Vectors Worked Examples (Medium)

Q: How do I show that three points are collinear?

Form two vectors along the line, such as AB→\overrightarrow{AB}AB and BC→\overrightarrow{BC}BC. If BC→=m AB→\overrightarrow{BC}=m\,\overrightarrow{AB}BC=mAB for one scalar mmm that works for both components, the points lie on a single straight line.

Q: Do I have to check both components when finding the unknown?

Yes. Use one component to get mmm, but the answer is only valid if the same mmm also satisfies the other component. A quick AC→\overrightarrow{AC}AC check confirms it.

What these examples cover

These medium Vectors examples move past single-step arithmetic into the reasoning SPM rewards most: scaling a unit vector to a required length, matching components to solve for two unknown scalars at once, and using the parallel condition to force three points onto one straight line. The numbers are still clean, but each question now asks you to hold two ideas together.

Use the set the same disciplined way, cover the solution, attempt the whole question on paper, and only then check line by line. When your answer differs, find the precise step where the two solutions part; that is where a medium question quietly teaches you the habit that makes the hard ones feel routine.

Keep components aligned and always finish with a substitution check.

Worked examples

Work through all three. Attempt each one fully before reading the solution, and watch how a unit vector, a pair of simultaneous equations, and the parallel condition each turn a wordy question into a short, checkable calculation.

Q1[4 marks]

The vector $\mathbf{p}=\binom{-8}{6}$ . Find (a) the magnitude $|\mathbf{p}|$ and the unit vector $\hat{\mathbf{p}}$ , and (b) the vector of magnitude $20$ in the direction of $\mathbf{p}$ .

Show worked solution

(a) First take the magnitude from the components, then divide $\mathbf{p}$ by it to get the unit vector:

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

\hat{\mathbf{p}}=\frac{1}{10}\binom{-8}{6}=\binom{-\tfrac{4}{5}}{\tfrac{3}{5}}

(b) A vector of magnitude $20$ in the same direction is $20$ times the unit vector, a unit vector points the way, and the scalar sets the length:

20\,\hat{\mathbf{p}}=20\binom{-\tfrac{4}{5}}{\tfrac{3}{5}}=\binom{-16}{12}

Answer

$|\mathbf{p}|=10$ , $\hat{\mathbf{p}}=\binom{-4/5}{3/5}$ , and the required vector is $\binom{-16}{12}$ . Check its magnitude: $\sqrt{(-16)^{2}+12^{2}}=\sqrt{256+144}=\sqrt{400}=20$ , exactly as asked.

Q2[4 marks]

Given $\mathbf{a}=\binom{2}{1}$ and $\mathbf{b}=\binom{-1}{3}$ , find the scalars $h$ and $k$ such that $h\mathbf{a}+k\mathbf{b}=\binom{7}{0}$ .

Show worked solution

Write the combination out and match the top and bottom components separately. This turns one vector equation into two ordinary equations:

h\binom{2}{1}+k\binom{-1}{3}=\binom{2h-k}{\,h+3k\,}=\binom{7}{0}

So $2h-k=7$ and $h+3k=0$ . The second equation gives $h=-3k$ ; substitute it into the first:

2(-3k)-k=7 \;\Rightarrow\; -7k=7 \;\Rightarrow\; k=-1

Then $h=-3k=-3(-1)=3$ .

Answer

$h=3$ and $k=-1$ . Check by substituting back: $3\binom{2}{1}-1\binom{-1}{3}=\binom{6}{3}+\binom{1}{-3}=\binom{7}{0}$ , as required.

Q3[4 marks]

The points $A(1,2)$ , $B(3,5)$ and $C(7,k)$ are collinear. Find the value of $k$ .

Show worked solution

Three points are collinear when the vectors along the line are parallel. Form $\overrightarrow{AB}$ and $\overrightarrow{BC}$ as position-vector differences:

\overrightarrow{AB}=\binom{3-1}{5-2}=\binom{2}{3},\qquad \overrightarrow{BC}=\binom{7-3}{k-5}=\binom{4}{k-5}

Because $A$ , $B$ , $C$ lie on one line and share the point $B$ , we need $\overrightarrow{BC}=m\,\overrightarrow{AB}$ . Compare the top components to find $m$ :

4=m(2)\;\Rightarrow\; m=2

Apply the same $m$ to the bottom components:

k-5=m(3)=2(3)=6 \;\Rightarrow\; k=11

Answer

$k=11$ . Check with $\overrightarrow{AC}=\binom{7-1}{11-2}=\binom{6}{9}=3\binom{2}{3}=3\,\overrightarrow{AB}$ ; since $\overrightarrow{AC}$ is parallel to $\overrightarrow{AB}$ through $A$ , the three points are collinear.

Q4[4 marks]

The points $A$ and $B$ have position vectors $\overrightarrow{OA}=\binom{1}{4}$ and $\overrightarrow{OB}=\binom{16}{-6}$ relative to the origin $O$ . The point $P$ lies on $AB$ such that $AP:PB=2:3$ .

Find the position vector of $P$ .

Show worked solution

$P$ divides $AB$ in the ratio $AP:PB=2:3$ , so $P$ is $\tfrac{2}{5}$ of the way from $A$ to $B$ . Use the ratio theorem:

\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{2}{5}\overrightarrow{AB}, \qquad \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\binom{16}{-6}-\binom{1}{4}=\binom{15}{-10}

Substitute and simplify:

\overrightarrow{OP}=\binom{1}{4}+\tfrac{2}{5}\binom{15}{-10}=\binom{1}{4}+\binom{6}{-4}=\binom{7}{0}

Answer

$\overrightarrow{OP}=\binom{7}{0}$ , so $P$ is the point $(7,0)$ . Check: $\overrightarrow{AP}=\binom{6}{-4}$ and $\overrightarrow{PB}=\binom{9}{-6}=\tfrac{3}{2}\overrightarrow{AP}$ , confirming $AP:PB=2:3$ .

Q5[3 marks]

In triangle $OAB$ , $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ . The point $N$ lies on $OB$ such that $ON:NB=1:2$ .

Express $\overrightarrow{AN}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Show worked solution

Since $ON:NB=1:2$ , point $N$ is $\tfrac13$ of the way along $OB$ from $O$ , so:

\overrightarrow{ON}=\tfrac13\overrightarrow{OB}=\tfrac13\mathbf{b}

To get from $A$ to $N$ , go from $A$ to $O$ (reversing $\mathbf{a}$ ) and then from $O$ to $N$ :

\overrightarrow{AN}=\overrightarrow{AO}+\overrightarrow{ON}=-\mathbf{a}+\tfrac13\mathbf{b}=\tfrac13\mathbf{b}-\mathbf{a}

Answer

$\overrightarrow{AN}=\tfrac13\mathbf{b}-\mathbf{a}$ . As a check, the same vector can be reached via $\overrightarrow{AN}=\overrightarrow{ON}-\overrightarrow{OA}=\tfrac13\mathbf{b}-\mathbf{a}$ , the identical result from a different pair of legs.

Q6[4 marks]

$ABCD$ is a parallelogram with $A(2,1)$ , $B(7,3)$ and $C(9,8)$ . Find the coordinates of $D$ .

Show worked solution

In parallelogram $ABCD$ , side $AB$ is parallel and equal to side $DC$ , so $\overrightarrow{AB}=\overrightarrow{DC}$ . Find $\overrightarrow{AB}$ first:

\overrightarrow{AB}=\binom{7-2}{3-1}=\binom{5}{2}

Since $\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}$ must equal $\overrightarrow{AB}$ , solve for $\overrightarrow{OD}$ :

\overrightarrow{OD}=\overrightarrow{OC}-\overrightarrow{AB}=\binom{9}{8}-\binom{5}{2}=\binom{4}{6}

Answer

$D=(4,6)$ . Check: the midpoint of diagonal $AC$ is $\left(\tfrac{2+9}{2},\tfrac{1+8}{2}\right)=(5.5,4.5)$ , and the midpoint of diagonal $BD$ is $\left(\tfrac{7+4}{2},\tfrac{3+6}{2}\right)=(5.5,4.5)$ too, the diagonals bisect each other, confirming $ABCD$ is a parallelogram.

Q7[3 marks]

Given $\mathbf{a}=4\mathbf{i}-3\mathbf{j}$ and $\mathbf{b}=-12\mathbf{i}+9\mathbf{j}$ , show that $\mathbf{a}$ and $\mathbf{b}$ are parallel, and find $|\mathbf{b}|$ .

Show worked solution

Write $\mathbf{b}$ as a scalar multiple of $\mathbf{a}$ by comparing coefficients of $\mathbf{i}$ and $\mathbf{j}$ :

\mathbf{b}=-12\mathbf{i}+9\mathbf{j}=-3(4\mathbf{i}-3\mathbf{j})=-3\mathbf{a}

Because $\mathbf{b}$ is a scalar multiple of $\mathbf{a}$ , the two vectors are parallel (the negative scalar means they point in opposite directions). Their magnitudes are then in the ratio $3:1$ :

|\mathbf{a}|=\sqrt{4^{2}+(-3)^{2}}=\sqrt{25}=5, \qquad |\mathbf{b}|=3|\mathbf{a}|=3(5)=15

Answer

$\mathbf{b}=-3\mathbf{a}$ , so $\mathbf{a}$ and $\mathbf{b}$ are parallel, and $|\mathbf{b}|=15$ . Check directly: $|\mathbf{b}|=\sqrt{(-12)^{2}+9^{2}}=\sqrt{144+81}=\sqrt{225}=15$ , matching.

Q8[4 marks]

A hiker walks from camp $P$ to a checkpoint $Q$ with displacement vector $\binom{12}{5}$ km, then from $Q$ to the destination $R$ with displacement vector $\binom{-3}{7}$ km. Find (a) the displacement vector $\overrightarrow{PR}$ , and (b) the direct distance from $P$ to $R$ .

Show worked solution

(a) The direct path from $P$ to $R$ is the sum of the two legs (triangle law of vector addition):

\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}=\binom{12}{5}+\binom{-3}{7}=\binom{9}{12}

(b) The direct distance is the magnitude of this resultant vector:

|\overrightarrow{PR}|=\sqrt{9^{2}+12^{2}}=\sqrt{81+144}=\sqrt{225}=15

Answer

$\overrightarrow{PR}=\binom{9}{12}$ km and the direct distance is $15$ km, a $9\text{-}12\text{-}15$ triple (three times $3\text{-}4\text{-}5$ ), a quick way to confirm the arithmetic.

Notice the shared engine across the three: a unit vector carries direction so a scalar can set length; matching components turns a vector equation into simultaneous equations; and the parallel condition $\mathbf{b}=m\mathbf{a}$ is exactly what pins points onto a line. Recognise which one a question is really asking for, and the working shortens fast.

Key method points

These medium examples connect the basic moves into short chains of reasoning. Keep these points to hand as the questions grow.

To build a vector of a set magnitude in a direction, scale the unit vector: $\text{(length)}\times\hat{\mathbf{a}}$ .
Matching components turns one vector equation into two equations, solve them as a normal simultaneous pair.
Three points are collinear when the joining vectors are parallel, $\overrightarrow{BC}=m\,\overrightarrow{AB}$ .
Find the scalar $m$ from one component, then apply the very same $m$ to the other.
Always close with a substitution or magnitude check, it is a fast, independent confirmation.
Because marking is analytic, laying out the component equations clearly earns method marks even if a later step slips.

How a teacher helps

Medium questions are where a student either builds real fluency or quietly develops a blind spot, solving for one scalar and forgetting the second, or checking only one component of a parallel condition. In a one-to-one lesson our teacher watches how you set the work out and steadies the exact step that wobbles, so the method becomes automatic under exam pressure.

Because our teachers are experienced, you learn the reasoning, not just a recipe. Lessons are taught in English, while SPM papers are set in both Malay and English, so the vector notation reads the same to you either way.

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Frequently asked questions

How do I make a vector with a specific magnitude in a given direction?

Find the unit vector in that direction, then multiply it by the magnitude you want. For $\mathbf{p}$ with $|\mathbf{p}|=10$ , a length- $20$ vector is $20\hat{\mathbf{p}}=\binom{-16}{12}$ .

Why does matching components work for

h\mathbf{a}+k\mathbf{b}=\mathbf{c}

Two vectors are equal only when their top components match and their bottom components match. That gives two equations in $h$ and $k$ , which you solve simultaneously.

How do I show that three points are collinear?

Form two vectors along the line, such as $\overrightarrow{AB}$ and $\overrightarrow{BC}$ . If $\overrightarrow{BC}=m\,\overrightarrow{AB}$ for one scalar $m$ that works for both components, the points lie on a single straight line.

Do I have to check both components when finding the unknown?

Yes. Use one component to get $m$ , but the answer is only valid if the same $m$ also satisfies the other component.

A quick $\overrightarrow{AC}$ check confirms it.

Source:SRC-DSKP-EN

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

Vectors, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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