The diagram shows a journey from O to D. OA = pmatrix 2 \ -1 pmatrix, AB = pmatrix -3 \ 5 pmatrix, BC = pmatrix 4 \ 2 pmatrix and CD = pmatrix -1 \ -3 pmatrix. Work out OD as a column vector.

Write down the vectors along the route: OA = pmatrix 2 \ -1 pmatrix, AB = pmatrix -3 \ 5 pmatrix, BC = pmatrix 4 \ 2 pmatrix, CD = pmatrix -1 \ -3 pmatrix State the rule for adding displacements: OD = OA + AB + BC + CD Substitute the column vectors: OD = pmatrix 2 \ -1 pmatrix + pmatrix -3 \ 5 pmatrix + pmatrix 4 \ 2 pmatrix + pmatrix -1 \ -3 pmatrix Add the top components: top: 2 - 3 + 4 - 1 = 2 Add the bottom components: bottom: -1 + 5 + 2 - 3 = 3 State the resultant as a column vector: OD = pmatrix 2 \ 3 pmatrix Check the resultant on the diagram: OD = pmatrix 2 \ 3 pmatrix Check by walking the route on the grid: pmatrix 2 \ 3 pmatrix = 2 right, 3 up Note what the reverse journey would be: DO = -OD = pmatrix -2 \ -3 pmatrix Note the mistake to avoid: OD DO Sense check the answer against the picture: OD = pmatrix 2 \ 3 pmatrix = 2 right, 3 up Work out how far apart the ends are: | OD | = √(13) = √(13) Note that any route between the two points gives the same answer: OD = OA + AB + BC + CD Summarise the method: add the legs arrow the direct route State the answer again as a column vector: OD = pmatrix 2 \ 3 pmatrix Answer: OD = pmatrix 2 \ 3 pmatrix

a = pmatrix -3 \ 4 pmatrix and b = pmatrix 6 \ -2 pmatrix. Which one of these column vectors is equal to 2a - (1)/(2)b?

Write down the vectors the question gives you: a = pmatrix -3 \ 4 pmatrix, b = pmatrix 6 \ -2 pmatrix Read the first vector as a movement: a = pmatrix -3 \ 4 pmatrix = 3 left, 4 up State the rule for combining column vectors: pmatrix p \ q pmatrix + pmatrix r \ s pmatrix = pmatrix p + r \ q + s pmatrix, pmatrix p \ q pmatrix - pmatrix r \ s pmatrix = pmatrix p - r \ q - s pmatrix, kpmatrix p \ q pmatrix = pmatrix kp \ kq pmatrix Substitute the column vectors into the expression: 2a - (1)/(2)b = 2pmatrix -3 \ 4 pmatrix - (1)/(2)pmatrix 6 \ -2 pmatrix Multiply out each scalar first: 2pmatrix -3 \ 4 pmatrix - (1)/(2)pmatrix 6 \ -2 pmatrix = pmatrix -6 \ 8 pmatrix - pmatrix 3 \ -1 pmatrix Work out the top component: top: -6 - 3 = -9 Work out the bottom component: bottom: 8 + 1 = 9 Work out the vector you are looking for: 2a - (1)/(2)b = pmatrix -9 \ 9 pmatrix Rule out the wrong option number 1: pmatrix -3 \ 7 pmatrix pmatrix -9 \ 9 pmatrix Rule out the wrong option number 2: pmatrix 9 \ -9 pmatrix pmatrix -9 \ 9 pmatrix Rule out the wrong option number 3: pmatrix -6 \ 5 pmatrix pmatrix -9 \ 9 pmatrix Rule out the wrong option number 4: pmatrix -12 \ 10 pmatrix pmatrix -9 \ 9 pmatrix Check the answer on a diagram: 2a - (1)/(2)b = pmatrix -9 \ 9 pmatrix Note the mistake to avoid: kpmatrix p \ q pmatrix = pmatrix kp \ kq pmatrix pmatrix kp \ q pmatrix Choose the option that matches: 2a - (1)/(2)b = pmatrix -9 \ 9 pmatrix Answer: pmatrix -9 \ 9 pmatrix

The diagram shows the points O, A, B, C and D. OA = a, OB = b, OC = 3a and OD = 3b. Express CD in terms of a and b.

Write down the vectors the diagram gives you: OA = a, OB = b, OC = 3a, OD = 3b State the rule for travelling between two points: PR = PQ + QR, QP = -PQ Choose a route between the two points: -OC + OD Write each leg in terms of the named vectors: CD = -3a + 3b Collect like terms: CD = -3a + 3b State the vector in terms of the named vectors: CD = -3a + 3b Check the answer on the diagram: CD = -3a + 3b Rule out the wrong option number 1: 3a - 3b -3a + 3b Rule out the wrong option number 2: 3a + 3b -3a + 3b Rule out the wrong option number 3: 3b -3a + 3b Rule out the wrong option number 4: -3a -3a + 3b Note the mistake to avoid: CD = -DC Note that the route chosen does not matter: CD = -3a + 3b Summarise the method: walk a route arrow add the arrows arrow collect terms Choose the matching option: CD = -3a + 3b Answer: -3a + 3b

The diagram shows the points O, A, B, C and D. OA = a, AB = 2b, OC = 3a and CD = 6b. Given that OD = k OB, work out the value of k.

Write down the vectors the diagram gives you: OA = a, AB = 2b, OC = 3a, CD = 6b State the rule for travelling between two points: PR = PQ + QR, QP = -PQ Work out the first vector: OB = a + 2b Work out the second vector: OD = 3a + 6b Compare the two vectors: 3a + 6b = 3(a + 2b) Check the scalar on the first named vector: 3 × 1 = 3 Check the scalar on the second named vector: 3 × 2 = 6 Note that only one scalar can work: ka + 2b = 3a + 6b k = 3 Say what this proves: OD = 3 OB OD OB Say what the scalar means for the lengths: | OD | = 3 | OB | See it on the diagram: OD = 3 OB Note the mistake to avoid: k = ODOB OBOD Summarise the method: write both vectors in terms of the named ones arrow divide Check the answer once more: 3(a + 2b) = 3a + 6b State the value of k: k = 3 Answer: k = 3

In the diagram M is the midpoint of OA and N is the midpoint of OB. OA = 2a, OB = 2b, OM = a and ON = b. Given that MN = k AB, work out the value of k.

Write down the vectors the diagram gives you: OA = 2a, OB = 2b, OM = a, ON = b State the rule for travelling between two points: PR = PQ + QR, QP = -PQ Work out the first vector: AB = -2a + 2b Work out the second vector: MN = -a + b Compare the two vectors: -a + b = (1)/(2)(-2a + 2b) Check the scalar on the first named vector: (1)/(2) × -2 = -1 Check the scalar on the second named vector: (1)/(2) × 2 = 1 Note that only one scalar can work: k-2a + 2b = -a + b k = (1)/(2) Say what this proves: MN = (1)/(2) AB MN AB Say what the scalar means for the lengths: | MN | = (1)/(2) | AB | See it on the diagram: MN = (1)/(2) AB Note the mistake to avoid: k = MNAB ABMN Summarise the method: write both vectors in terms of the named ones arrow divide Check the answer once more: (1)/(2)(-2a + 2b) = -a + b State the value of k: k = (1)/(2) Answer: k = (1)/(2)

Hard GCSE Vector arithmetic Questions

Challenging, exam-style GCSE Vector arithmetic questions with worked solutions. Stretch yourself on the hardest three vectors, linear combination, fractional scalars, sign handling problems.

three vectorslinear combinationfractional scalarssign handlingmagnitude of a resultantsubtraction

GCSE Foundation34 questionsStep-by-step solutions

Question 1

5 markschallenging

The diagram shows a journey from

O

D

\vec{OA} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}

\vec{AB} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}

\vec{BC} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}

and

\vec{CD} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}

. Work out

\vec{OD}

as a column vector.

Show worked solution

Worked solution

Write down the vectors along the route
$\vec{OA} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \vec{AB} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}, \quad \vec{BC} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}, \quad \vec{CD} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$
Each arrow on the diagram is one leg of the journey, given as a column vector.
State the rule for adding displacements
$\vec{OD} = \vec{OA} + \vec{AB} + \vec{BC} + \vec{CD}$
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.
Substitute the column vectors
$\vec{OD} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \begin{pmatrix} -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -1 \\ -3 \end{pmatrix}$
Replace each leg by its column vector and add them all — top row with top row, bottom row with bottom row.
Add the top components
$\text{top: } 2 - 3 + 4 - 1 = 2$
The whole journey moves $2$ across.
Add the bottom components
$\text{bottom: } -1 + 5 + 2 - 3 = 3$
The whole journey moves $3$ up.
State the resultant as a column vector
$\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
The direct route from $O$ to $D$ is $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ .
Check the resultant on the diagram
$\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
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.
Check by walking the route on the grid
$\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \text{2 right, 3 up}$
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.
Note what the reverse journey would be
$\vec{DO} = -\vec{OD} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}$
Reversing a displacement negates both of its components. It is the same arrow, turned round.
Note the mistake to avoid
$\vec{OD} \ne \vec{DO}$
Going the other way round the route reverses the vector: $\vec{DO} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ , which is the negative of the answer. Always travel the arrows in the direction the question asks for.
Sense check the answer against the picture
$\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \text{2 right, 3 up}$
The resultant should be a movement of 2 right, 3 up, and that is exactly where the last point sits relative to the first.
Work out how far apart the ends are
$\left| \vec{OD} \right| = \sqrt{13} = \sqrt{13}$
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.
Note that any route between the two points gives the same answer
$\vec{OD} = \vec{OA} + \vec{AB} + \vec{BC} + \vec{CD}$
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.
Summarise the method
$\text{add the legs} \rightarrow \text{the direct route}$
Add the column vectors of the legs, in the order the route is travelled. The sum IS the direct route.
State the answer again as a column vector
$\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$
The resultant is $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ .

Answer

\vec{OD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}

Question 2

6 markschallenging

\mathbf{a} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}

. Which one of these column vectors is equal to

2\mathbf{a} - \frac{1}{2}\mathbf{b}

Show worked solution

Worked solution

Write down the vectors the question gives you
$\mathbf{a} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$
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.
Read the first vector as a movement
$\mathbf{a} = \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \text{3 left, 4 up}$
$\mathbf{a}$ means 3 left, 4 up. Reading a column vector as a movement is what ties the column form to the arrow on the diagram — the two representations must agree.
State the rule for combining column vectors
$\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}, \quad k\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} kp \\ kq \end{pmatrix}$
Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.
Substitute the column vectors into the expression
$2\mathbf{a} - \frac{1}{2}\mathbf{b} = 2\begin{pmatrix} -3 \\ 4 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} 6 \\ -2 \end{pmatrix}$
Replace each named vector by its column. Keep every sign exactly where the question put it — that is where the marks are lost.
Multiply out each scalar first
$2\begin{pmatrix} -3 \\ 4 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix}$
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.
Work out the top component
$\text{top: } -6 - 3 = -9$
Combining the top row gives $-9$ .
Work out the bottom component
$\text{bottom: } 8 + 1 = 9$
Combining the bottom row gives $9$ .
Work out the vector you are looking for
$2\mathbf{a} - \frac{1}{2}\mathbf{b} = \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
Do the arithmetic first and only then look at the options — that way a plausible-looking wrong answer cannot pull you off course.
Rule out the wrong option number 1
$\begin{pmatrix} -3 \\ 7 \end{pmatrix} \ne \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
$\begin{pmatrix} -3 \\ 7 \end{pmatrix}$ is what you get if every sign after the first has been flipped — this is what you get by adding when the question says subtract, or subtracting when it says add. Recomputing $2\mathbf{a} - \frac{1}{2}\mathbf{b}$ properly gives $\begin{pmatrix} -9 \\ 9 \end{pmatrix}$ , so that option is wrong.
Rule out the wrong option number 2
$\begin{pmatrix} 9 \\ -9 \end{pmatrix} \ne \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
$\begin{pmatrix} 9 \\ -9 \end{pmatrix}$ is what you get if this is the answer with both components negated, which is what subtracting the wrong way round gives you. Recomputing $2\mathbf{a} - \frac{1}{2}\mathbf{b}$ properly gives $\begin{pmatrix} -9 \\ 9 \end{pmatrix}$ , so that option is wrong.
Rule out the wrong option number 3
$\begin{pmatrix} -6 \\ 5 \end{pmatrix} \ne \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
$\begin{pmatrix} -6 \\ 5 \end{pmatrix}$ is what you get if only the second vector has been scaled; the first one has been used just as it stands. Recomputing $2\mathbf{a} - \frac{1}{2}\mathbf{b}$ properly gives $\begin{pmatrix} -9 \\ 9 \end{pmatrix}$ , so that option is wrong.
Rule out the wrong option number 4
$\begin{pmatrix} -12 \\ 10 \end{pmatrix} \ne \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
$\begin{pmatrix} -12 \\ 10 \end{pmatrix}$ is what you get if only the first vector has been scaled; the second one has been used just as it stands. Recomputing $2\mathbf{a} - \frac{1}{2}\mathbf{b}$ properly gives $\begin{pmatrix} -9 \\ 9 \end{pmatrix}$ , so that option is wrong.
Check the answer on a diagram
$2\mathbf{a} - \frac{1}{2}\mathbf{b} = \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
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.
Note the mistake to avoid
$k\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} kp \\ kq \end{pmatrix} \ne \begin{pmatrix} kp \\ q \end{pmatrix}$
The commonest slip is scaling only the top component and copying the bottom one down unchanged. The scalar multiplies BOTH components.
Choose the option that matches
$2\mathbf{a} - \frac{1}{2}\mathbf{b} = \begin{pmatrix} -9 \\ 9 \end{pmatrix}$
The only option equal to $\begin{pmatrix} -9 \\ 9 \end{pmatrix}$ is the correct one; every other option differs in at least one component.

Answer

\begin{pmatrix} -9 \\ 9 \end{pmatrix}

Question 3

6 markschallenging

The diagram shows the points

O

A

B

C

and

D

\vec{OA} = \mathbf{a}

\vec{OB} = \mathbf{b}

\vec{OC} = 3\mathbf{a}

and

\vec{OD} = 3\mathbf{b}

. Express

\vec{CD}

in terms of

\mathbf{a}

and

\mathbf{b}

Show worked solution

Worked solution

Write down the vectors the diagram gives you
$\vec{OA} = \mathbf{a}, \quad \vec{OB} = \mathbf{b}, \quad \vec{OC} = 3\mathbf{a}, \quad \vec{OD} = 3\mathbf{b}$
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.
State the rule for travelling between two points
$\vec{PR} = \vec{PQ} + \vec{QR}, \quad \vec{QP} = -\vec{PQ}$
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.
Choose a route between the two points
$-\vec{OC} + \vec{OD}$
This route uses only arrows the question gives. A leg written with a minus sign is a leg travelled backwards.
Write each leg in terms of the named vectors
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
Substitute what the question says each arrow is, keeping the minus sign on any leg that is travelled backwards.
Collect like terms
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
Collecting the terms gives $-3\mathbf{a} + 3\mathbf{b}$ . This is an exact answer for EVERY choice of the two vectors — it does not depend on what they happen to be.
State the vector in terms of the named vectors
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
So $\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$ .
Check the answer on the diagram
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
Drawn on the grid, the answer is the arrow that closes the route — and it lands exactly where the algebra says it should. The diagram and the algebra have to agree.
Rule out the wrong option number 1
$3\mathbf{a} - 3\mathbf{b} \ne -3\mathbf{a} + 3\mathbf{b}$
$3\mathbf{a} - 3\mathbf{b}$ is what you get if this is the route walked backwards, so every sign is flipped. Walking the route again gives $-3\mathbf{a} + 3\mathbf{b}$ , so that option is wrong.
Rule out the wrong option number 2
$3\mathbf{a} + 3\mathbf{b} \ne -3\mathbf{a} + 3\mathbf{b}$
$3\mathbf{a} + 3\mathbf{b}$ is what you get if the two vectors have been added when the route runs AGAINST one of the arrows, so one of them has to be subtracted. Walking the route again gives $-3\mathbf{a} + 3\mathbf{b}$ , so that option is wrong.
Rule out the wrong option number 3
$3\mathbf{b} \ne -3\mathbf{a} + 3\mathbf{b}$
$3\mathbf{b}$ is what you get if the first leg has been left out of the route altogether. Walking the route again gives $-3\mathbf{a} + 3\mathbf{b}$ , so that option is wrong.
Rule out the wrong option number 4
$-3\mathbf{a} \ne -3\mathbf{a} + 3\mathbf{b}$
$-3\mathbf{a}$ is what you get if the second leg has been left out of the route altogether. Walking the route again gives $-3\mathbf{a} + 3\mathbf{b}$ , so that option is wrong.
Note the mistake to avoid
$\vec{CD} = -\vec{DC}$
Travelling the route the wrong way round negates the answer. Check which point you are starting from before you write anything down.
Note that the route chosen does not matter
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
Any route through the given arrows gives the same answer, because a displacement depends only on where the two points are, not on how you travel between them.
Summarise the method
$\text{walk a route} \rightarrow \text{add the arrows} \rightarrow \text{collect terms}$
Walk from the first point to the second through the given arrows, adding an arrow travelled forwards and subtracting one travelled backwards, then collect like terms.
Choose the matching option
$\vec{CD} = -3\mathbf{a} + 3\mathbf{b}$
The only option equal to $-3\mathbf{a} + 3\mathbf{b}$ is the correct one.

Answer

-3\mathbf{a} + 3\mathbf{b}

Question 4

5 markschallenging

The diagram shows the points

O

A

B

C

and

D

\vec{OA} = \mathbf{a}

\vec{AB} = 2\mathbf{b}

\vec{OC} = 3\mathbf{a}

and

\vec{CD} = 6\mathbf{b}

. Given that

\vec{OD} = k\,\vec{OB}

, work out the value of

k

Show worked solution

Worked solution

Write down the vectors the diagram gives you
$\vec{OA} = \mathbf{a}, \quad \vec{AB} = 2\mathbf{b}, \quad \vec{OC} = 3\mathbf{a}, \quad \vec{CD} = 6\mathbf{b}$
Every arrow is given in terms of the two named vectors, so every route through the figure can be written in terms of them too.
State the rule for travelling between two points
$\vec{PR} = \vec{PQ} + \vec{QR}, \quad \vec{QP} = -\vec{PQ}$
Walk a route through the given arrows: forwards along an arrow adds it, backwards along it subtracts it.
Work out the first vector
$\vec{OB} = \mathbf{a} + 2\mathbf{b}$
Walking from $O$ to $B$ through the given arrows gives $\mathbf{a} + 2\mathbf{b}$ .
Work out the second vector
$\vec{OD} = 3\mathbf{a} + 6\mathbf{b}$
The same method from $O$ to $D$ gives $3\mathbf{a} + 6\mathbf{b}$ .
Compare the two vectors
$3\mathbf{a} + 6\mathbf{b} = 3\left(\mathbf{a} + 2\mathbf{b}\right)$
The second vector is exactly $3$ times the first: the same scalar works on the $\mathbf{a}$ term and on the $\mathbf{b}$ term.
Check the scalar on the first named vector
$3 \times 1 = 3$
The coefficient of $\mathbf{a}$ multiplies up correctly.
Check the scalar on the second named vector
$3 \times 2 = 6$
The coefficient of $\mathbf{b}$ multiplies up by the SAME scalar. Both must agree, or no single $k$ exists.
Note that only one scalar can work
$k\mathbf{a} + 2\mathbf{b} = 3\mathbf{a} + 6\mathbf{b} \Rightarrow k = 3$
Both coefficients have to give the same $k$ . They do, so the two vectors really are parallel and $k$ is pinned down exactly.
Say what this proves
$\vec{OD} = 3\,\vec{OB} \Rightarrow \vec{OD} \parallel \vec{OB}$
Writing one vector as a scalar multiple of another PROVES the two are parallel. This is the whole engine of a vector proof: find the two vectors, show one is $k$ times the other.
Say what the scalar means for the lengths
$\left| \vec{OD} \right| = 3 \left| \vec{OB} \right|$
The scalar also fixes the ratio of the lengths: $\vec{OD}$ is $3$ times as long as $\vec{OB}$ , and points the same way.
See it on the diagram
$\vec{OD} = 3\,\vec{OB}$
On the grid the two arrows lie along parallel lines, and one is a scaled copy of the other. That is what a scalar multiple looks like.
Note the mistake to avoid
$k = \frac{\vec{OD}}{\vec{OB}} \ne \frac{\vec{OB}}{\vec{OD}}$
Divide the vector you are asked about by the one it is compared with, not the other way round — the reciprocal is a very easy mark to throw away.
Summarise the method
$\text{write both vectors in terms of the named ones} \rightarrow \text{divide}$
Write both vectors in terms of the same two named vectors, then read off the scalar that turns one into the other.
Check the answer once more
$3\left(\mathbf{a} + 2\mathbf{b}\right) = 3\mathbf{a} + 6\mathbf{b} \ \checkmark$
Multiplying $\vec{OB}$ by $3$ reproduces $\vec{OD}$ exactly, so the scalar is right.
State the value of k
$k = 3$
So $k = 3$ .

Answer

k = 3

Question 5

6 markschallenging

In the diagram

M

is the midpoint of

OA

and

N

is the midpoint of

OB

\vec{OA} = 2\mathbf{a}

\vec{OB} = 2\mathbf{b}

\vec{OM} = \mathbf{a}

and

\vec{ON} = \mathbf{b}

. Given that

\vec{MN} = k\,\vec{AB}

, work out the value of

k

Show worked solution

Worked solution

Write down the vectors the diagram gives you
$\vec{OA} = 2\mathbf{a}, \quad \vec{OB} = 2\mathbf{b}, \quad \vec{OM} = \mathbf{a}, \quad \vec{ON} = \mathbf{b}$
Every arrow is given in terms of the two named vectors, so every route through the figure can be written in terms of them too.
State the rule for travelling between two points
$\vec{PR} = \vec{PQ} + \vec{QR}, \quad \vec{QP} = -\vec{PQ}$
Walk a route through the given arrows: forwards along an arrow adds it, backwards along it subtracts it.
Work out the first vector
$\vec{AB} = -2\mathbf{a} + 2\mathbf{b}$
Walking from $A$ to $B$ through the given arrows gives $-2\mathbf{a} + 2\mathbf{b}$ .
Work out the second vector
$\vec{MN} = -\mathbf{a} + \mathbf{b}$
The same method from $M$ to $N$ gives $-\mathbf{a} + \mathbf{b}$ .
Compare the two vectors
$-\mathbf{a} + \mathbf{b} = \frac{1}{2}\left(-2\mathbf{a} + 2\mathbf{b}\right)$
The second vector is exactly $\frac{1}{2}$ times the first: the same scalar works on the $\mathbf{a}$ term and on the $\mathbf{b}$ term.
Check the scalar on the first named vector
$\frac{1}{2} \times -2 = -1$
The coefficient of $\mathbf{a}$ multiplies up correctly.
Check the scalar on the second named vector
$\frac{1}{2} \times 2 = 1$
The coefficient of $\mathbf{b}$ multiplies up by the SAME scalar. Both must agree, or no single $k$ exists.
Note that only one scalar can work
$k-2\mathbf{a} + 2\mathbf{b} = -\mathbf{a} + \mathbf{b} \Rightarrow k = \frac{1}{2}$
Both coefficients have to give the same $k$ . They do, so the two vectors really are parallel and $k$ is pinned down exactly.
Say what this proves
$\vec{MN} = \frac{1}{2}\,\vec{AB} \Rightarrow \vec{MN} \parallel \vec{AB}$
Writing one vector as a scalar multiple of another PROVES the two are parallel. This is the whole engine of a vector proof: find the two vectors, show one is $k$ times the other.
Say what the scalar means for the lengths
$\left| \vec{MN} \right| = \frac{1}{2} \left| \vec{AB} \right|$
The scalar also fixes the ratio of the lengths: $\vec{MN}$ is $\frac{1}{2}$ times as long as $\vec{AB}$ , and points the same way.
See it on the diagram
$\vec{MN} = \frac{1}{2}\,\vec{AB}$
On the grid the two arrows lie along parallel lines, and one is a scaled copy of the other. That is what a scalar multiple looks like.
Note the mistake to avoid
$k = \frac{\vec{MN}}{\vec{AB}} \ne \frac{\vec{AB}}{\vec{MN}}$
Divide the vector you are asked about by the one it is compared with, not the other way round — the reciprocal is a very easy mark to throw away.
Summarise the method
$\text{write both vectors in terms of the named ones} \rightarrow \text{divide}$
Write both vectors in terms of the same two named vectors, then read off the scalar that turns one into the other.
Check the answer once more
$\frac{1}{2}\left(-2\mathbf{a} + 2\mathbf{b}\right) = -\mathbf{a} + \mathbf{b} \ \checkmark$
Multiplying $\vec{AB}$ by $\frac{1}{2}$ reproduces $\vec{MN}$ exactly, so the scalar is right.
State the value of k
$k = \frac{1}{2}$
So $k = \frac{1}{2}$ .

Answer

k = \frac{1}{2}

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