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How do you use column vectors, add and subtract vectors, multiply by a scalar and use vectors in geometric proofs?

Use column vector notation, add and subtract vectors and multiply by a scalar, and use vectors to find resultants and to prove that lines are parallel or points are collinear (Higher tier).

A focused answer to the WJEC GCSE Mathematics geometry content on vectors, covering column vector notation, vector addition and subtraction, multiplication by a scalar, finding resultants and using vectors in geometric proofs of parallel lines and collinearity.

Generated by Claude Opus 4.812 min answer

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

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  1. What this dot point is asking
  2. Column vectors and notation
  3. Adding, subtracting and scaling
  4. Vector paths
  5. Parallel lines and collinearity
  6. Why this matters

What this dot point is asking

Vectors are a Higher-tier-only topic. WJEC asks you to use column vector notation, to add and subtract vectors and multiply them by a scalar, and to use vectors in geometric arguments, for example to find a resultant or to prove that two lines are parallel or three points are collinear. A vector has both magnitude and direction, and the power of the topic is that geometric relationships become algebra: a path around a shape is a sum of vectors. The proof questions reward expressing a journey in terms of given vectors and then drawing a conclusion.

Column vectors and notation

A column vector records movement as two components.

The magnitude (length) of (xy)\binom{x}{y} is x2+y2\sqrt{x^2 + y^2} by Pythagoras, linking vectors back to the right-angled triangle.

Adding, subtracting and scaling

Vector arithmetic works component by component.

To add (x1y1)+(x2y2)=(x1+x2y1+y2)\binom{x_1}{y_1} + \binom{x_2}{y_2} = \binom{x_1 + x_2}{y_1 + y_2}, and to subtract do the same with minus signs. To multiply by a scalar kk, multiply both components: k(xy)=(kxky)k\binom{x}{y} = \binom{kx}{ky}. Geometrically, adding vectors places them nose to tail, so the resultant AC⃗=AB⃗+BC⃗\vec{AC} = \vec{AB} + \vec{BC} is the single vector from start to finish. Scaling by a positive number keeps the direction and changes the length; scaling by a negative number reverses it.

Vector paths

Any journey across a diagram is a sum of vectors.

This "path-finding" is the heart of vector questions: choose a route through known vectors, then add them with the correct signs.

Parallel lines and collinearity

Vectors prove geometric facts about direction.

Two vectors are parallel exactly when one is a scalar multiple of the other; if such parallel vectors also share a common point, the points lie on a straight line (collinear).

Why this matters

Vectors are a Higher-tier topic that rewards a different kind of thinking: turning a geometric diagram into vector algebra. The arithmetic (adding, subtracting and scaling column vectors) is routine, but the proof questions, expressing a journey in terms of given vectors and then arguing that vectors are parallel or points collinear, are demanding AO3 problems worth several marks. The skill of building a path with correct signs, and of recognising a scalar multiple as the signature of parallel lines, is what the marks reward.

Exam-style practice questions

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

WJEC 20193 marksGiven a=(3βˆ’1)\mathbf{a} = \binom{3}{-1} and b=(24)\mathbf{b} = \binom{2}{4}, work out the column vector 2aβˆ’b2\mathbf{a} - \mathbf{b}. (Higher, Unit 1, non-calculator.)
Show worked answer β†’

Multiply a\mathbf{a} by the scalar 22: 2a=(6βˆ’2)2\mathbf{a} = \binom{6}{-2}.

Subtract b\mathbf{b} component by component: (6βˆ’2)βˆ’(24)=(6βˆ’2βˆ’2βˆ’4)=(4βˆ’6)\binom{6}{-2} - \binom{2}{4} = \binom{6-2}{-2-4} = \binom{4}{-6}.

Markers award a mark for 2a2\mathbf{a}, a mark for the subtraction and a mark for the answer (4βˆ’6)\binom{4}{-6}. Subtracting in the wrong order, or only operating on one component, are the common slips.

WJEC 20214 marksIn a diagram, OA⃗=a\vec{OA} = \mathbf{a} and OB⃗=b\vec{OB} = \mathbf{b}. M is the midpoint of AB. Express OM⃗\vec{OM} in terms of a\mathbf{a} and b\mathbf{b}. (Higher, Unit 1, non-calculator.)
Show worked answer β†’

First find ABβƒ—=AOβƒ—+OBβƒ—=βˆ’a+b=bβˆ’a\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}.

M is the midpoint, so AMβƒ—=12(bβˆ’a)\vec{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}).

Then OMβƒ—=OAβƒ—+AMβƒ—=a+12(bβˆ’a)=12a+12b=12(a+b)\vec{OM} = \vec{OA} + \vec{AM} = \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 ABβƒ—\vec{AB}, for halving to reach M, and for the simplified 12(a+b)\tfrac{1}{2}(\mathbf{a} + \mathbf{b}). Sign errors in AOβƒ—=βˆ’a\vec{AO} = -\mathbf{a} are the usual cause of lost marks.

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