What is 10.3: Vector arithmetic?

Vector addition combines successive displacements: placing the tail of b at the head of a makes the resultant from the first tail to the final head equal to a+b.

How many marks are 10.3 questions worth?

10.3 practice drills on this page are worth 1–6 marks each; marks in a real exam vary by question.

Which past exam papers has 10.3 appeared in?

We have not yet indexed a verified past-paper appearance for 10.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

10.3

Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Draft — not yet indexed

Vector arithmetic

Worked answers and methods for 10.3 on Edexcel A-level Maths 9MA0.

Explanation

Vector addition combines successive displacements: placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$ makes the resultant from the first tail to the final head equal to $\mathbf{a}+\mathbf{b}$ .
Add corresponding components and multiply every component by a scalar; subtraction is addition of the opposite vector.
The vector $k\mathbf{a}$ is parallel to $\mathbf{a}$ , has magnitude $|k||\mathbf{a}|$ , and points in the reverse direction when $k<0$ .
A common error is to multiply only one component by a scalar or to draw vectors head-to-head instead of using a head-to-tail or parallelogram construction.

The head-to-tail construction places

\mathbf v

at the head of

2\mathbf u

; the direct arrow is their resultant.

Worked example

Let $\mathbf{u}=(4,1)$ and $\mathbf{v}=(-2,5)$ . Find $2\mathbf{u}+\mathbf{v}$ and describe a head-to-tail construction for this resultant.

1.Calculate $2\mathbf{u}=(8,2)$ , then add $\mathbf{v}$ to obtain $(8-2,2+5)=(6,7)$ .
2.Diagrammatically, translate the vectors without rotating them and place the second $\mathbf{u}$ after the first, followed by $\mathbf{v}$ ; the direct closing vector is the sum.

Answer: $2\mathbf{u}+\mathbf{v}=(6,7)$ .; Place two copies of $\mathbf{u}$ and then one copy of $\mathbf{v}$ head-to-tail; the resultant joins the initial tail to the final head.

Common mistakes

Don't reverse a displacement vector in the diagram but leave its algebraic sign unchanged.
Don't add vector diagrams tail-to-tail without completing the parallelogram or head-to-tail construction.

Exam tip

For a resultant, scale each vector first and join them head-to-tail in the order represented algebraically.

Worked practice

Tier 1 · Easy

Given

\mathbf{u}=(2,-3)

and

\mathbf{v}=(-5,4)

, find

\mathbf{u}+\mathbf{v}

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$(-3,1)$	1
Notes
Add corresponding components: $\mathbf{u}+\mathbf{v}=(2+(-5),-3+4)=(-3,1)$ .

(1 mark)

Tier 2 · Standard

In triangle

ABC

\overrightarrow{AB}=\mathbf{u}

and

\overrightarrow{AC}=\mathbf{v}

. The point

M

is the midpoint of

BC

. Express

\overrightarrow{AM}

in terms of

\mathbf{u}

and

\mathbf{v}

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\overrightarrow{AM}=\dfrac12(\mathbf{u}+\mathbf{v})$	3
Notes
Take $A$ as the origin. Then the position vectors of $B$ and $C$ are $\mathbf{u}$ and $\mathbf{v}$ . The midpoint has position vector equal to their average, so $\overrightarrow{AM}=\tfrac12(\mathbf{u}+\mathbf{v})$ .

(3 marks)

Tier 3 · Hard

The vectors

\mathbf{a}=(1,2)

and

\mathbf{b}=(3,-1)

combine to give

\mathbf{w}=(11,1)

. Find

\alpha

and

\beta

such that

\mathbf{w}=\alpha\mathbf{a}+\beta\mathbf{b}

, and interpret the result geometrically.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\alpha=2$ and $\beta=3$ , so $\mathbf{w}=2\mathbf{a}+3\mathbf{b}$ .	4
Notes
Equating components gives $\alpha+3\beta=11$ and $2\alpha-\beta=1$ . From the second, $\beta=2\alpha-1$ ; substitution gives $7\alpha=14$ , hence $\alpha=2$ and $\beta=3$ . Geometrically, two copies of $\mathbf{a}$ and three copies of $\mathbf{b}$ placed head-to-tail have resultant $\mathbf{w}$ .

(4 marks)

Tier 1 · Easy

Given

\mathbf{u}=(3,-1)

and

\mathbf{w}=(-6,2)

, express

\mathbf{w}

as a scalar multiple of

\mathbf{u}

and describe its direction and magnitude relative to

\mathbf{u}

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\mathbf{w}=-2\mathbf{u}$ , so it points in the opposite direction and has twice the magnitude of $\mathbf{u}$ .	2
Notes
Multiplying both components of $\mathbf{u}$ by $-2$ gives $(-6,2)=\mathbf{w}$ . A negative scalar reverses direction, and its absolute value $2$ doubles the magnitude.

(2 marks)

Tier 2 · Standard

Three successive displacements are

(4,-1)

(-2,5)

and

(-3,-2)

. Find their resultant and the single displacement that would return directly to the starting point. Describe how the four vectors form a closed head-to-tail diagram.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	The resultant is $(-1,2)$ and the return displacement is $(1,-2)$ ; placed head-to-tail, the return vector closes the path.	3
Notes
Add corresponding components: $(4,-1)+(-2,5)+(-3,-2)=(-1,2)$ . The displacement back to the start is the negative of this resultant, $(1,-2)$ . Drawing this return vector from the final head to the initial tail makes the four-vector path closed.

(3 marks)

Tier 3 · Hard

Let

\mathbf{u}=(2,1)

and

\mathbf{v}=(1,-2)

. The vector

\mathbf{r}=2\mathbf{u}+\lambda\mathbf{v}

is parallel to

(1,3)

and points in the same direction. Find

\lambda

, express

\mathbf{r}

as a scalar multiple of

(1,3)

, and interpret the result geometrically.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\lambda=-2$ and $\mathbf{r}=2(1,3)=(2,6)$ ; the resultant is parallel to $(1,3)$ , in the same direction and twice its magnitude.	4
Notes
$\mathbf{r}=(4+\lambda,2-2\lambda)$ . If $\mathbf{r}=k(1,3)$ , then $4+\lambda=k$ and $2-2\lambda=3k$ . Eliminating $k$ gives $2-2\lambda=12+3\lambda$ , so $\lambda=-2$ and $k=2$ . Thus $\mathbf{r}=(2,6)=2(1,3)$ , which proves the stated parallel, same-direction relationship and the factor-two scaling.

(4 marks)

Tier 2 · Standard

Vectors

\overrightarrow{OA}=\mathbf{a}=(5,2)

and

\overrightarrow{OB}=\mathbf{b}=(1,-3)

are drawn from the common point

O

. Find

\mathbf{a}-\mathbf{b}

and identify the directed segment between

A

and

B

that this vector represents.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\mathbf{a}-\mathbf{b}=(4,5)=\overrightarrow{BA}$	3
Notes
$\mathbf{a}-\mathbf{b}=(5,2)-(1,-3)=(4,5)$ . Since $\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$ , the difference is the directed segment from $B$ to $A$ .

(3 marks)

Tier 3 · Hard

In triangle

ABC

\overrightarrow{AB}=\mathbf{a}

and

\overrightarrow{AC}=\mathbf{b}

. Point

D

divides

AB

internally in the ratio

AD:DB=2:1

, and point

E

divides

AC

internally in the ratio

AE:EC=2:1

. Express

\overrightarrow{DE}

in terms of

\mathbf{a}

and

\mathbf{b}

. Hence show that

DE

is parallel to

BC

and find

DE:BC

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\overrightarrow{DE}=\dfrac23(\mathbf{b}-\mathbf{a})=\dfrac23\overrightarrow{BC}$ , so $DE$ is parallel to $BC$ and $DE:BC=2:3$ .	5
Notes
$\overrightarrow{AD}=\frac23\mathbf{a}$ and $\overrightarrow{AE}=\frac23\mathbf{b}$ . Therefore $\overrightarrow{DE}=\overrightarrow{AE}-\overrightarrow{AD}=\frac23(\mathbf{b}-\mathbf{a})$ . Also $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ , so $\overrightarrow{DE}=\frac23\overrightarrow{BC}$ . The positive scalar multiple proves that the segments are parallel in the same direction, and their lengths are in the ratio $2:3$ .

(5 marks)

Tier 3 · Hard

In triangle

OAB

, the position vectors of

A

and

B

are

\mathbf{a}

and

\mathbf{b}

. Point

M

is the midpoint of

AB

, and point

N

lies on

OM

with

ON:NM=2:1

. Point

P

is the midpoint of

OB

. Find the position vector of

N

. Hence prove that

A

N

and

P

are collinear and find

AN:NP

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\overrightarrow{ON}=\dfrac13(\mathbf{a}+\mathbf{b})$ ; $A,N,P$ are collinear and $AN:NP=2:1$ .	6
Notes
Since $M$ is the midpoint of $AB$ , $\overrightarrow{OM}=(\mathbf{a}+\mathbf{b})/2$ . The ratio $ON:NM=2:1$ gives $\overrightarrow{ON}=(2/3)\overrightarrow{OM}=(\mathbf{a}+\mathbf{b})/3$ . Also $\overrightarrow{OP}=\mathbf{b}/2$ . Therefore $\overrightarrow{AN}=\overrightarrow{ON}-\overrightarrow{OA}=(\mathbf{b}-2\mathbf{a})/3$ , while $\overrightarrow{NP}=\overrightarrow{OP}-\overrightarrow{ON}=(\mathbf{b}-2\mathbf{a})/6$ . Thus $\overrightarrow{AN}=2\overrightarrow{NP}$ , proving that $A,N,P$ are collinear in that order and $AN:NP=2:1$ .

(6 marks)

Q10

Tier 3 · Hard

10.

The position vectors of the vertices

A

B

C

and

D

of a quadrilateral are

\mathbf{a}

\mathbf{b}

\mathbf{c}

and

\mathbf{d}

, where

AC

and

BD

are not parallel. Points

P

Q

R

and

S

are the midpoints of

AB

BC

CD

and

DA

, respectively. Use vectors to prove that

PQRS

is a parallelogram.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\overrightarrow{PQ}=\overrightarrow{SR}=\dfrac12(\mathbf{c}-\mathbf{a})$ and $\overrightarrow{QR}=\overrightarrow{PS}=\dfrac12(\mathbf{d}-\mathbf{b})$ , so $PQRS$ is a parallelogram.	5
Notes
The midpoint position vectors are $(\mathbf{a}+\mathbf{b})/2$ , $(\mathbf{b}+\mathbf{c})/2$ , $(\mathbf{c}+\mathbf{d})/2$ and $(\mathbf{d}+\mathbf{a})/2$ . Hence $\overrightarrow{PQ}=(\mathbf{c}-\mathbf{a})/2$ and $\overrightarrow{SR}=(\mathbf{c}-\mathbf{a})/2$ . Similarly, $\overrightarrow{QR}=(\mathbf{d}-\mathbf{b})/2$ and $\overrightarrow{PS}=(\mathbf{d}-\mathbf{b})/2$ . Both pairs of opposite directed sides are equal, so $PQRS$ is a parallelogram.

(5 marks)

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