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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 Eduqas GCSE Mathematics geometry content on vectors, covering vector notation, adding subtracting and scaling vectors with column vectors, and using vectors to construct geometric arguments and proofs at Higher tier.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Vector notation and arithmetic
  3. Finding a path across a diagram
  4. Using vectors in proofs (Higher)
  5. Why vectors matter

What this dot point is asking

The Eduqas geometry content asks you to use vector notation, to add and subtract vectors and multiply them by a scalar, and at Higher tier to use vectors to construct geometric arguments and proofs. A vector has both magnitude and direction, written as a column vector or with an arrow notation such as AB→\overrightarrow{AB}. The arithmetic is straightforward; the challenge, and the Higher-tier marks, lie in vector geometry: finding a path across a diagram in terms of named vectors and proving that lines are parallel or that points are collinear.

Vector notation and arithmetic

A vector is written as a column vector or named with a direction.

Adding vectors places them nose to tail: (23)+(4βˆ’1)=(62)\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}. Subtracting is adding the reverse. Multiplying by a scalar stretches the vector: 3(2βˆ’1)=(6βˆ’3)3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}, keeping the same direction (or reversing it for a negative scalar).

Finding a path across a diagram

Vector geometry questions give a diagram with some vectors named and ask for another in terms of them. The method is to travel from the start to the end along known vectors.

So if OAβ†’=a\overrightarrow{OA} = \mathbf{a} and OBβ†’=b\overrightarrow{OB} = \mathbf{b}, then ABβ†’=AOβ†’+OBβ†’=βˆ’a+b\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}. Breaking a journey into known legs is the core technique, and midpoints and ratios along a line are handled by taking the appropriate fraction of a vector.

Using vectors in proofs (Higher)

At Higher tier, vectors prove geometric facts such as parallelism and collinearity.

The same idea proves three points are collinear: if AB→\overrightarrow{AB} and BC→\overrightarrow{BC} are scalar multiples of one another and share point B, then A, B and C lie on a straight line.

Why vectors matter

Vectors connect geometry to algebra and underpin the more advanced mathematics of mechanics and further study. Within GCSE, they let you prove results that would be awkward with angle-chasing alone, which is why Eduqas reserves the proof questions for Higher tier and weights the reasoning heavily. The arithmetic is quick once the notation is secure, so the marks reward a clear, well-justified route across the diagram and a precise concluding statement.

Exam-style practice questions

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

Eduqas 20193 marksa=(31)\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} and b=(βˆ’24)\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}. Work out 2aβˆ’b2\mathbf{a} - \mathbf{b} as a column vector. (Foundation, Component 1, non-calculator.)
Show worked answer β†’

First scale a\mathbf{a} by 2: 2a=(62)2\mathbf{a} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.

Then subtract b\mathbf{b} component by component: (62)βˆ’(βˆ’24)=(6βˆ’(βˆ’2)2βˆ’4)=(8βˆ’2)\begin{pmatrix} 6 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 - (-2) \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -2 \end{pmatrix}.

Markers award a mark for scaling a\mathbf{a}, a mark for the subtraction method, and a mark for the answer. Mishandling the double negative in 6βˆ’(βˆ’2)6 - (-2) is the standard slip.

Eduqas 20224 marksIn a triangle OAB, OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b}. M is the midpoint of AB. Find OM→\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}, showing your reasoning. (Higher, Component 1, non-calculator.)
Show worked answer β†’

First find ABβ†’\overrightarrow{AB} by going from A back to O then O to B: ABβ†’=βˆ’a+b=bβˆ’a\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}.

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

Travel from O to A then A to M: OMβ†’=a+12(bβˆ’a)=12a+12b=12(a+b)\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}).

Markers give marks for AB→\overrightarrow{AB}, for using the midpoint, for the route to M, and for the simplified result. The reasoning marks depend on showing each path, not just stating the answer.

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