UniGeniusGCSE Maths · Edexcel 1MA1

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Column vectors and vector arithmetic

8 exam-style questions, grades 2 to 6. Worked solutions and the marks are on the last page.

  1. Question 1Non-calculator · 4 marks

    AA is the point (1,4)(1, 4) and BB is the point (5,1)(5, 1).

    (a) Write AB→\overrightarrow{AB} as a column vector. (1)

    (b) Write BA→\overrightarrow{BA} as a column vector. (1)

    (c) AC→=2AB→\overrightarrow{AC} = 2\overrightarrow{AB}. Find the coordinates of CC. (2)

  2. Question 2Non-calculator · 4 marks

    a=(3−1)\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} and b=(−24)\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}.

    (a) Work out a+b\mathbf{a} + \mathbf{b} as a column vector. (1)

    (b) Work out 3a3\mathbf{a} as a column vector. (1)

    (c) Work out a−b\mathbf{a} - \mathbf{b} as a column vector. (2)

  3. Question 3Non-calculator · 4 marks

    Vectors a and b are (2,−3)(2,-3) and (−2,3)(-2,3).

    (a) Find 2a - b as an ordered pair of components. (2)

    (b) Find a vector c such that a + c = b. (2)

  4. Question 4Non-calculator · 4 marks

    A is (0,0)(0,0), B is (3,−3)(3,-3) and C is (1,1)(1,1).

    (a) Find vector BC. (2)

    (b) ABCD is a parallelogram in that order. Find D. (2)

  5. Question 5Non-calculator · 4 marks

    A particle is translated by (4,−3)(4,-3), then by (−2,5)(-2,5).

    (a) Find the resultant translation. (2)

    (b) Find the single translation that returns the particle to its starting point. (2)

  6. Question 6Non-calculator · 6 marks

    p=(3−2)\mathbf{p} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and q=(−14)\mathbf{q} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}. Give vector answers as (x, y).

    (a) Work out 2p+3q2\mathbf{p} + 3\mathbf{q}. (2)

    (b) Find the vector r such that p + r = q. (2)

    (c) Is 2p+3q2\mathbf{p} + 3\mathbf{q} parallel to (6, 16)? Give a reason. (2)

  7. Question 7Non-calculator · 4 marks

    a=(4−1)\mathbf{a} = \begin{pmatrix} 4 \\ -1 \end{pmatrix} and b=(−23)\mathbf{b} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}.

    (a) Work out 2a−3b2\mathbf{a} - 3\mathbf{b} as a column vector. (2)

    (b) The vector c\mathbf{c} is such that a+c=b\mathbf{a} + \mathbf{c} = \mathbf{b}. Find c\mathbf{c} as a column vector. (2)

  8. Question 8Non-calculator · 5 marks

    p(21)+q(1−3)=(70)p\begin{pmatrix} 2 \\ 1 \end{pmatrix} + q\begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix}, where pp and qq are numbers.

    (a) Find the values of pp and qq. Give your answer as (p,q)(p, q). (3)

    (b) The vector (k6)\begin{pmatrix} k \\ 6 \end{pmatrix} is parallel to (2−3)\begin{pmatrix} 2 \\ -3 \end{pmatrix}. Find the value of kk. (2)

Worked solutions and marks

Question 1

(a) (4−3)\begin{pmatrix} 4 \\ -3 \end{pmatrix}

  1. From AA to BB: 5−1=45 - 1 = 4 across and 1−4=−31 - 4 = -3 up.
  • B1 (4,−3)(4, -3) as a column vector.

(b) (−43)\begin{pmatrix} -4 \\ 3 \end{pmatrix}

  1. Going the other way reverses both signs: BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}.
  • B1 (−4,3)(-4, 3) as a column vector.

(c) (9,−2)(9, -2)

  1. 2AB→=(8−6)2\overrightarrow{AB} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}
  2. Start at A(1,4)A(1, 4): C=(1+8,4−6)=(9,−2)C = (1 + 8, 4 - 6) = (9, -2).
  • M1 Doubling AB→\overrightarrow{AB} to get (8,−6)(8, -6).
  • A1 Correct answer: (9,−2)(9, -2).

Question 2

(a) (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix}

  1. Add the top numbers and the bottom numbers: 3+(−2)=13 + (-2) = 1 and −1+4=3-1 + 4 = 3.
  • B1 (1,3)(1, 3) as a column vector.

(b) (9−3)\begin{pmatrix} 9 \\ -3 \end{pmatrix}

  1. Multiply both components by 3.
  • B1 (9,−3)(9, -3) as a column vector.

(c) (5−5)\begin{pmatrix} 5 \\ -5 \end{pmatrix}

  1. Subtract component by component.
    (3−1)−(−24)=(3−(−2)−1−4)\begin{pmatrix} 3 \\ -1 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 - (-2) \\ -1 - 4 \end{pmatrix}
  2. So a−b=(5−5)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}.
  • M1 Subtracting each component, with 3−(−2)3 - (-2) and −1−4-1 - 4 shown.
  • A1 Correct answer: (5,−5)(5, -5).

Question 3

(a) (6,−9)(6,-9)

  1. Double both components of a, then subtract the matching components of b.
    2×2−(−2)2\times 2-(-2)
  2. Therefore (6,−9)(6,-9).
  • M1 Double both components of a, then subtract the matching components of b.
  • A1 Correct answer: (6,−9)(6,-9)

(b) (−4,6)(-4,6)

  1. Rearrange to c = b - a, component by component.
    3−(−3)3-(-3)
  2. Therefore (−4,6)(-4,6).
  • M1 Rearrange to c = b - a, component by component.
  • A1 Correct answer: (−4,6)(-4,6)

Question 4

(a) (−2,4)(-2,4)

  1. Subtract B’s position from C’s position.
    (1)−(3)(1)-(3)
  2. Therefore (−2,4)(-2,4).
  • M1 Subtract B’s position from C’s position.
  • A1 Correct answer: (−2,4)(-2,4)

(b) (−2,4)(-2,4)

  1. The displacement AD equals BC.
    y=4y=4
  2. Therefore (−2,4)(-2,4).
  • M1 The displacement AD equals BC.
  • A1 Correct answer: (−2,4)(-2,4)

Question 5

(a) (2,2)(2,2)

  1. Add matching components of the two vectors.
    4−24-2
  2. Therefore (2,2)(2,2).
  • M1 Add matching components of the two vectors.
  • A1 Correct answer: (2,2)(2,2)

(b) (−2,−2)(-2,-2)

  1. Negate both components of the resultant.
    −(2)-(2)
  2. Therefore (−2,−2)(-2,-2).
  • M1 Negate both components of the resultant.
  • A1 Correct answer: (−2,−2)(-2,-2)

Question 6

(a) (3,8)(3,8)

  1. Multiply each vector by its number, then add the components.
    2×3+3×(−1)2\times 3+3\times (-1)
  2. Therefore (3,8)(3,8).
  • M1 Multiply each vector by its number, then add the components.
  • A1 Correct answer: (3,8)(3,8)

(b) (−4,6)(-4,6)

  1. Subtract p from q.
    −1−3-1-3
  2. Therefore (−4,6)(-4,6).
  • M1 Subtract p from q.
  • A1 Correct answer: (−4,6)(-4,6)

(c) Yes: (6, 16) = 2 × (3, 8), so it is a multiple of the same vector.

  1. Compare the ratios of the components.
    6/3=16/86/3=16/8
  2. Yes: (6, 16) = 2 × (3, 8), so it is a multiple of the same vector.
  • M1 Compare the ratios of the components.
  • C1 Correct conclusion with supporting reasoning: Yes: (6, 16) = 2 × (3, 8), so it is a multiple of the same vector.

Question 7

(a) (14−11)\begin{pmatrix} 14 \\ -11 \end{pmatrix}

  1. 2(4−1)−3(−23)=(8−2)−(−69)2\begin{pmatrix} 4 \\ -1 \end{pmatrix} - 3\begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ -2 \end{pmatrix} - \begin{pmatrix} -6 \\ 9 \end{pmatrix}
  2. =(8+6−2−9)=(14−11)= \begin{pmatrix} 8 + 6 \\ -2 - 9 \end{pmatrix} = \begin{pmatrix} 14 \\ -11 \end{pmatrix}
  • M1 2a=(8,−2)2\mathbf{a} = (8, -2) and 3b=(−6,9)3\mathbf{b} = (-6, 9), or one correct component of the answer.
  • A1 Correct answer: (14,−11)(14, -11).

(b) (−64)\begin{pmatrix} -6 \\ 4 \end{pmatrix}

  1. Rearrange first.
    c=b−a\mathbf{c} = \mathbf{b} - \mathbf{a}
  2. c=(−2−43−(−1))=(−64)\mathbf{c} = \begin{pmatrix} -2 - 4 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} -6 \\ 4 \end{pmatrix}
  • P1 Rearranging to c=b−a\mathbf{c} = \mathbf{b} - \mathbf{a}.
  • A1 Correct answer: (−6,4)(-6, 4).

Question 8

(a) p=3p = 3, q=1q = 1

  1. Top components and bottom components give two equations.
    2p+q=7,p−3q=02p + q = 7, \qquad p - 3q = 0
  2. From the second, p=3qp = 3q.
  3. Substitute.
    6q+q=7  ⇒  q=1,  p=36q + q = 7 \;\Rightarrow\; q = 1,\; p = 3
  • P1 Writing both component equations.
  • P1 Solving the pair to find one of the values.
  • A1 p=3p = 3 and q=1q = 1.

(b) k=−4k = -4

  1. Parallel vectors are multiples of each other. The bottom number −3-3 must be multiplied by −2-2 to give 6.
  2. So the top is 2×(−2)=−42 \times (-2) = -4.
  • M1 Finding the multiplier −2-2 (from 6÷(−3)6 \div (-3)).
  • A1 Correct answer: k=−4k = -4.

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