Vectors Worked Examples (Easy)

Q: What makes a vector a unit vector?

A unit vector has magnitude exactly 111. To build one in the direction of a\mathbf{a}a, divide a\mathbf{a}a by its own magnitude: a^=1∣a∣a\hat{\mathbf{a}}=\frac{1}{|\mathbf{a}|}\mathbf{a}a^=∣a∣1a.

Q: Why is \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} and not the other way round?

Going from AAA to BBB is the same as going back to the origin, then out to BBB: AB→=AO→+OB→=−OA→+OB→\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}AB=AO+OB=−OA+OB. In short, end point minus start point.

What these examples cover

These easy Vectors examples drill the four moves the whole chapter is built on: finding the magnitude of a vector from its components, writing the unit vector in a given direction, adding and subtracting vectors written as columns, and testing when two vectors are parallel. Every question uses small, clean numbers so you can follow each line by hand, without leaning on a calculator.

Use the set the honest way, cover the solution, attempt the question in full on paper, and only then check line by line against our working. Where your answer differs, hunt for the exact step where the two solutions split; that single line is usually where the real learning sits.

Vectors reward tidy bookkeeping, so line up your $\mathbf{i}$ and $\mathbf{j}$ components, or your top and bottom entries, and the marks come steadily.

Worked examples

Work through all four. Attempt each one fully before reading the matching solution, and notice how the same discipline, keep components aligned, take the magnitude out cleanly, and check by an independent route, runs through every question.

Q1[3 marks]

The vector $\mathbf{a}$ is given by $\mathbf{a}=3\mathbf{i}+4\mathbf{j}$ . Find (a) the magnitude $|\mathbf{a}|$ , and (b) the unit vector in the direction of $\mathbf{a}$ .

Show worked solution

(a) The magnitude is the length of the vector, read from its components with $|x\mathbf{i}+y\mathbf{j}|=\sqrt{x^{2}+y^{2}}$ . Here $x=3$ and $y=4$ :

|\mathbf{a}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5

(b) The unit vector in the direction of $\mathbf{a}$ is $\mathbf{a}$ divided by its own magnitude, $\hat{\mathbf{a}}=\frac{1}{|\mathbf{a}|}\mathbf{a}$ . Divide each component by $5$ :

\hat{\mathbf{a}}=\frac{1}{5}(3\mathbf{i}+4\mathbf{j})=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}

Answer

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

Q2[3 marks]

Two vectors are given by $\mathbf{a}=\binom{2}{-1}$ and $\mathbf{b}=\binom{3}{5}$ . Find (a) $2\mathbf{a}+\mathbf{b}$ , and (b) $\mathbf{b}-\mathbf{a}$ .

Show worked solution

(a) Multiply $\mathbf{a}$ by the scalar $2$ first, this multiplies each entry, then add the matching entries of $\mathbf{b}$ :

2\mathbf{a}=\binom{4}{-2},\qquad 2\mathbf{a}+\mathbf{b}=\binom{4+3}{-2+5}=\binom{7}{3}

(b) Subtract entry by entry, taking care with the double negative in the second row:

\mathbf{b}-\mathbf{a}=\binom{3-2}{5-(-1)}=\binom{1}{6}

Answer

$2\mathbf{a}+\mathbf{b}=\binom{7}{3}$ and $\mathbf{b}-\mathbf{a}=\binom{1}{6}$ . Check part (b) by adding $\mathbf{a}$ back: $\binom{1}{6}+\binom{2}{-1}=\binom{3}{5}=\mathbf{b}$ , as required.

Q3[3 marks]

The vector $\mathbf{a}=2\mathbf{i}+3\mathbf{j}$ . A second vector $\mathbf{b}=6\mathbf{i}+k\mathbf{j}$ is parallel to $\mathbf{a}$ .

Find the value of $k$ .

Show worked solution

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, so $\mathbf{b}=m\mathbf{a}$ for some number $m$ . Compare the $\mathbf{i}$ components to find $m$ :

6=m(2)\;\Rightarrow\; m=3

The same scalar must work for the $\mathbf{j}$ components, so $k=m(3)$ :

k=3(3)=9

Answer

$k=9$ . Then $\mathbf{b}=6\mathbf{i}+9\mathbf{j}=3(2\mathbf{i}+3\mathbf{j})=3\mathbf{a}$ , which is clearly parallel to $\mathbf{a}$ .

Q4[4 marks]

The position vectors of points $A$ and $B$ are $\overrightarrow{OA}=\binom{2}{3}$ and $\overrightarrow{OB}=\binom{8}{11}$ . Find (a) $\overrightarrow{AB}$ , and (b) the distance $|\overrightarrow{AB}|$ .

Show worked solution

(a) The vector from $A$ to $B$ is the end point's position vector minus the start point's, $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ :

\overrightarrow{AB}=\binom{8-2}{11-3}=\binom{6}{8}

(b) The distance $AB$ is the magnitude of $\overrightarrow{AB}$ :

|\overrightarrow{AB}|=\sqrt{6^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10

Answer

$\overrightarrow{AB}=\binom{6}{8}$ and $|\overrightarrow{AB}|=10$ . Check part (a) with the triangle law: $\overrightarrow{OA}+\overrightarrow{AB}=\binom{2}{3}+\binom{6}{8}=\binom{8}{11}=\overrightarrow{OB}$ , as it should.

Q5[3 marks]

Points $P$ , $Q$ and $R$ are joined so that $\overrightarrow{PQ}=2\mathbf{i}+5\mathbf{j}$ and $\overrightarrow{QR}=3\mathbf{i}+7\mathbf{j}$ . Find (a) $\overrightarrow{PR}$ , and (b) the distance $PR$ .

Show worked solution

(a) The triangle law lets you chain vectors head to tail: going from $P$ to $Q$ then $Q$ to $R$ is the same as going straight from $P$ to $R$ , so $\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}$ . Add the matching components:

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

(b) The distance $PR$ is the magnitude of $\overrightarrow{PR}$ :

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

Answer

$\overrightarrow{PR}=5\mathbf{i}+12\mathbf{j}$ and $PR=13$ . $5,12,13$ is a clean Pythagorean triple, a quick sign that the component addition in part (a) was done correctly.

Q6[3 marks]

Points $A$ and $B$ have position vectors $\overrightarrow{OA}=\binom{2}{6}$ and $\overrightarrow{OB}=\binom{8}{2}$ relative to the origin $O$ . $M$ is the midpoint of $AB$ .

Find the position vector $\overrightarrow{OM}$ .

Show worked solution

The position vector of the midpoint of a line joining two points is the average of their position vectors, $\overrightarrow{OM}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})$ . Add the position vectors first:

\overrightarrow{OA}+\overrightarrow{OB}=\binom{2}{6}+\binom{8}{2}=\binom{10}{8}

Then halve each component:

\overrightarrow{OM}=\frac{1}{2}\binom{10}{8}=\binom{5}{4}

Answer

$\overrightarrow{OM}=\binom{5}{4}$ . Check: $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\binom{3}{-2}$ and $\overrightarrow{MB}=\overrightarrow{OB}-\overrightarrow{OM}=\binom{3}{-2}$ , equal, confirming $M$ really is halfway between $A$ and $B$ .

Q7[3 marks]

Two vectors $\mathbf{p}=(x+3)\mathbf{i}+7\mathbf{j}$ and $\mathbf{q}=9\mathbf{i}+(2y-1)\mathbf{j}$ are equal vectors. Find the values of $x$ and $y$ .

Show worked solution

Two vectors are equal exactly when their $\mathbf{i}$ components match and their $\mathbf{j}$ components match. Compare the $\mathbf{i}$ components first:

x+3=9\;\Rightarrow\; x=6

Now compare the $\mathbf{j}$ components:

7=2y-1\;\Rightarrow\; 2y=8\;\Rightarrow\; y=4

Answer

$x=6$ and $y=4$ . Check by substituting back: $\mathbf{p}=9\mathbf{i}+7\mathbf{j}$ and $\mathbf{q}=9\mathbf{i}+7\mathbf{j}$ , the same vector, as required.

Q8[3 marks]

A vector is given by $\mathbf{v}=x\mathbf{i}+8\mathbf{j}$ , where $x>0$ . Given that $|\mathbf{v}|=17$ , find the value of $x$ .

Show worked solution

Square both sides of the magnitude formula $|\mathbf{v}|=\sqrt{x^{2}+8^{2}}$ to clear the square root:

17^{2}=x^{2}+8^{2}

Substitute and solve for $x^{2}$ , then take the positive square root since $x>0$ :

x^{2}=289-64=225\;\Rightarrow\; x=15

Answer

$x=15$ . Check: $|\mathbf{v}|=\sqrt{15^{2}+8^{2}}=\sqrt{225+64}=\sqrt{289}=17$ , which matches the given magnitude.

Different as they look, these four questions share one habit: keep the components lined up, do one operation at a time, and confirm with a second route. That tidy bookkeeping is what turns Vectors into a dependable block of marks rather than a place for small slips.

Key method points

These four examples rehearse the everyday skills that open almost every Vectors question in Add Math. Keep the following points in mind as you practise more.

The magnitude of a vector comes from its components: $|x\mathbf{i}+y\mathbf{j}|=\sqrt{x^{2}+y^{2}}$ .
The unit vector in a direction is the vector divided by its own magnitude, $\hat{\mathbf{a}}=\frac{1}{|\mathbf{a}|}\mathbf{a}$ ; it always has length $1$ .
Scalar multiplication and vector addition act component by component, line up $\mathbf{i}$ with $\mathbf{i}$ and $\mathbf{j}$ with $\mathbf{j}$ .
Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, $\mathbf{b}=m\mathbf{a}$ .
For points, $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ , and the distance $AB$ is $|\overrightarrow{AB}|$ .
Because marking is analytic, a clear component or magnitude line can still earn method marks even if the final arithmetic slips.

How a teacher helps

When a student drops a mark on questions like these, it is almost always a small, fixable habit, a sign lost in a subtraction, or the square root taken before the components were squared. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it hardens into a routine.

Because our teachers are experienced, you work with someone who explains the reasoning behind each step, not only the mechanics. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the vector notation reads the same to you in either version.

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Frequently asked questions

How do I find the magnitude of a vector?

Square each component, add the squares, and take the positive square root: $|x\mathbf{i}+y\mathbf{j}|=\sqrt{x^{2}+y^{2}}$ . For $3\mathbf{i}+4\mathbf{j}$ this gives $\sqrt{9+16}=5$ .

What makes a vector a unit vector?

A unit vector has magnitude exactly $1$ . To build one in the direction of $\mathbf{a}$ , divide $\mathbf{a}$ by its own magnitude: $\hat{\mathbf{a}}=\frac{1}{|\mathbf{a}|}\mathbf{a}$ .

How can I tell whether two vectors are parallel?

They are parallel when one is a scalar multiple of the other, $\mathbf{b}=m\mathbf{a}$ . Compare one pair of components to find $m$ , then check that the same $m$ works for the other pair.

Why is

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}

and not the other way round?

Going from $A$ to $B$ is the same as going back to the origin, then out to $B$ : $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}$ . In short, end point minus start point.

Source:SRC-DSKP-EN

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

Vectors, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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