GCSE Vector arithmetic Practice Questions

Free GCSE Vector arithmetic practice questions with full step-by-step worked solutions. Covers adding column vectors, negative components, subtracting column vectors, sign handling. Practise exam-style problems and check your method.

adding column vectorsnegative componentssubtracting column vectorssign handlingorder of subtractionscalar multiplication
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
a=(31)\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} and b=(24)\mathbf{b} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}. Work out a+b\mathbf{a} + \mathbf{b} as a column vector.
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Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    a+b=(31)+(24)=(3+21+4)=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 + 2 \\ 1 + 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (55)\begin{pmatrix} 5 \\ 5 \end{pmatrix}.

  3. State the answer as a column vector

    a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}

    So a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}.

Answer
a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}
Question 2
2 markseasy
The diagram shows the triangle OABOAB. OA⃗=a\vec{OA} = \mathbf{a} and AB⃗=b\vec{AB} = \mathbf{b}. Express OB⃗\vec{OB} in terms of a\mathbf{a} and b\mathbf{b}.
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Worked solution

  1. State the rule for travelling between two points

    PR⃗=PQ⃗+QR⃗,QP⃗=−PQ⃗\vec{PR} = \vec{PQ} + \vec{QR}, \quad \vec{QP} = -\vec{PQ}

    To get from one point to another, walk any route through the given arrows. Going the right way along an arrow adds it; going backwards along an arrow subtracts it.

  2. Write each leg in terms of the named vectors

    OB⃗=a+b\vec{OB} = \mathbf{a} + \mathbf{b}

    Substitute what the question says each arrow is, keeping the minus sign on any leg that is travelled backwards.

  3. Choose the matching option

    OB⃗=a+b\vec{OB} = \mathbf{a} + \mathbf{b}

    The only option equal to a+b\mathbf{a} + \mathbf{b} is the correct one.

Answer
a+b\mathbf{a} + \mathbf{b}
Question 3
2 marksintermediate
OABCOABC is a parallelogram. OA⃗=a\vec{OA} = \mathbf{a}, OC⃗=c\vec{OC} = \mathbf{c} and AB⃗=c\vec{AB} = \mathbf{c}. Express AC⃗\vec{AC} in terms of a\mathbf{a} and c\mathbf{c}.
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Worked solution

  1. Write down the vectors the diagram gives you

    OA⃗=a,OC⃗=c,AB⃗=c\vec{OA} = \mathbf{a}, \quad \vec{OC} = \mathbf{c}, \quad \vec{AB} = \mathbf{c}

    Every arrow in the figure is given in terms of the named vectors. Nothing else may be assumed — the route has to be built out of these arrows alone.

  2. State the rule for travelling between two points

    PR⃗=PQ⃗+QR⃗,QP⃗=−PQ⃗\vec{PR} = \vec{PQ} + \vec{QR}, \quad \vec{QP} = -\vec{PQ}

    To get from one point to another, walk any route through the given arrows. Going the right way along an arrow adds it; going backwards along an arrow subtracts it.

  3. Choose a route between the two points

    −OA⃗+OC⃗-\vec{OA} + \vec{OC}

    This route uses only arrows the question gives. A leg written with a minus sign is a leg travelled backwards.

  4. Write each leg in terms of the named vectors

    AC⃗=−a+c\vec{AC} = -\mathbf{a} + \mathbf{c}

    Substitute what the question says each arrow is, keeping the minus sign on any leg that is travelled backwards.

  5. Collect like terms

    AC⃗=−a+c\vec{AC} = -\mathbf{a} + \mathbf{c}

    Collecting the terms gives −a+c-\mathbf{a} + \mathbf{c}. This is an exact answer for EVERY choice of the two vectors — it does not depend on what they happen to be.

  6. Choose the matching option

    AC⃗=−a+c\vec{AC} = -\mathbf{a} + \mathbf{c}

    The only option equal to −a+c-\mathbf{a} + \mathbf{c} is the correct one.

Answer
−a+c-\mathbf{a} + \mathbf{c}
Question 4
4 markshard
a=(−1−2)\mathbf{a} = \begin{pmatrix} -1 \\ -2 \end{pmatrix} and b=(3−4)\mathbf{b} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}. Work out −3a−2b-3\mathbf{a} - 2\mathbf{b} as a column vector.
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Worked solution

  1. Write down the vectors the question gives you

    a=(−1−2),b=(3−4)\mathbf{a} = \begin{pmatrix} -1 \\ -2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}

    Each column vector says how far to move across (top) and how far to move up (bottom). Everything that follows works on those two rows separately.

  2. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s),k(pq)=(kpkq)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}, \quad k\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} kp \\ kq \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  3. Substitute the column vectors into the expression

    −3a−2b=−3(−1−2)−2(3−4)-3\mathbf{a} - 2\mathbf{b} = -3\begin{pmatrix} -1 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ -4 \end{pmatrix}

    Replace each named vector by its column. Keep every sign exactly where the question put it — that is where the marks are lost.

  4. Multiply out each scalar first

    −3(−1−2)−2(3−4)=−(−3−6)−(6−8)-3\begin{pmatrix} -1 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ -4 \end{pmatrix} = -\begin{pmatrix} -3 \\ -6 \end{pmatrix} - \begin{pmatrix} 6 \\ -8 \end{pmatrix}

    Each scalar multiplies BOTH components of its vector. Do all the scaling before you add or subtract anything, and the signs stay in the joining symbols where you can see them.

  5. Work out the top component

    top: 3−6=−3\text{top: } 3 - 6 = -3

    Combining the top row gives −3-3.

  6. Work out the bottom component

    bottom: 6+8=14\text{bottom: } 6 + 8 = 14

    Combining the bottom row gives 1414.

  7. State the answer as a column vector

    −3a−2b=(−314)-3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} -3 \\ 14 \end{pmatrix}

    So −3a−2b=(−314)-3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} -3 \\ 14 \end{pmatrix}.

  8. Check the answer on a diagram

    −3a−2b=(−314)-3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} -3 \\ 14 \end{pmatrix}

    Laying the vectors head to tail on the grid, the direct route from the start of the first arrow to the end of the last one is the answer — and the arrow it draws is exactly the column vector the arithmetic gave. The diagram and the column form must agree.

  9. Check the answer by working backwards

    (−314)+2(3−4)=(36)=−3a\begin{pmatrix} -3 \\ 14 \end{pmatrix} + 2\begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix} = -3\mathbf{a}

    Taking the last term back off the answer must return −3a=(36)-3\mathbf{a} = \begin{pmatrix} 3 \\ 6 \end{pmatrix} — and it does, so the signs were handled correctly.

  10. Sense check the direction of the answer

    (−314)=3 left, 14 up\begin{pmatrix} -3 \\ 14 \end{pmatrix} = \text{3 left, 14 up}

    The answer is a movement of 3 left, 14 up, which is the way the resultant arrow points on the diagram. If the arrow and the column disagree, one of them is wrong.

Answer
−3a−2b=(−314)-3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} -3 \\ 14 \end{pmatrix}
Question 5
5 markschallenging
The diagram shows a journey from OO to DD. OA⃗=(2−1)\vec{OA} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, AB⃗=(−35)\vec{AB} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}, BC⃗=(42)\vec{BC} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} and CD⃗=(−1−3)\vec{CD} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}. Work out OD⃗\vec{OD} as a column vector.
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Worked solution

  1. Write down the vectors along the route

    OA⃗=(2−1),AB⃗=(−35),BC⃗=(42),CD⃗=(−1−3)\vec{OA} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \vec{AB} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}, \quad \vec{BC} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}, \quad \vec{CD} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}

    Each arrow on the diagram is one leg of the journey, given as a column vector.

  2. State the rule for adding displacements

    OD⃗=OA⃗+AB⃗+BC⃗+CD⃗\vec{OD} = \vec{OA} + \vec{AB} + \vec{BC} + \vec{CD}

    Displacements add: travelling the legs one after another gets you to exactly the same place as the direct route. That single fact is the triangle law.

  3. Substitute the column vectors

    OD⃗=(2−1)+(−35)+(42)+(−1−3)\vec{OD} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \begin{pmatrix} -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -1 \\ -3 \end{pmatrix}

    Replace each leg by its column vector and add them all — top row with top row, bottom row with bottom row.

  4. Add the top components

    top: 2−3+4−1=2\text{top: } 2 - 3 + 4 - 1 = 2

    The whole journey moves 22 across.

  5. Add the bottom components

    bottom: −1+5+2−3=3\text{bottom: } -1 + 5 + 2 - 3 = 3

    The whole journey moves 33 up.

  6. State the resultant as a column vector

    OD⃗=(23)\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

    The direct route from OO to DD is (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}.

  7. Check the resultant on the diagram

    OD⃗=(23)\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

    The resultant is the arrow that closes the path: it runs straight from the start of the first leg to the end of the last one. The arrow on the grid and the column vector must be the same thing.

  8. Check by walking the route on the grid

    (23)=2 right, 3 up\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \text{2 right, 3 up}

    Counting squares along the path on the diagram gives 2 right, 3 up — the same as the arithmetic. If the two disagree, one of them has a sign wrong.

  9. Note what the reverse journey would be

    DO⃗=−OD⃗=(−2−3)\vec{DO} = -\vec{OD} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}

    Reversing a displacement negates both of its components. It is the same arrow, turned round.

  10. Note the mistake to avoid

    OD⃗≠DO⃗\vec{OD} \ne \vec{DO}

    Going the other way round the route reverses the vector: DO⃗=(−2−3)\vec{DO} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}, which is the negative of the answer. Always travel the arrows in the direction the question asks for.

  11. Sense check the answer against the picture

    OD⃗=(23)=2 right, 3 up\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \text{2 right, 3 up}

    The resultant should be a movement of 2 right, 3 up, and that is exactly where the last point sits relative to the first.

  12. Work out how far apart the ends are

    ∣OD⃗∣=13=13\left| \vec{OD} \right| = \sqrt{13} = \sqrt{13}

    The magnitude of the resultant is the straight-line distance between the start and the finish — usually much less than the distance actually travelled along the legs.

  13. Note that any route between the two points gives the same answer

    OD⃗=OA⃗+AB⃗+BC⃗+CD⃗\vec{OD} = \vec{OA} + \vec{AB} + \vec{BC} + \vec{CD}

    The displacement between two points does not depend on how you get from one to the other — only on where they are. Any route through the given arrows gives the same resultant.

  14. Summarise the method

    add the legs→the direct route\text{add the legs} \rightarrow \text{the direct route}

    Add the column vectors of the legs, in the order the route is travelled. The sum IS the direct route.

  15. State the answer again as a column vector

    OD⃗=(23)\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

    The resultant is (23)\begin{pmatrix} 2 \\ 3 \end{pmatrix}.

Answer
OD⃗=(23)\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

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