Given a=2i + 5j and b=4i - j, find a+b.

Write each vector in component form: a=2i + 5j, b=4i - j Add the i components and the j components: (2+4)i+(5+(-1))j Simplify: 6i+4j Answer: 6i+4j

Given a=2i-3j and b=i+7j, find |a+b|.

Add the vectors first: a+b=(2+1)i+(-3+7)j=3i+4j Apply the magnitude formula: |a+b|=√(3^2+4^2) Evaluate: √(9+16)=√(25)=5 Answer: 5

OABC is a parallelogram with OA=a and OC=c (so OB=a+c). M is the midpoint of BC. The line OM meets the diagonal AC at X. Find the ratio OX:XM.

Understand what is being asked: a_1+td_1=a_2+sd_2 Recall the key result we will use: a_1+td_1=a_2+sd_2 Set up base vectors: OA=a, OC=c Position vector of B: OB=a+c Find M, the midpoint of BC: OM=12(OB+OC)=12a+c Describe line OM: OX=(12a+c) Describe diagonal AC: OX=a+(c-a)=(1-)a+c Equate coefficients of a: 2=1- Equate coefficients of c: = Solve: 2=1- 32=1 =23 Why this method works: a_1+td_1=a_2+sd_2 Link to earlier work: a_1+td_1=a_2+sd_2 Write down what we are working with: list the given vectors / points before substituting Check the answer: OX:XM=2:1 Interpret the result: OX:XM=2:1 Watch out for a common slip: a_1+td_1=a_2+sd_2 Answer: OX:XM=2:1

A-Level Vector geometry Practice Questions

Free A-Level Vector geometry practice questions with full step-by-step worked solutions. Covers vector-addition, components, vector-subtraction, column-vectors. Practise exam-style problems and check your method.

vector-additioncomponentsvector-subtractioncolumn-vectorsscalar-multiplelinear-combination

A-Level70 questionsStep-by-step solutions

Question 1

2 markseasy

Given

\mathbf{a}=2\mathbf{i} + 5\mathbf{j}

and

\mathbf{b}=4\mathbf{i} - \mathbf{j}

, find

\mathbf{a}+\mathbf{b}

Show worked solution

Worked solution

Write each vector in component form
$\mathbf{a}=2\mathbf{i} + 5\mathbf{j},\quad \mathbf{b}=4\mathbf{i} - \mathbf{j}$
We line the vectors up by their $\mathbf{i}$ (across) and $\mathbf{j}$ (up) parts so we can add matching parts together.
Add the $\mathbf{i}$ components and the $\mathbf{j}$ components
$(2+4)\mathbf{i}+(5+(-1))\mathbf{j}$
Adding vectors just means adding the across-parts together and the up-parts together separately.
Simplify
$6\mathbf{i}+4\mathbf{j}$
This single vector is the result of the addition.

Answer

6\mathbf{i}+4\mathbf{j}

Question 2

3 markseasy

Given

\mathbf{a}=2\mathbf{i}-3\mathbf{j}

and

\mathbf{b}=\mathbf{i}+7\mathbf{j}

, find

|\mathbf{a}+\mathbf{b}|

Show worked solution

Worked solution

Add the vectors first
$\mathbf{a}+\mathbf{b}=(2+1)\mathbf{i}+(-3+7)\mathbf{j}=3\mathbf{i}+4\mathbf{j}$
We must combine the vectors before finding the length of the result.
Apply the magnitude formula
$|\mathbf{a}+\mathbf{b}|=\sqrt{3^2+4^2}$
Now take the length of the single resultant vector using Pythagoras.
Evaluate
$\sqrt{9+16}=\sqrt{25}=5$
The resultant has length 5.

Answer

5

Question 3

3 marksintermediate

The point

P(4,5)

lies on

AB

with

A(2,3)

and

B(10,11)

. Find

AP:PB

Show worked solution

Worked solution

Understand what is being asked
$\overrightarrow{OP}=\mathbf{a}+\tfrac{m}{m+n}\overrightarrow{AB}$
Before calculating, we identify the goal and choose the method: the section (ratio) formula along the line. Having a plan stops us getting lost part-way through.
Recall the key result we will use
$\overrightarrow{OP}=\mathbf{a}+\tfrac{m}{m+n}\overrightarrow{AB}$
A point dividing $AB$ in the ratio $m:n$ is that fraction of the way from $A$ to $B$ .
Find $\overrightarrow{AP}$
$\overrightarrow{AP}=\begin{pmatrix} 4 \\ 5 \end{pmatrix}-\begin{pmatrix} 2 \\ 3 \end{pmatrix}=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$
Step from $A$ to $P$ .
Find $\overrightarrow{PB}$
$\overrightarrow{PB}=\begin{pmatrix} 10 \\ 11 \end{pmatrix}-\begin{pmatrix} 4 \\ 5 \end{pmatrix}=\begin{pmatrix} 6 \\ 6 \end{pmatrix}$
Step from $P$ to $B$ .
Compare
$\overrightarrow{PB}=3\,\overrightarrow{AP}$
$\overrightarrow{PB}$ is three times $\overrightarrow{AP}$ , so the length ratio is $1:3$ .
State the ratio
$AP:PB=1:3$
So $P$ divides $AB$ in the ratio $1:3$ .
Why this method works
$\overrightarrow{OP}=\mathbf{a}+\tfrac{m}{m+n}\overrightarrow{AB}$
Turning a ratio into a fraction of the whole makes it easy to step the right distance along the line.

Answer

AP:PB=1:3

Question 4

4 markshard

A line passes through

A(1,1)

and

B(5,9)

. Find the coordinates of the point on this line whose

x

-coordinate is

3

Show worked solution

Worked solution

Understand what is being asked
$\mathbf{r}=\mathbf{a}+t\mathbf{d}$
Before calculating, we identify the goal and choose the method: writing each line in the form $\mathbf{r}=\mathbf{a}+t\mathbf{d}$ . Having a plan stops us getting lost part-way through.
Recall the key result we will use
$\mathbf{r}=\mathbf{a}+t\mathbf{d}$
A line is a base point plus any multiple of its direction vector.
Find the direction vector
$\overrightarrow{AB}=\begin{pmatrix} 5 \\ 9 \end{pmatrix}-\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 4 \\ 8 \end{pmatrix}$
The direction of the line is the vector from $A$ to $B$ ; we can simplify it to $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ .
Write the line in vector form
$\mathbf{r}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}+t\begin{pmatrix} 1 \\ 2 \end{pmatrix}$
Using $A$ as the base point and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ as direction.
Set the $x$ -coordinate to 3
$1+t=3\ \Rightarrow\ t=2$
The top component gives the $x$ -coordinate; solve for the parameter $t$ .
Find the $y$ -coordinate
$y=1+2t=1+2(2)=5\ \Rightarrow\ (3,\ 5)$
Substitute $t=2$ into the bottom component to get $y$ .
Why this method works
$\mathbf{r}=\mathbf{a}+t\mathbf{d}$
At an intersection both lines give the same point, so we equate and solve.
Link to earlier work
$\mathbf{r}=\mathbf{a}+t\mathbf{d}$
This uses simultaneous equations from GCSE and AS algebra. Connecting new work to things you already know makes it far easier to remember.
Write down what we are working with
$\text{list the given vectors / points before substituting}$
Carefully noting the given information first is a habit that prevents careless substitution errors.
Check the answer
$(3,\ 5)$
Substitute the parameters back into both lines to confirm the same point. Checking your work is how you catch small mistakes before they cost marks.
Interpret the result
$(3,\ 5)$
The solution is the single point common to both lines.
Watch out for a common slip
$\mathbf{r}=\mathbf{a}+t\mathbf{d}$
A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

Answer

(3,\ 5)

Question 5

8 markschallenging

OABC

is a parallelogram with

\overrightarrow{OA}=\mathbf{a}

and

\overrightarrow{OC}=\mathbf{c}

(so

\overrightarrow{OB}=\mathbf{a}+\mathbf{c}

M

is the midpoint of

BC

. The line

OM

meets the diagonal

AC

X

. Find the ratio

OX:XM

Show worked solution

Worked solution

Understand what is being asked
$\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2$
Before calculating, we identify the goal and choose the method: equating the two line equations and solving. Having a plan stops us getting lost part-way through.
Recall the key result we will use
$\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2$
Set the two position vectors equal to find where the lines meet.
Set up base vectors
$\overrightarrow{OA}=\mathbf{a},\quad \overrightarrow{OC}=\mathbf{c}$
Use the two sides from $O$ as non-parallel base vectors.
Position vector of $B$
$\overrightarrow{OB}=\mathbf{a}+\mathbf{c}$
The opposite corner of the parallelogram is the sum of the two sides.
Find $M$ , the midpoint of $BC$
$\overrightarrow{OM}=\tfrac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC})=\tfrac{1}{2}\mathbf{a}+\mathbf{c}$
Average the position vectors of $B$ and $C$ : $\tfrac{1}{2}((\mathbf{a}+\mathbf{c})+\mathbf{c})$ .
Describe line $OM$
$\overrightarrow{OX}=\lambda\left(\tfrac{1}{2}\mathbf{a}+\mathbf{c}\right)$
Any point on $OM$ is a scalar multiple $\lambda$ of $\overrightarrow{OM}$ .
Describe diagonal $AC$
$\overrightarrow{OX}=\mathbf{a}+\mu(\mathbf{c}-\mathbf{a})=(1-\mu)\mathbf{a}+\mu\mathbf{c}$
A point on $AC$ is a fraction $\mu$ of the way from $A$ to $C$ .
Equate coefficients of $\mathbf{a}$
$\tfrac{\lambda}{2}=1-\mu$
Match the $\mathbf{a}$ -parts of the two expressions.
Equate coefficients of $\mathbf{c}$
$\lambda=\mu$
Match the $\mathbf{c}$ -parts; this gives $\lambda=\mu$ directly.
Solve
$\tfrac{\lambda}{2}=1-\lambda\ \Rightarrow\ \tfrac{3\lambda}{2}=1\ \Rightarrow\ \lambda=\tfrac{2}{3}$
Substitute $\mu=\lambda$ and solve. Then $OX:XM=\lambda:(1-\lambda)=\tfrac{2}{3}:\tfrac{1}{3}=2:1$ .
Why this method works
$\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2$
Equating gives two equations in two unknowns, which we solve simultaneously.
Link to earlier work
$\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2$
Relies on solving simultaneous equations. Connecting new work to things you already know makes it far easier to remember.
Write down what we are working with
$\text{list the given vectors / points before substituting}$
Carefully noting the given information first is a habit that prevents careless substitution errors.
Check the answer
$OX:XM=2:1$
Check the found point satisfies both original equations. Checking your work is how you catch small mistakes before they cost marks.
Interpret the result
$OX:XM=2:1$
The answer is the crossing point of the two lines.
Watch out for a common slip
$\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2$
A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

Answer

OX:XM=2:1

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