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Exam questions · Maths · Vectors, Constructions and Loci

Vector Geometry and Proof

  • 6 exam questions
  • 18 marks
  • 9 quick checks
  1. 1 Write down [2 marks]

    In the triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). Write down (a) \(\overrightarrow{BO}\), (b) \(\overrightarrow{BA}\). [2 marks]

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    Model answer

    (a) \(\overrightarrow{BO} = -\mathbf{b}\). (b) \(\overrightarrow{BA} = \overrightarrow{BO} + \overrightarrow{OA} = -\mathbf{b} + \mathbf{a} = \mathbf{a} - \mathbf{b}\).

    Mark scheme

    • (a) \(-\mathbf{b}\) — B1
    • (b) \(\mathbf{a} - \mathbf{b}\) — B1
  2. 2 Calculate [4 marks]

    \(OAB\) is a triangle with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) is the point on \(AB\) such that \(AP : PB = 2 : 1\). (a) Write down \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [1 mark] (b) Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Give your answer in its simplest form. [3 marks]

    Triangle OAB with OA equal to a and OB equal to b, and the point P on AB with AP : PB equal to 2 : 1.
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    Model answer

    (a) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). (b) \(AP = \dfrac{2}{3}AB\), so \(\overrightarrow{AP} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\). Then \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\).

    Mark scheme

    • (a) \(\mathbf{b} - \mathbf{a}\) — B1
    • (b) \(\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB}\) — M1
    • (b) \(\mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\) — M1
    • (b) \(\dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\) — A1
  3. 3 Prove [3 marks]

    \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 4\mathbf{b} - 3\mathbf{a}\). Prove that \(A\), \(B\) and \(C\) lie on a straight line. [3 marks]

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    Model answer

    \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BC} = -\mathbf{b} + 4\mathbf{b} - 3\mathbf{a} = 3\mathbf{b} - 3\mathbf{a} = 3\overrightarrow{AB}\). The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.

    Mark scheme

    • \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) — M1
    • \(\overrightarrow{BC} = 3\mathbf{b} - 3\mathbf{a}\) or \(3\overrightarrow{AB}\) — M1
    • Parallel with a common point, so collinear — A1
  4. 4 Find [3 marks]

    \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(Q\) is the point on \(OA\) such that \(OQ : QA = 1 : 3\). Find \(\overrightarrow{BQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [3 marks]

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    Model answer

    \(OQ = \dfrac{1}{4}OA\), so \(\overrightarrow{OQ} = \dfrac{1}{4}\mathbf{a}\). Then \(\overrightarrow{BQ} = \overrightarrow{BO} + \overrightarrow{OQ} = -\mathbf{b} + \dfrac{1}{4}\mathbf{a} = \dfrac{1}{4}\mathbf{a} - \mathbf{b}\).

    Mark scheme

    • \(\overrightarrow{OQ} = \dfrac{1}{4}\mathbf{a}\) — M1
    • \(\overrightarrow{BQ} = \overrightarrow{BO} + \overrightarrow{OQ}\) or \(-\mathbf{b} + \dfrac{1}{4}\mathbf{a}\) — M1
    • \(\dfrac{1}{4}\mathbf{a} - \mathbf{b}\) — A1
  5. 5 Show that [3 marks]

    \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(X\) is the point such that \(\overrightarrow{OX} = 2\mathbf{a}\), and \(Y\) is the point such that \(\overrightarrow{OY} = 2\mathbf{b}\). Show that \(XY\) is parallel to \(AB\). [3 marks]

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    Model answer

    \(\overrightarrow{XY} = -2\mathbf{a} + 2\mathbf{b} = 2(\mathbf{b} - \mathbf{a}) = 2\overrightarrow{AB}\). It is a multiple of \(\overrightarrow{AB}\), so the lines are parallel.

    Mark scheme

    • \(\overrightarrow{XY} = 2\mathbf{b} - 2\mathbf{a}\) — M1
    • \(2\overrightarrow{AB}\) or \(2(\mathbf{b} - \mathbf{a})\) — M1
    • A multiple, so parallel — A1
  6. 6 Calculate [3 marks]

    \(\overrightarrow{PQ} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\) and \(\overrightarrow{QR} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\). (a) Show that \(P\), \(Q\) and \(R\) lie on a straight line. [2 marks] (b) Calculate the length of \(PR\). [1 mark]

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    Model answer

    (a) \(\begin{pmatrix} 6 \\ 8 \end{pmatrix} = 2 \times \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), so \(\overrightarrow{PQ} = 2\overrightarrow{QR}\). The vectors are parallel and share \(Q\), so the points are collinear. (b) \(\overrightarrow{PR} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}\), and the length is \(\sqrt{81 + 144} = \sqrt{225} = 15\).

    Mark scheme

    • (a) \(\overrightarrow{PQ} = 2\overrightarrow{QR}\) — M1
    • (a) Parallel with a common point, so collinear — A1
    • (b) 15 — B1

Quick check

  1. 1

    \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). What is \(\overrightarrow{AB}\)?

    1. A\(\mathbf{a} - \mathbf{b}\)
    2. B\(\mathbf{a} + \mathbf{b}\)
    3. C\(\mathbf{b} - \mathbf{a}\)
    4. D\(-\mathbf{a} - \mathbf{b}\)
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    C: \(\mathbf{b} - \mathbf{a}\)

    Go from \(A\) to \(O\) (\(-\mathbf{a}\)) and then to \(B\) (\(\mathbf{b}\)).

  2. 2

    \(\overrightarrow{OA} = \mathbf{a}\). What is \(\overrightarrow{AO}\)?

    1. A\(\mathbf{a}\)
    2. B\(-\mathbf{a}\)
    3. C\(2\mathbf{a}\)
    4. D\(\dfrac{1}{2}\mathbf{a}\)
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    B: \(-\mathbf{a}\)

    Going backwards reverses the vector.

  3. 3

    \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). What is \(\overrightarrow{OM}\)?

    1. A\(\dfrac{1}{2}(\mathbf{a} + \mathbf{b})\)
    2. B\(\dfrac{1}{2}(\mathbf{b} - \mathbf{a})\)
    3. C\(\mathbf{a} + \mathbf{b}\)
    4. D\(\dfrac{1}{2}\mathbf{a}\)
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    A: \(\dfrac{1}{2}(\mathbf{a} + \mathbf{b})\)

    \(\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\).

  4. 4

    \(P\) is on \(AB\) with \(AP : PB = 1 : 2\). What fraction of \(AB\) is \(AP\)?

    1. A\(\dfrac{1}{2}\)
    2. B\(\dfrac{2}{3}\)
    3. C\(\dfrac{1}{4}\)
    4. D\(\dfrac{1}{3}\)
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    D: \(\dfrac{1}{3}\)

    There are \(1 + 2 = 3\) parts, and \(AP\) is 1 of them.

  5. 5

    \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), and \(P\) is on \(AB\) with \(AP : PB = 1 : 2\). What is \(\overrightarrow{OP}\)?

    1. A\(\dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
    2. B\(\dfrac{1}{3}(\mathbf{b} - \mathbf{a})\)
    3. C\(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
    4. D\(\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
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    C: \(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)

    \(\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\).

  6. 6

    How do you show that two lines are parallel using vectors?

    1. AShow that the vectors have the same length
    2. BShow that one vector is a multiple of the other
    3. CShow that the vectors add to zero
    4. DShow that the vectors meet at a point
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    B: Show that one vector is a multiple of the other

    Parallel vectors are scalar multiples of each other.

  7. 7

    \(\overrightarrow{AB} = 2\overrightarrow{BC}\). What does this show?

    1. A\(A\), \(B\) and \(C\) are on a straight line
    2. B\(B\) is the midpoint of \(AC\)
    3. C\(AB\) is perpendicular to \(BC\)
    4. D\(ABC\) is an isosceles triangle
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    A: \(A\), \(B\) and \(C\) are on a straight line

    The vectors are parallel and share the point \(B\), so the three points are collinear.

  8. 8

    \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). What is \(\overrightarrow{BC}\)?

    1. A\(3\mathbf{b} - 2\mathbf{a}\)
    2. B\(2\mathbf{a} - 2\mathbf{b}\)
    3. C\(4\mathbf{b} - 2\mathbf{a}\)
    4. D\(2\mathbf{b} - 2\mathbf{a}\)
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    D: \(2\mathbf{b} - 2\mathbf{a}\)

    \(\overrightarrow{BC} = -\mathbf{b} + 3\mathbf{b} - 2\mathbf{a} = 2\mathbf{b} - 2\mathbf{a}\).

  9. 9

    What is the length of the vector \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\)?

    1. A\(17\)
    2. B\(7\)
    3. C\(13\)
    4. D\(60\)
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    C: \(13\)

    \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\).