmetic
- 6 exam questions
- 17 marks
- 9 quick checks
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1 Work out [2 marks]
Work out \(\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \end{pmatrix}\). [2 marks]
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Model answer
\(\begin{pmatrix} 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}\).
Mark scheme
- Adds the top numbers or the bottom numbers — M1
- \(\begin{pmatrix} 5 \\ 5 \end{pmatrix}\) — A1
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2 Work out [4 marks]
The vectors \(\mathbf{r}\) and \(\mathbf{s}\) are drawn on the grid. (a) Write \(\mathbf{r}\) and \(\mathbf{s}\) as column vectors. [2 marks] (b) Work out \(3\mathbf{r} - \mathbf{s}\). [2 marks]
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Model answer
(a) \(\mathbf{r}\) goes 2 right and 4 up, so \(\mathbf{r} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\). \(\mathbf{s}\) goes 3 right and 1 down, so \(\mathbf{s} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\). (b) \(3\mathbf{r} = \begin{pmatrix} 6 \\ 12 \end{pmatrix}\), so \(3\mathbf{r} - \mathbf{s} = \begin{pmatrix} 3 \\ 13 \end{pmatrix}\).
Mark scheme
- (a) \(\mathbf{r} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\) — B1
- (a) \(\mathbf{s} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\) — B1
- (b) \(3\mathbf{r} = \begin{pmatrix} 6 \\ 12 \end{pmatrix}\) or a correct method — M1
- (b) \(\begin{pmatrix} 3 \\ 13 \end{pmatrix}\) — A1 (follow through from (a))
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3 Work out [3 marks]
\(P\) is the point \((2, -1)\) and \(Q\) is the point \((-3, 4)\). (a) Write \(\overrightarrow{PQ}\) as a column vector. [2 marks] (b) Write \(\overrightarrow{QP}\) as a column vector. [1 mark]
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Model answer
(a) \((-3 - 2, 4 - (-1)) = (-5, 5)\), so \(\overrightarrow{PQ} = \begin{pmatrix} -5 \\ 5 \end{pmatrix}\). (b) \(\overrightarrow{QP} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}\).
Mark scheme
- (a) \(-3 - 2\) or \(4 - (-1)\) — M1
- (a) \(\begin{pmatrix} -5 \\ 5 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 5 \\ -5 \end{pmatrix}\) — B1 (follow through from (a))
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4 Work out [3 marks]
\(\mathbf{a} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}\). (a) Work out \(2\mathbf{a} - \mathbf{b}\). [2 marks] (b) Work out \(\mathbf{a} + \mathbf{b}\). [1 mark]
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Model answer
(a) \(2\mathbf{a} = \begin{pmatrix} -4 \\ 10 \end{pmatrix}\), so \(2\mathbf{a} - \mathbf{b} = \begin{pmatrix} -8 \\ 9 \end{pmatrix}\). (b) \(\begin{pmatrix} 2 \\ 6 \end{pmatrix}\).
Mark scheme
- (a) \(2\mathbf{a} = \begin{pmatrix} -4 \\ 10 \end{pmatrix}\) or a correct method — M1
- (a) \(\begin{pmatrix} -8 \\ 9 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 2 \\ 6 \end{pmatrix}\) — B1
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5 Write down [2 marks]
Write down a vector that is parallel to \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) and four times as long. [2 marks]
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Model answer
Multiply by 4: \(4 \times \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ -12 \end{pmatrix}\).
Mark scheme
- Multiplies both numbers by 4 — M1
- \(\begin{pmatrix} 4 \\ -12 \end{pmatrix}\) — A1
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6 Show that [3 marks]
\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\). Show that \(A\), \(B\) and \(C\) lie on a straight line. [3 marks]
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Model answer
\(\overrightarrow{BC} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} = 2 \times \begin{pmatrix} 2 \\ 3 \end{pmatrix} = 2\overrightarrow{AB}\). The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.
Mark scheme
- \(\overrightarrow{BC} = 2\overrightarrow{AB}\) — M1
- Parallel — A1
- Common point \(B\), so collinear — Q1
Quick check
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1
What does the column vector \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\) mean?
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B: 2 left and 5 up
The top number is the horizontal move, and the bottom number is the vertical move.
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2
\(A\) is \((2, 5)\) and \(B\) is \((6, 2)\). What is \(\overrightarrow{AB}\)?
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A: \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\)
Subtract the start from the end: \((6 - 2, 2 - 5) = (4, -3)\).
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3
What is \(\begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)?
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D: \(\begin{pmatrix} 4 \\ 6 \end{pmatrix}\)
Add the top numbers and add the bottom numbers.
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4
What is \(\begin{pmatrix} 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)?
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C: \(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\)
\((3 - 1, 2 - 4) = (2, -2)\).
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5
What is \(3\begin{pmatrix} 2 \\ -1 \end{pmatrix}\)?
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B: \(\begin{pmatrix} 6 \\ -3 \end{pmatrix}\)
Multiply both numbers by 3.
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6
\(\mathbf{p} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\). What is \(2\mathbf{p} - \mathbf{q}\)?
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A: \(\begin{pmatrix} 5 \\ 2 \end{pmatrix}\)
\(2\mathbf{p} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\), then \((4 - (-1), 6 - 4) = (5, 2)\).
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7
Which vector is parallel to \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\)?
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D: \(\begin{pmatrix} 6 \\ 9 \end{pmatrix}\)
\(\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 3 \end{pmatrix}\), so it is parallel.
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8
\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\). What is \(\overrightarrow{BA}\)?
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C: \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\)
\(\overrightarrow{BA} = -\overrightarrow{AB}\), so both signs change.
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9
What is the length of the vector \(\begin{pmatrix} 3 \\ 4 \end{pmatrix}\)?
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B: \(5\)
The length is \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\).