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Maths · Vectors, Constructions and Loci

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Vector Geometry and Proof

Finding vectors by routes around a diagram, dividing lines in a ratio, and proving lines parallel or points collinear.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Write one vector in terms of others using paths around a diagram.
  2. 2Find the vector to a midpoint, and (Higher tier) to a point that divides a line in a given ratio.
  3. 3Prove that lines are parallel, and that three points are on a straight line (Higher tier).
  4. 4Find the magnitude of a vector in a geometric problem (Higher tier).

Proving things with vectors

In vector geometry you are not given the numbers, only letters such as \(\mathbf{a}\) and \(\mathbf{b}\). The skill is to get from one point to another along a path of known vectors, adding them up with the right signs, and then to compare two results. If one result is a multiple of the other, the lines are parallel, and if they also share a point, the three points are on a straight line. Finding a vector by a route in a diagram appears at both tiers, while dividing lines in a ratio and proving that lines are parallel or that points are collinear are Higher tier. The questions are written so that the algebra is short, but each step needs a reason.

Finding vectors by routes

To find a vector, go from the start to the finish along paths you know, adding vectors that go the right way and subtracting those that go the wrong way.

  • Any route

    \(\overrightarrow{AB}\) can be reached by going from \(A\) to \(O\) and then from \(O\) to \(B\), so \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}\).

  • Going backwards

    If \(\overrightarrow{OA} = \mathbf{a}\), then \(\overrightarrow{AO} = -\mathbf{a}\).

  • A key result

    If \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), then \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

  • Simplify at the end

    Collect the terms in \(\mathbf{a}\) and the terms in \(\mathbf{b}\).

Vectors in a triangle

\(OAB\) is a triangle with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

Show the solutionHide the solution
  1. 1 Find \(\overrightarrow{AB}\) \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}\).
  2. 2 Half way \(\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})\).
  3. 3 Go via \(A\) \(\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\).
  4. 4 Simplify \(\overrightarrow{OM} = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\).

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

Points that divide a line in a ratio (Higher tier)

A point that divides a line in the ratio \(m : n\) is \(\dfrac{m}{m + n}\) of the way along.

  • Ratio to fraction

    A point \(P\) with \(AP : PB = 2 : 3\) is \(\dfrac{2}{5}\) of the way from \(A\) to \(B\).

  • Vector to the point

    \(\overrightarrow{AP} = \dfrac{2}{5}\overrightarrow{AB}\).

  • Then add the route

    \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\).

  • Check with the midpoint

    The ratio \(1 : 1\) gives \(\dfrac{1}{2}\), which matches the midpoint.

Proving lines are parallel or points are collinear (Higher tier)

Compare two vectors written in the same letters.

  • Parallel

    If \(\overrightarrow{PQ} = k\overrightarrow{RS}\) for a number \(k\), then \(PQ\) and \(RS\) are parallel.

  • Collinear

    Three points \(A\), \(B\) and \(C\) are on a straight line if \(\overrightarrow{AB} = k\overrightarrow{BC}\), because the vectors are parallel and share the point \(B\).

  • Say it in words

    "\(\overrightarrow{AB}\) is a multiple of \(\overrightarrow{BC}\), so they are parallel, and \(B\) is a common point, so \(A\), \(B\) and \(C\) are collinear."

  • Length

    The length of a vector in column form is \(\sqrt{x^2 + y^2}\), as in Pythagoras.

Proving points are collinear (Higher tier)

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). Prove that \(A\), \(B\) and \(C\) lie on a straight line.

Show the solutionHide the solution
  1. 1 Find \(\overrightarrow{AB}\) \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}\).
  2. 2 Find \(\overrightarrow{BC}\) \(\overrightarrow{BC} = -\mathbf{b} + 3\mathbf{b} - 2\mathbf{a} = 2\mathbf{b} - 2\mathbf{a}\).
  3. 3 Compare \(\overrightarrow{BC} = 2(\mathbf{b} - \mathbf{a}) = 2\overrightarrow{AB}\).
  4. 4 Conclude The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.

Answer\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BC} = 2\mathbf{b} - 2\mathbf{a} = 2\overrightarrow{AB}\), so they are collinear

Test yourself

  1. 1

    If \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), what is \(\overrightarrow{AB}\)?

    Show answerHide answer

    \(\mathbf{b} - \mathbf{a}\).

  2. 2

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

    Show answerHide answer

    \(-\mathbf{a}\).

  3. 3

    What fraction of \(AB\) is \(AP\) if \(AP : PB = 2 : 3\)?

    Show answerHide answer

    \(\dfrac{2}{5}\).

  4. 4

    How do you show two lines are parallel?

    Show answerHide answer

    Show that one vector is a multiple of the other.

  5. 5

    How do you show three points are collinear?

    Show answerHide answer

    Show that two of the vectors are multiples, with a common point.

Exam technique: vector proofs

Show every step, because the answer is given or obvious.

  • Draw a diagram

    Mark all the known vectors, with arrows.

  • Choose a route

    Go from the start to the finish along known vectors.

  • Simplify fully

    Collect the \(\mathbf{a}\) terms and the \(\mathbf{b}\) terms.

  • Write the conclusion

    Say that the vectors are parallel and that there is a common point.

Summary and exam focus

  • Find a vector by adding known vectors along a route, reversing the sign when going backwards.
  • A point dividing \(AB\) in the ratio \(m : n\) is \(\dfrac{m}{m + n}\) of the way from \(A\) to \(B\).
  • Parallel vectors are multiples of each other.
  • Points are collinear if two vectors between them are parallel and share a point.

Exam focus

\(OAB\) is a triangle with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) is the point on \(AB\) with \(AP : PB = 1 : 2\). Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). (3 marks) (3 marks)

\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), and \(AP\) is \(\dfrac{1}{3}\) of it. Then \(\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\). Show the route \(\overrightarrow{OA} + \overrightarrow{AP}\) for the method mark.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Vector proof
A proof that uses vectors to show a geometric fact.
Route
A path from one point to another made from known vectors.
Collinear
Lying on the same straight line.
Midpoint
The point halfway along a line.
Ratio
A comparison of two parts of a line, such as \(2 : 3\).
Position vector
The vector from the origin to a point.
Scalar multiple
A vector multiplied by a number.
Magnitude
The length of a vector.
Parallel
Having the same direction, so one vector is a multiple of the other.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Write down 2 marks Easier

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

Mark scheme — 2 marks available

  • (a) \(-\mathbf{b}\) — B1
  • (b) \(\mathbf{a} - \mathbf{b}\) — B1

Model answer

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

2. Exam question Calculate 4 marks Core

\(OAB\) 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.

Mark scheme — 4 marks available

  • (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

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

3. Exam question Prove 3 marks Stretch

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

Mark scheme — 3 marks available

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

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.

4. Exam question Find 3 marks Stretch

\(\overrightarrow{OA} = \mathbf{a}\) 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]

Mark scheme — 3 marks available

  • \(\overrightarrow{OQ} = \dfrac{1}{4}\mathbf{a}\) — M1
  • \(\overrightarrow{BQ} = \overrightarrow{BO} + \overrightarrow{OQ}\) or \(-\mathbf{b} + \dfrac{1}{4}\mathbf{a}\) — M1
  • \(\dfrac{1}{4}\mathbf{a} - \mathbf{b}\) — A1

Model answer

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

5. Exam question Show that 3 marks Stretch

\(\overrightarrow{OA} = \mathbf{a}\) 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]

Mark scheme — 3 marks available

  • \(\overrightarrow{XY} = 2\mathbf{b} - 2\mathbf{a}\) — M1
  • \(2\overrightarrow{AB}\) or \(2(\mathbf{b} - \mathbf{a})\) — M1
  • A multiple, so parallel — A1

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.

6. Exam question Calculate 3 marks Stretch

\(\overrightarrow{PQ} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\) 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]

Mark scheme — 3 marks available

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

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 \(\sqrt{81 + 144} = \sqrt{225} = 15\).

7. Multiple choice 1 mark Core

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

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

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

8. Multiple choice 1 mark Easier

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

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

Why: Going backwards reverses the vector.

9. Multiple choice 1 mark Core

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

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

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

10. Multiple choice 1 mark Core

\(P\) 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 \(\dfrac{1}{3}\) Correct

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

11. Multiple choice 1 mark Stretch

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), 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 \(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\) Correct
  4. D \(\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)

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

12. Multiple choice 1 mark Core

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

  1. A Show that the vectors have the same length
  2. B Show that one vector is a multiple of the other Correct
  3. C Show that the vectors add to zero
  4. D Show that the vectors meet at a point

Why: Parallel vectors are scalar multiples of each other.

13. Multiple choice 1 mark Core

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

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

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

14. Multiple choice 1 mark Stretch

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) 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 \(2\mathbf{b} - 2\mathbf{a}\) Correct

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

15. Multiple choice 1 mark Core

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

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

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