Mathematics · AQA 7357 · Topic J

Vectors

Vectors in two and three dimensions: magnitude, direction, position vectors and geometric problems.

5 sub-topics
Taught in Year 12
72 practice questions
5 animated lessons

Practice questions (72) Lessons Worksheets All AQA Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

J1Vectors in two and three dimensionsY13
J2Magnitude, direction and component formY12
J3Adding vectors and multiplying by scalarsY12
J4Position vectors and distancesY13
J5Vectors in pure mathematics and in contextY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 72 questions in this topic, free, no account needed.

Vectors

Easy

Magnitude and unit vector · Year 12 · 2 marks

\mathbf{a}

{}= 3\mathbf{i} + 4\mathbf{j}

Find $|\mathbf{a}|$ and a unit vector in the direction of $\mathbf{a}$ .
(2)

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Pythagoras. $|\mathbf{a}|$ ${}= \sqrt{3^2 + 4^2}$ ${}= 5$
Divide by the magnitude to get length 1. $\tfrac{1}{5}(3\mathbf{i} + 4\mathbf{j})$

Total2

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Vectors

Medium

Parallel vectors · Year 12 · 2 marks

Find $k$ such that $2\mathbf{i} + k\mathbf{j}$ is parallel to $3\mathbf{i} - 6\mathbf{j}$ .
(2)

Check your answer before you look at the solution

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Parallel vectors are multiples of each other. $3\mathbf{i} - 6\mathbf{j}$ ${}= 3(\mathbf{i} - 2\mathbf{j})$ , so we need $2(\mathbf{i} - 2\mathbf{j})$
$k = -4$

Total2

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Vectors

Hard

Vectors in a parallelogram · Year 12 · 6 marks

OABC

is a parallelogram with

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OC} = \mathbf{c}

M

is the midpoint of

AB

, and

OM

meets

AC

X

Find $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .
(6)

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$\overrightarrow{AB}$ ${}= \overrightarrow{OC}$ ${}= \mathbf{c}$ . $\overrightarrow{OM}$ ${}= \mathbf{a}$ ${}+ \tfrac{1}{2}\mathbf{c}$
$X$ is on $OM$ . $\overrightarrow{OX}$ ${}= \lambda\left(\mathbf{a} + \tfrac{1}{2}\mathbf{c}\right)$
$X$ is also on $AC$ . $\overrightarrow{OX}$ ${}= \mathbf{a}$ ${}+ \mu(\mathbf{c} - \mathbf{a})$
Compare coefficients of $\mathbf{a}$ and $\mathbf{c}$ . $\lambda = 1 - \mu$ and $\tfrac{1}{2}\lambda = \mu$
$\lambda = \tfrac{2}{3}$ , so $\overrightarrow{OX}$ ${}= \tfrac{2}{3}\mathbf{a}$ ${}+ \tfrac{1}{3}\mathbf{c}$

Total6

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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What the specification says

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