Mathematics · OCR H240 · Topic 1.10

Vectors

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

5 sub-topics
Taught in Year 13
72 practice questions
5 animated lessons

Practice questions (72) Lessons Worksheets coming soon All OCR Mathematics topics

What the specification says

Every sub-topic OCR 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.

1.10a-bVectorsY13
1.10cMagnitude and direction of vectorsY13
1.10dBasic operations on vectorsY13
1.10e-fPosition vectors and distance between pointsY13
1.10g-hProblem solving using vectorsY13

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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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