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How do you add, subtract and scale vectors, and use them in geometric proofs?

Use vector notation; add, subtract and multiply vectors by a scalar; and use vectors to construct geometric arguments and proofs (proof at Higher tier).

A focused answer to the OCR GCSE Mathematics geometry content on vectors, covering vector notation, addition, subtraction and scalar multiplication, and using vectors in geometric arguments and proofs at Higher tier.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Vector notation and arithmetic
  3. Vectors around a figure
  4. Vector proof (Higher)
  5. Why vectors matter

What this dot point is asking

OCR references G24 and G25 cover vectors: the notation, adding, subtracting and scaling vectors, and at Higher tier using them to build geometric arguments and proofs. A vector has both magnitude and direction, which makes it the natural tool for translations and for describing paths around a figure. Vector geometry, especially proof, is a demanding Higher-tier topic that tests AO2 reasoning and is a reliable source of the hardest marks on the paper.

Vector notation and arithmetic

A vector describes a movement, not a position.

To add or subtract, combine components: (52)+(βˆ’13)=(45)\begin{pmatrix} 5 \\ 2 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}. To scale, multiply each component: 3(2βˆ’1)=(6βˆ’3)3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}. The magnitude of a column vector is found with Pythagoras: ∣(34)∣=32+42=5\left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = \sqrt{3^2 + 4^2} = 5.

Vectors around a figure

Vector geometry adds vectors nose to tail along a path.

So in a triangle OABOAB with OAβ†’=a\overrightarrow{OA} = \mathbf{a} and OBβ†’=b\overrightarrow{OB} = \mathbf{b}, the side ABβ†’\overrightarrow{AB} is bβˆ’a\mathbf{b} - \mathbf{a}. Midpoints and points dividing a line in a ratio are reached by scaling: the midpoint MM of ABAB gives AMβ†’=12(bβˆ’a)\overrightarrow{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}). Drawing the figure and marking each known vector keeps the route clear.

When a point divides a line in a given ratio rather than at the midpoint, scale by the matching fraction. If PP divides ABAB so that AP:PB=1:2AP : PB = 1 : 2, then APβ†’=13ABβ†’=13(bβˆ’a)\overrightarrow{AP} = \tfrac{1}{3}\overrightarrow{AB} = \tfrac{1}{3}(\mathbf{b} - \mathbf{a}), because PP is one third of the way along. Then OPβ†’=a+13(bβˆ’a)\overrightarrow{OP} = \mathbf{a} + \tfrac{1}{3}(\mathbf{b} - \mathbf{a}), which simplifies to 23a+13b\tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}. Setting up the ratio fraction correctly is the step that most often goes wrong, so always check that the fractions of the two parts add to one.

Vector proof (Higher)

The power of vectors is proving geometric facts.

Why vectors matter

Vectors connect geometry to algebra and to the transformations topic, and they give a clean way to prove parallelism, collinearity and ratios that would be awkward with angles alone. OCR's vector-proof questions are among the most discriminating on the Higher paper, and they reward a clear path argument with each step justified. Recognising that "parallel" means "a scalar multiple" is the single most useful insight.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20192 marksVectors a=(3βˆ’1)\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} and b=(24)\mathbf{b} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}. Work out a+2b\mathbf{a} + 2\mathbf{b} as a column vector. (Higher, Paper 4, calculator.)
Show worked answer β†’

First find 2b2\mathbf{b} by multiplying each component by 22: 2b=(48)2\mathbf{b} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}.

Then add component by component: a+2b=(3+4βˆ’1+8)=(77)\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 + 4 \\ -1 + 8 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \end{pmatrix}.

Markers award a mark for 2b2\mathbf{b} and a mark for the correct sum. Adding the components across instead of down, or forgetting to double both components of b\mathbf{b}, are the usual errors.

OCR 20214 marksIn a triangle OABOAB, OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b}. MM is the midpoint of ABAB. Find OM→\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}, showing your reasoning. (Higher, Paper 5, non-calculator.)
Show worked answer β†’

First find ABβ†’\overrightarrow{AB}: travelling from AA to BB goes back along a\mathbf{a} then out along b\mathbf{b}, so ABβ†’=bβˆ’a\overrightarrow{AB} = \mathbf{b} - \mathbf{a}.

MM is the midpoint, so AMβ†’=12(bβˆ’a)\overrightarrow{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}).

Then OMβ†’=OAβ†’+AMβ†’=a+12(bβˆ’a)=12a+12b\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}.

Markers give marks for ABβ†’=bβˆ’a\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, for halving it, for adding to a\mathbf{a}, and for the simplified 12(a+b)\tfrac{1}{2}(\mathbf{a} + \mathbf{b}). Getting the direction of ABβ†’\overrightarrow{AB} backwards is the standard error.

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