What is 10.4: Position vectors?

The position vector of a point A is vector OA from a fixed origin O; if these vectors are a and b, then vector AB=b-a.

How many marks are 10.4 questions worth?

10.4 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 10.4 appeared in?

We have not yet indexed a verified past-paper appearance for 10.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

10.4

Understand and use position vectors; calculate the distance between two points represented by position vectors.

Draft — not yet indexed

Position vectors

Worked answers and methods for 10.4 on Edexcel A-level Maths 9MA0.

Explanation

The position vector of a point $A$ is $\overrightarrow{OA}$ from a fixed origin $O$ ; if these vectors are $\mathbf{a}$ and $\mathbf{b}$ , then $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .
Find a displacement by subtracting position vectors in final-minus-initial order, then find distance by taking the magnitude of that displacement.
A point dividing $AB$ internally in the fraction $t$ from $A$ to $B$ has position vector $\mathbf{a}+t(\mathbf{b}-\mathbf{a})$ .
Distance is a non-negative scalar, not a vector; a common error is to quote $\mathbf{b}-\mathbf{a}$ as the distance without calculating its magnitude.

Worked example

The position vectors of $A$ and $B$ are $(-1,2,4)$ and $(3,-4,7)$ . Calculate the exact distance $AB$ .

1. $\overrightarrow{AB}=(3-(-1),-4-2,7-4)=(4,-6,3)$ .
2.Hence $AB=|\overrightarrow{AB}|=\sqrt{4^2+(-6)^2+3^2}=\sqrt{61}$ .

Answer: $AB=\sqrt{61}$

Common mistakes

Don't square the coordinates of the two points separately instead of squaring their component differences.
Don't find the difference of position vectors but forget to take its Euclidean magnitude.

Exam tip

Subtract the endpoint position vectors in a consistent order, then square and sum every component for the distance.

Worked practice

Tier 1 · Easy

Points

A

and

B

have position vectors

(2,-1)

and

(7,3)

. Find

\overrightarrow{AB}

and the exact distance

AB

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\overrightarrow{AB}=(5,4)$ and $AB=\sqrt{41}$	3
Notes
Subtract the position vector of $A$ from that of $B$ : $\overrightarrow{AB}=(7-2,3-(-1))=(5,4)$ . Therefore $AB=\sqrt{5^2+4^2}=\sqrt{41}$ .

(3 marks)

Tier 2 · Standard

The position vectors of

A

and

B

are

(-2,1,5)

and

(6,5,-3)

. Find the midpoint of

AB

and the point

P

that divides

AB

internally in the ratio

AP:PB=3:1

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Midpoint $(2,3,1)$ $P=(4,4,-1)$	4
Notes
The midpoint is the componentwise average: $((-2+6)/2,(1+5)/2,(5-3)/2)=(2,3,1)$ . Also $\overrightarrow{AB}=(8,4,-8)$ . Since $AP$ is three quarters of $AB$ , $\overrightarrow{OP}=(-2,1,5)+\tfrac34(8,4,-8)=(4,4,-1)$ .

(4 marks)

Tier 3 · Hard

Points

A

and

B

have position vectors

(1,-2,3)

and

(9,6,-1)

. The point

M

divides

AB

internally in the ratio

AM:MB=3:1

. Find the position vector of

M

and the exact distance

OM

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\overrightarrow{OM}=(7,4,0)$ and $OM=\sqrt{65}$	5
Notes
Since $AM$ is three of the four equal ratio parts, $\overrightarrow{OM}=\overrightarrow{OA}+\frac34(\overrightarrow{OB}-\overrightarrow{OA})$ . Now $\overrightarrow{OB}-\overrightarrow{OA}=(8,8,-4)$ , so $\overrightarrow{OM}=(1,-2,3)+(6,6,-3)=(7,4,0)$ . Thus $OM=\sqrt{7^2+4^2}=\sqrt{65}$ .

(5 marks)

Tier 1 · Easy

Point

A

has position vector

(3,-2,1)

and

\overrightarrow{AB}=(-5,4,2)

. Find the position vector of

B

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\overrightarrow{OB}=(-2,2,3)$	2
Notes
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=(3,-2,1)+(-5,4,2)=(-2,2,3)$ .

(2 marks)

Tier 2 · Standard

Points

A

and

B

have position vectors

(1,2,-1)

and

(5,-2,3)

. The point

P

has position vector

(t,0,1)

and is equidistant from

A

and

B

. Find

t

and the exact distance

AP

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$t=3$ and $AP=2\sqrt3$	4
Notes
$AP^2=(t-1)^2+(0-2)^2+(1-(-1))^2=(t-1)^2+8$ , while $BP^2=(t-5)^2+(0+2)^2+(1-3)^2=(t-5)^2+8$ . Equating these gives $(t-1)^2=(t-5)^2$ , hence $t=3$ . Then $AP=\sqrt{(3-1)^2+8}=\sqrt{12}=2\sqrt3$ .

(4 marks)

Tier 3 · Hard

The position vectors of

A

B

and

C

are

(1,0,2)

(3,1,4)

and

(-1,2,3)

respectively. By calculating exact distances, show that triangle

ABC

is right-angled. Hence find the position vector of the midpoint

M

of the hypotenuse and verify that

MA=MB=MC

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$AB=AC=3$ and $BC=3\sqrt2$ , so the triangle is right-angled at $A$ . $\overrightarrow{OM}=(1,\tfrac32,\tfrac72)$ and $MA=MB=MC=\dfrac{3\sqrt2}{2}$ .	6
Notes
$\overrightarrow{AB}=(2,1,2)$ and $\overrightarrow{AC}=(-2,2,1)$ , so $AB=AC=\sqrt9=3$ . Also $\overrightarrow{BC}=(-4,1,-1)$ , so $BC=\sqrt{18}=3\sqrt2$ . Since $AB^2+AC^2=9+9=18=BC^2$ , the right angle is at $A$ and $BC$ is the hypotenuse. Its midpoint is $M=((3-1)/2,(1+2)/2,(4+3)/2)=(1,3/2,7/2)$ . Now $\overrightarrow{AM}=(0,3/2,3/2)$ has magnitude $3\sqrt2/2$ , while $MB=MC=BC/2=3\sqrt2/2$ .

(6 marks)

Tier 2 · Standard

Points

A

and

B

have position vectors

(t,2,-1)

and

(3,-2,5)

respectively. Given that

AB=2\sqrt{17}

, find all possible values of

t

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$t=-1$ or $t=7$	4
Notes
$\overrightarrow{AB}=(3-t,-4,6)$ , so $AB^2=(3-t)^2+16+36$ . Since $AB=2\sqrt{17}$ , $(3-t)^2+52=68$ , giving $(3-t)^2=16$ . Hence $3-t=\pm4$ , so $t=-1$ or $t=7$ .

(4 marks)

Tier 3 · Hard

Points

A

and

B

have position vectors

(1,-2,3)

and

(4,4,-3)

. Point

P

lies beyond

B

on the ray from

A

through

B

, and

AP:PB=2:1

. Find the position vector of

P

and verify the ratio by calculating the exact distances

AP

and

PB

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\overrightarrow{OP}=(7,10,-9)$ , with $AP=18$ and $PB=9$ , so $AP:PB=2:1$ .	5
Notes
$\overrightarrow{AB}=(3,6,-6)$ . Since $P$ lies beyond $B$ and $AP=2PB$ , $\overrightarrow{AP}=2\overrightarrow{AB}=(6,12,-12)$ . Hence $\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=(7,10,-9)$ . Now $AP=\sqrt{6^2+12^2+(-12)^2}=18$ , while $\overrightarrow{BP}=(3,6,-6)$ gives $PB=9$ . Thus the required ratio is verified.

(5 marks)

Tier 3 · Hard

The position vectors of three points are

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

\mathbf{b}=(4,6,1)

and

\mathbf{c}=(10,14,5)

. Show that their points

A

B

and

C

are collinear with

B

between the other two. Calculate the exact distances

AB

BC

and

AC

, and hence find

AB:BC

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\overrightarrow{AB}=(3,4,2)$ and $\overrightarrow{BC}=2\overrightarrow{AB}$ , so the points are collinear with $B$ between $A$ and $C$ . $AB=\sqrt{29}$ , $BC=2\sqrt{29}$ , $AC=3\sqrt{29}$ , so $AB:BC=1:2$ .	5
Notes
$\overrightarrow{AB}=(4,6,1)-(1,2,-1)=(3,4,2)$ and $\overrightarrow{BC}=(10,14,5)-(4,6,1)=(6,8,4)=2\overrightarrow{AB}$ . The positive scalar multiple shows that the directions agree, so $A,B,C$ are collinear with $B$ between the other two points. The distances are $AB=\sqrt{3^2+4^2+2^2}=\sqrt{29}$ , $BC=\sqrt{6^2+8^2+4^2}=2\sqrt{29}$ and $AC=\sqrt{9^2+12^2+6^2}=3\sqrt{29}$ . Therefore $AB:BC=1:2$ .

(5 marks)

Q10

Tier 3 · Hard

10.

In three-dimensional coordinate space, let

A=(0,0,0)

B=(4,0,0)

and

C=(1,k,2)

. Find every value of

k

for which triangle

ABC

is isosceles. State the equal sides in each case.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$k=\pm\sqrt{11}$ gives $AB=AC$ , and $k=\pm\sqrt3$ gives $AB=BC$ .	6
Notes
The squared side lengths are $AB^2=16$ , $AC^2=1+k^2+4=k^2+5$ and $BC^2=(-3)^2+k^2+2^2=k^2+13$ . The condition $AB=AC$ gives $16=k^2+5$ , so $k=\pm\sqrt{11}$ . The condition $AB=BC$ gives $16=k^2+13$ , so $k=\pm\sqrt3$ . The equation $AC=BC$ would require $k^2+5=k^2+13$ , which is impossible. These four values are therefore all the possibilities.

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 10.4. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 10 Vectors

Want help turning this into marks?

Bring 10.4 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call