Vectors Practice Questions

Q: How do I find the vector \overrightarrow{AB} from position vectors?

Use end minus start: AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}AB=OB−OA. Writing it the other way round gives the vector pointing from BBB to AAA instead, which flips every sign.

Q: What makes two vectors parallel?

One must be a scalar multiple of the other, b=λa\mathbf{b}=\lambda\mathbf{a}b=λa. In practice you match components and solve for λ\lambdaλ; if a single λ\lambdaλ fits both components, the vectors are parallel.

How to use these practice questions

The six questions below rise in difficulty across the whole Vectors chapter, from adding two vectors up to finding a point that divides a segment. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, a method line stating the rule you are using, a clear substitution in components, then the final answer.

Resist the urge to peek. Only once you have committed to a full answer should you open the solution and mark yourself line by line.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a correct rule and a correct substitution still earn credit even when the final arithmetic slips, so always write the components in full before you simplify.

Treat this page as a rehearsal, not a test, the point is to find weak steps now, while there is still time to fix them.

Six practice questions

Q1[3 marks]

Two vectors are $\mathbf{a}=3\mathbf{i}+2\mathbf{j}$ and $\mathbf{b}=\mathbf{i}-4\mathbf{j}$ . Find (a) $\mathbf{a}+\mathbf{b}$ , and (b) $2\mathbf{a}-\mathbf{b}$ .

Show worked solution

(a) Add the vectors component by component, the $\mathbf{i}$ parts together, the $\mathbf{j}$ parts together:

\mathbf{a}+\mathbf{b}=(3+1)\mathbf{i}+(2-4)\mathbf{j}=4\mathbf{i}-2\mathbf{j}

(b) First scale $\mathbf{a}$ by $2$ , then subtract $\mathbf{b}$ . Keep the signs of $\mathbf{b}$ carefully; subtracting $-4\mathbf{j}$ adds $4\mathbf{j}$ :

2\mathbf{a}=6\mathbf{i}+4\mathbf{j}

2\mathbf{a}-\mathbf{b}=(6-1)\mathbf{i}+(4-(-4))\mathbf{j}=5\mathbf{i}+8\mathbf{j}

Answer

$\mathbf{a}+\mathbf{b}=4\mathbf{i}-2\mathbf{j}$ and $2\mathbf{a}-\mathbf{b}=5\mathbf{i}+8\mathbf{j}$ . The most common slip in (b) is treating $-\mathbf{b}$ as $-\mathbf{i}-4\mathbf{j}$ ; the correct negative is $-\mathbf{i}+4\mathbf{j}$ .

Q2[3 marks]

A vector is $\mathbf{v}=6\mathbf{i}-8\mathbf{j}$ . Find (a) the magnitude $|\mathbf{v}|$ , and (b) the unit vector in the direction of $\mathbf{v}$ .

Show worked solution

(a) The magnitude of $x\mathbf{i}+y\mathbf{j}$ is $\sqrt{x^{2}+y^{2}}$ . Square each component, add, then take the root:

|\mathbf{v}|=\sqrt{6^{2}+(-8)^{2}}=\sqrt{36+64}=\sqrt{100}=10

(b) The unit vector is the vector divided by its own magnitude, $\hat{\mathbf{v}}=\dfrac{1}{|\mathbf{v}|}\,\mathbf{v}$ :

\hat{\mathbf{v}}=\frac{1}{10}\left(6\mathbf{i}-8\mathbf{j}\right)=\frac{3}{5}\mathbf{i}-\frac{4}{5}\mathbf{j}

Answer

$|\mathbf{v}|=10$ and $\hat{\mathbf{v}}=\tfrac{3}{5}\mathbf{i}-\tfrac{4}{5}\mathbf{j}$ . Check the unit vector has length $1$ : $\sqrt{(\tfrac{3}{5})^{2}+(\tfrac{4}{5})^{2}}=\sqrt{\tfrac{9+16}{25}}=\sqrt{1}=1$ , as it must.

Q3[4 marks]

The position vectors of $A$ and $B$ relative to the origin $O$ are $\overrightarrow{OA}=2\mathbf{i}+3\mathbf{j}$ and $\overrightarrow{OB}=7\mathbf{i}+15\mathbf{j}$ . Find (a) the vector $\overrightarrow{AB}$ , and (b) its magnitude $|\overrightarrow{AB}|$ .

Show worked solution

(a) Travel from $A$ to $B$ by going back to $O$ then out to $B$ , so $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ (end minus start):

\overrightarrow{AB}=(7-2)\mathbf{i}+(15-3)\mathbf{j}=5\mathbf{i}+12\mathbf{j}

(b) Now take the magnitude of the result:

|\overrightarrow{AB}|=\sqrt{5^{2}+12^{2}}=\sqrt{25+144}=\sqrt{169}=13

Answer

$\overrightarrow{AB}=5\mathbf{i}+12\mathbf{j}$ and $|\overrightarrow{AB}|=13$ . Order matters: $\overrightarrow{AB}$ is end minus start, $\overrightarrow{OB}-\overrightarrow{OA}$ ; reversing it would give $-5\mathbf{i}-12\mathbf{j}$ , the vector pointing the wrong way.

Q4[4 marks]

The vectors $\mathbf{a}=4\mathbf{i}+6\mathbf{j}$ and $\mathbf{b}=k\mathbf{i}+9\mathbf{j}$ are parallel, where $k$ is a constant. Find the value of $k$ .

Show worked solution

Two vectors are parallel when one is a scalar multiple of the other, so $\mathbf{b}=\lambda\mathbf{a}$ for some number $\lambda$ . Match the components:

k=4\lambda \quad\text{and}\quad 9=6\lambda

Solve the $\mathbf{j}$ -component equation for $\lambda$ first, then substitute into the $\mathbf{i}$ -component:

\lambda=\frac{9}{6}=\frac{3}{2}\;\Rightarrow\; k=4\times\frac{3}{2}=6

Answer

$k=6$ . Check the parallel condition: $\mathbf{b}=6\mathbf{i}+9\mathbf{j}=\tfrac{3}{2}(4\mathbf{i}+6\mathbf{j})=\tfrac{3}{2}\mathbf{a}$ , so $\mathbf{b}$ really is a scalar multiple of $\mathbf{a}$ .

Q5[5 marks]

Relative to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors $\overrightarrow{OA}=2\mathbf{i}+\mathbf{j}$ , $\overrightarrow{OB}=5\mathbf{i}+7\mathbf{j}$ and $\overrightarrow{OC}=9\mathbf{i}+15\mathbf{j}$ . Show that $A$ , $B$ and $C$ are collinear, and find the ratio $AB:BC$ .

Show worked solution

Find the two vectors along the path, using end minus start each time:

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=3\mathbf{i}+6\mathbf{j}

\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=4\mathbf{i}+8\mathbf{j}

Test whether one is a scalar multiple of the other. Factor $\overrightarrow{BC}$ :

\overrightarrow{BC}=4\mathbf{i}+8\mathbf{j}=\frac{4}{3}\left(3\mathbf{i}+6\mathbf{j}\right)=\frac{4}{3}\,\overrightarrow{AB}

Since $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$ and they share the point $B$ , the three points lie on one straight line. For the ratio, compare magnitudes:

|\overrightarrow{AB}|=\sqrt{3^{2}+6^{2}}=\sqrt{45}=3\sqrt{5},\quad |\overrightarrow{BC}|=\sqrt{4^{2}+8^{2}}=\sqrt{80}=4\sqrt{5}

Answer

Because $\overrightarrow{BC}=\tfrac{4}{3}\overrightarrow{AB}$ with the common point $B$ , the points $A$ , $B$ , $C$ are collinear. The ratio is $AB:BC=3\sqrt{5}:4\sqrt{5}=3:4$ .

Q6[5 marks]

Relative to the origin $O$ , points $A$ and $B$ have position vectors $\overrightarrow{OA}=\mathbf{i}+2\mathbf{j}$ and $\overrightarrow{OB}=7\mathbf{i}+8\mathbf{j}$ . The point $P$ lies on $AB$ such that $AP:PB=1:2$ .

Find $\overrightarrow{OP}$ and its magnitude $|\overrightarrow{OP}|$ .

Show worked solution

Point $P$ is $\tfrac{1}{3}$ of the way from $A$ to $B$ , so $\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{1}{3}\overrightarrow{AB}$ . First find $\overrightarrow{AB}$ :

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=6\mathbf{i}+6\mathbf{j}

Now build $\overrightarrow{OP}$ , adding one third of $\overrightarrow{AB}$ to $\overrightarrow{OA}$ :

\overrightarrow{OP}=(\mathbf{i}+2\mathbf{j})+\frac{1}{3}(6\mathbf{i}+6\mathbf{j})=(\mathbf{i}+2\mathbf{j})+(2\mathbf{i}+2\mathbf{j})=3\mathbf{i}+4\mathbf{j}

Take the magnitude of $\overrightarrow{OP}$ :

|\overrightarrow{OP}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5

Answer

$\overrightarrow{OP}=3\mathbf{i}+4\mathbf{j}$ and $|\overrightarrow{OP}|=5$ . Check with the section formula $\overrightarrow{OP}=\tfrac{2\,\overrightarrow{OA}+1\,\overrightarrow{OB}}{3}=\tfrac{2(\mathbf{i}+2\mathbf{j})+(7\mathbf{i}+8\mathbf{j})}{3}=\tfrac{9\mathbf{i}+12\mathbf{j}}{3}=3\mathbf{i}+4\mathbf{j}$ , which agrees.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final vector. When you check your own script, award yourself credit the way a marker would: look for the correct rule, the right components in place, and a clean final statement.

Method mark: did you write the correct rule, end minus start for $\overrightarrow{AB}$ , $\sqrt{x^{2}+y^{2}}$ for magnitude, or $\mathbf{b}=\lambda\mathbf{a}$ for parallel vectors?
Substitution mark: are the components put in the right slots, with the sign of every $\mathbf{j}$ term kept, especially when subtracting a vector?
Answer mark: is the final vector or value stated clearly, and does it survive a check by an independent route?
For collinearity, you only earn full marks by showing the scalar multiple and naming the common point.
If your final answer is wrong but the rule and substitution lines are right, give yourself those marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, $\overrightarrow{AB}$ written as start minus end, a sign dropped when subtracting a vector, or collinearity claimed without naming the common point, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

How long should each of these questions take me?

Aim for roughly five to eight minutes each, rising with the mark value. If a question takes far longer, note it and bring it to a lesson, the time it steals in the exam is often the real problem, not the topic itself.

How do I find the vector

\overrightarrow{AB}

from position vectors?

Use end minus start: $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ . Writing it the other way round gives the vector pointing from $B$ to $A$ instead, which flips every sign.

What makes two vectors parallel?

One must be a scalar multiple of the other, $\mathbf{b}=\lambda\mathbf{a}$ . In practice you match components and solve for $\lambda$ ; if a single $\lambda$ fits both components, the vectors are parallel.

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct rule line and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always show full working in components.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Vectors, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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