What is 10.1: Vectors in 2D and 3D?

A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors i, j and k.

How many marks are 10.1 questions worth?

10.1 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 10.1 appeared in?

We have not yet indexed a verified past-paper appearance for 10.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

10.1

Use vectors in two dimensions and in three dimensions.

Draft — not yet indexed

Vectors in 2D and 3D

Worked answers and methods for 10.1 on Edexcel A-level Maths 9MA0.

Explanation

A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ .
Work component by component, keeping the $x$ , $y$ and $z$ entries aligned; subtract the initial point from the final point to form a displacement vector.
For example, from $A(1,2,3)$ to $B(4,0,5)$ the displacement is $\overrightarrow{AB}=(3,-2,2)$ .
A vector has no fixed location, whereas a point does; a common error is to confuse the coordinates of an endpoint with the components of the displacement leading to it.

Worked example

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $2\mathbf{a}-\mathbf{b}$ .

1.First $2\mathbf{a}=(4,-2,6)$ .
2.Subtract corresponding components of $\mathbf{b}$ to get $(4-(-1),-2-4,6-2)=(5,-6,4)$ .

Answer: $2\mathbf{a}-\mathbf{b}=(5,-6,4)$

Common mistakes

Don't change the coordinate order between vectors, so components from different axes are combined.
Don't combine vector components inconsistently, applying a scalar to only one coordinate.

Exam tip

Perform each vector operation component by component and preserve the coordinate order throughout.

Worked practice

Tier 1 · Easy

The point

A

has coordinates

(-2,5)

and the point

B

has coordinates

(4,1)

. Find

\overrightarrow{AB}

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\overrightarrow{AB}=(6,-4)$	2
Notes
Subtract the coordinates of $A$ from those of $B$ : $\overrightarrow{AB}=(4-(-2),1-5)=(6,-4)$ .

(2 marks)

Tier 2 · Standard

Given

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

\mathbf{b}=(2,-1,3)

and

\mathbf{c}=(5,0,5)

, find scalars

\lambda

and

\mu

such that

\lambda\mathbf{a}+\mu\mathbf{b}=\mathbf{c}

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\lambda=1$ , $\mu=2$	4
Notes
Comparing the first two components gives $\lambda+2\mu=5$ and $2\lambda-\mu=0$ . The second equation gives $\mu=2\lambda$ , so the first gives $5\lambda=5$ and $\lambda=1$ , $\mu=2$ . The third component checks: $-\lambda+3\mu=-1+6=5$ .

(4 marks)

Tier 3 · Hard

Let

\mathbf{p}=(1,1,2)

and

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

. By writing

(u,v,w)=\lambda\mathbf{p}+\mu\mathbf{q}

, show that

(u,v,w)

lies in the span of

\mathbf{p}

and

\mathbf{q}

if and only if

u+v-w=0

. Hence find the value of

k

for which

(7,1,k)

lies in this span and the value of

k

for which

(5,k,7)

lies in this span. Can the same value of

k

satisfy both conditions?

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	The spanning condition is $u+v-w=0$ . $(7,1,k)$ requires $k=8$ , whereas $(5,k,7)$ requires $k=2$ ; no single value of $k$ satisfies both.	4
Notes
A linear combination has components $(u,v,w)=(\lambda+2\mu,\lambda-\mu,2\lambda+\mu)$ , so $u+v-w=0$ . Conversely, if $u+v-w=0$ , choose $\lambda=(u+2v)/3$ and $\mu=(u-v)/3$ ; these give the stated first two components and $2\lambda+\mu=u+v=w$ . Thus the condition is necessary and sufficient. For $(7,1,k)$ it gives $7+1-k=0$ , so $k=8$ . For $(5,k,7)$ it gives $5+k-7=0$ , so $k=2$ . Since $8\ne2$ , there is no common value.

(4 marks)

Tier 1 · Easy

Given

\mathbf{a}=-4\mathbf{i}+7\mathbf{j}-2\mathbf{k}

, write

\mathbf{a}

in component form and find

3\mathbf{a}

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\mathbf{a}=(-4,7,-2)$ and $3\mathbf{a}=(-12,21,-6)$	2
Notes
The coefficients of $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ are the three components, so $\mathbf{a}=(-4,7,-2)$ . Multiplying every component by $3$ gives $3\mathbf{a}=(-12,21,-6)$ .

(2 marks)

Tier 2 · Standard

Point

A

has coordinates

(2,-1,4)

and

\overrightarrow{AB}=(-3,5,2)

. Find the coordinates of

B

and find

\overrightarrow{BA}

. Verify that

\overrightarrow{AB}+\overrightarrow{BA}=\mathbf{0}

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$B=(-1,4,6)$ and $\overrightarrow{BA}=(3,-5,-2)$ ; their sum is $(0,0,0)$ .	3
Notes
Add the displacement to the coordinates of $A$ : $B=(2,-1,4)+(-3,5,2)=(-1,4,6)$ . Reversing a displacement changes its sign, so $\overrightarrow{BA}=(3,-5,-2)$ . Componentwise addition gives $(-3,5,2)+(3,-5,-2)=(0,0,0)$ .

(3 marks)

Tier 3 · Hard

A point starts at

A=(1,-2,5)

and undergoes, in order, displacements

\mathbf{u}=(3,1,-2)

\lambda\mathbf{v}

where

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

, and

\mathbf{w}=(2,-3,-4)

. The final point lies in the plane

z=0

. Find

\lambda

, the final point and the single displacement from

A

to the final point.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\lambda=1$ , the final point is $(5,-2,0)$ and the single displacement is $(4,0,-5)$ .	5
Notes
The final $z$ -coordinate is $5-2+\lambda-4=\lambda-1$ . Since it is zero, $\lambda=1$ . The total displacement is $(3,1,-2)+(-1,2,1)+(2,-3,-4)=(4,0,-5)$ . Adding this to $A$ gives the final point $(1,-2,5)+(4,0,-5)=(5,-2,0)$ .

(5 marks)

Tier 2 · Standard

Given

\mathbf{a}=(p,2q,-3)

\mathbf{b}=(4,-1,r)

and

\mathbf{a}+\mathbf{b}=(7,5,2)

, find

p

q

and

r

. Hence find

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

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$p=3$ , $q=3$ , $r=5$ and $2\mathbf{a}-\mathbf{b}=(2,13,-11)$	4
Notes
Equality of corresponding components gives $p+4=7$ , $2q-1=5$ and $-3+r=2$ , so $p=3$ , $q=3$ and $r=5$ . Hence $\mathbf{a}=(3,6,-3)$ and $\mathbf{b}=(4,-1,5)$ , giving $2\mathbf{a}-\mathbf{b}=(6,12,-6)-(4,-1,5)=(2,13,-11)$ .

(4 marks)

Tier 3 · Hard

The vectors

\mathbf{a}

and

\mathbf{b}

satisfy

2\mathbf{a}+\mathbf{b}=(7,1,5)

and

\mathbf{a}-2\mathbf{b}=(1,-12,10)

. Find

\mathbf{a}

and

\mathbf{b}

, and verify both vector equations.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\mathbf{a}=(3,-2,4)$ and $\mathbf{b}=(1,5,-3)$ Check: $2(3,-2,4)+(1,5,-3)=(7,1,5)$ and $(3,-2,4)-2(1,5,-3)=(1,-12,10)$ .	4
Notes
Write $\mathbf{u}=(7,1,5)$ and $\mathbf{v}=(1,-12,10)$ . Doubling $2\mathbf{a}+\mathbf{b}=\mathbf{u}$ and adding $\mathbf{a}-2\mathbf{b}=\mathbf{v}$ gives $5\mathbf{a}=2\mathbf{u}+\mathbf{v}=(15,-10,20)$ , so $\mathbf{a}=(3,-2,4)$ . Then $\mathbf{b}=\mathbf{u}-2\mathbf{a}=(1,5,-3)$ . Checking, $2\mathbf{a}+\mathbf{b}=(7,1,5)$ and $\mathbf{a}-2\mathbf{b}=(1,-12,10)$ .

(4 marks)

Tier 3 · Hard

Let

\mathbf{u}=(1,1,0)

\mathbf{v}=(0,1,1)

and

\mathbf{w}=(1,0,1)

. A displacement

(7,8,9)

is formed as

p\mathbf{u}+q\mathbf{v}+r\mathbf{w}

. Find

p

q

and

r

, and verify the displacement component by component.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$p=3$ , $q=5$ and $r=4$ .	5
Notes
Comparing components in $p\mathbf{u}+q\mathbf{v}+r\mathbf{w}=(p+r,p+q,q+r)$ gives $p+r=7$ , $p+q=8$ and $q+r=9$ . Adding the first two equations and subtracting the third gives $2p=6$ , so $p=3$ . It follows that $q=5$ and $r=4$ . The check is $3(1,1,0)+5(0,1,1)+4(1,0,1)=(3+4,3+5,5+4)=(7,8,9)$ .

(5 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{a}=(p,q,r)

\mathbf{b}=(q,r,p)

and

\mathbf{c}=(r,p,q)

satisfy

\mathbf{a}+\mathbf{b}+\mathbf{c}=(13,13,13)

and

\mathbf{a}-\mathbf{b}=(1,-3,2)

. Find

\mathbf{a}

\mathbf{b}

and

\mathbf{c}

, and verify both vector equations.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\mathbf{a}=(4,3,6)$ , $\mathbf{b}=(3,6,4)$ and $\mathbf{c}=(6,4,3)$ .	5
Notes
The sum equation gives $p+q+r=13$ . The difference equation gives $p-q=1$ , $q-r=-3$ and $r-p=2$ . Thus $p=q+1$ and $r=q+3$ , so $(q+1)+q+(q+3)=13$ and $q=3$ . Hence $p=4$ and $r=6$ . Therefore $\mathbf{a}=(4,3,6)$ , $\mathbf{b}=(3,6,4)$ and $\mathbf{c}=(6,4,3)$ . Their sum is $(13,13,13)$ and $\mathbf{a}-\mathbf{b}=(1,-3,2)$ , as required.

(5 marks)

Verified exam appearances

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