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 Calculate [2 marks]
Calculate \(\begin{pmatrix} 4 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix}\). [2 marks]
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Model answer
\(\begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}\).
Mark scheme
- Subtracts the top numbers or the bottom numbers — M1
- \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\) — A1
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2 Calculate [4 marks]
The vectors \(\mathbf{m}\) and \(\mathbf{n}\) are drawn on the grid. (a) Write \(\mathbf{m}\) and \(\mathbf{n}\) as column vectors. [2 marks] (b) Calculate \(\mathbf{m} + 2\mathbf{n}\). [2 marks]
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Model answer
(a) \(\mathbf{m}\) goes 3 right and 1 up, so \(\mathbf{m} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\). \(\mathbf{n}\) goes 2 left and 3 down, so \(\mathbf{n} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\). (b) \(2\mathbf{n} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}\), so \(\mathbf{m} + 2\mathbf{n} = \begin{pmatrix} -1 \\ -5 \end{pmatrix}\).
Mark scheme
- (a) \(\mathbf{m} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\) — B1
- (a) \(\mathbf{n} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\) — B1
- (b) \(2\mathbf{n} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}\) or a correct method — M1
- (b) \(\begin{pmatrix} -1 \\ -5 \end{pmatrix}\) — A1 (follow through from (a))
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3 Calculate [3 marks]
\(A\) is the point \((5, -2)\) and \(B\) is the point \((-1, 4)\). (a) Calculate \(\overrightarrow{AB}\) as a column vector. [2 marks] (b) Write down \(\overrightarrow{BA}\) as a column vector. [1 mark]
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Model answer
(a) \((-1 - 5, 4 - (-2)) = (-6, 6)\), so \(\overrightarrow{AB} = \begin{pmatrix} -6 \\ 6 \end{pmatrix}\). (b) \(\overrightarrow{BA} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}\).
Mark scheme
- (a) \(-1 - 5\) or \(4 - (-2)\) — M1
- (a) \(\begin{pmatrix} -6 \\ 6 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 6 \\ -6 \end{pmatrix}\) — B1 (follow through from (a))
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4 Calculate [3 marks]
\(\mathbf{u} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}\). (a) Calculate \(3\mathbf{u} - \mathbf{v}\). [2 marks] (b) Calculate \(2\mathbf{u} + \mathbf{v}\). [1 mark]
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Model answer
(a) \(3\mathbf{u} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}\), so \(3\mathbf{u} - \mathbf{v} = \begin{pmatrix} 0 \\ -11 \end{pmatrix}\). (b) \(2\mathbf{u} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}\), so \(2\mathbf{u} + \mathbf{v} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\).
Mark scheme
- (a) \(3\mathbf{u} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}\) or a correct method — M1
- (a) \(\begin{pmatrix} 0 \\ -11 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 5 \\ 1 \end{pmatrix}\) — B1
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5 Show that [2 marks]
Show that the vectors \(\begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) are parallel. [2 marks]
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Model answer
\(\begin{pmatrix} -4 \\ 6 \end{pmatrix} = -2 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\). One vector is a multiple of the other, so they are parallel.
Mark scheme
- \(-2 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) — M1
- A multiple, so parallel — A1
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6 Calculate [3 marks]
(a) Calculate the length of the vector \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\). [2 marks] (b) Calculate the length of the vector \(\begin{pmatrix} 8 \\ 6 \end{pmatrix}\). [1 mark]
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Model answer
(a) \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\). (b) \(\sqrt{8^2 + 6^2} = \sqrt{100} = 10\).
Mark scheme
- (a) \(\sqrt{5^2 + 12^2}\) or \(\sqrt{169}\) — M1
- (a) 13 — A1
- (b) 10 — B1
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 \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\).