Magpie Tuition · Higher Maths

Unit vectors: targeted practice

Allow 15–20 minutes. No calculator needed. Give exact answers throughout.

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Overview

The minimum you need

A unit vector has magnitude (length) 1. For a non-zero vector $\mathbf v$ , the unit vector in the same direction is

\boxed{\widehat{\mathbf v}=\frac{\mathbf v}{|\mathbf v|}}.

For $\mathbf v=(a,b)$ , $|\mathbf v|=\sqrt{a^2+b^2}$ . For $\mathbf v=(a,b,c)$ , $|\mathbf v|=\sqrt{a^2+b^2+c^2}$ . Square negative components using brackets. Divide every component by the positive magnitude, not its square.

Check two things separately: length 1 and the requested direction. A positive multiplier preserves direction; a negative multiplier reverses it. The opposite unit vector is $-\widehat{\mathbf v}$ .

A magnitude is a number; a unit vector has components. They are different types of answer. The zero vector has no direction and cannot be normalised by this formula.

Here component triples such as $\mathbf v=(a,b,c)$ mean vectors. In basis notation, $a\mathbf i+b\mathbf j+c\mathbf k$ is the same vector; $\mathbf i,\mathbf j,\mathbf k$ are the three positive coordinate unit vectors.

Worked example

One worked example

Find a unit vector in the direction of $\mathbf v=(6,-8)$ .

|\mathbf v|=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=10, \qquad \widehat{\mathbf v}=\frac1{10}(6,-8)=\boxed{(3/5,-4/5)}.

Length check: $\sqrt{(3/5)^2+(-4/5)^2}=\sqrt{9/25+16/25}=1$ . Direction check: the multiplier $1/10$ is positive. The scalar $10$ is the original magnitude, not the requested unit vector.

Practice

Your Turn

Practice only — these are not your fresh-check questions. In Q1–8, find the magnitude and a unit vector in the same direction. Then check length and direction. Surd denominators or equivalent rationalised exact forms are acceptable.

$\mathbf a=(5,12)$
$\mathbf b=(-8,6)$
$\mathbf c=(0,-9)$
$\mathbf d=(3,4,12)$
$\mathbf e=(-2,3,6)$
$\mathbf f=(1,-1,0)$
$\mathbf g=(2,0,-3)$
$\mathbf h=-4\mathbf i+4\mathbf j+2\mathbf k$ . Give the final unit vector in $\mathbf i,\mathbf j,\mathbf k$ form.
For $\mathbf v=(0,3,-4)$ , five proposed answers to “find a unit vector in the direction of $\mathbf v$ ” are shown below. Identify the correct answer and explain why each other answer is unsuitable.

Answer	Proposed result
A	$5$
B	$(0,3/25,-4/25)$
C	$(0,-3/5,4/5)$
D	$(0,3,-4)$
E	$(0,3/5,-4/5)$

Original exam-style application. A device moves in the direction $\mathbf d=(-3,6,2)$ . (a) Find a unit vector in its direction. (b) Find a displacement of length 14 in that direction. Use the fact that multiplying a unit vector by a positive number $L$ produces a vector of length $L$ . (c) Check the length of your displacement and explain why it has the correct direction.

Answers

Full worked answers

Show the answers

1

$|\mathbf a|=\sqrt{5^2+12^2}=\sqrt{169}=\boxed{13}$ .

$\widehat{\mathbf a}=\boxed{(5/13,12/13)}$ . Its squared length is $(25+144)/169=1$ , so its length is 1. The multiplier $1/13>0$ preserves direction.

2

$|\mathbf b|=\sqrt{(-8)^2+6^2}=\sqrt{100}=\boxed{10}$ . $\widehat{\mathbf b}=(-8/10,6/10)=\boxed{(-4/5,3/5)}$ . Its squared length is $(16+9)/25=1$ . The multiplier $1/10>0$ , so its negative first component is correctly preserved.

3

$|\mathbf c|=\sqrt{0^2+(-9)^2}=\boxed{9}$ , not $-9$ : magnitude is non-negative.

$\widehat{\mathbf c}=\boxed{(0,-1)}$ . Its length is $\sqrt{0+1}=1$ , and it still points along the negative second axis because the multiplier is $1/9>0$ .

4

$|\mathbf d|=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\boxed{13}$ .

$\widehat{\mathbf d}=\boxed{(3/13,4/13,12/13)}$ . Squared length: $(9+16+144)/169=1$ . Multiplier: $1/13>0$ .

5

$|\mathbf e|=\sqrt{(-2)^2+3^2+6^2}=\sqrt{4+9+36}=\boxed{7}$ .

$\widehat{\mathbf e}=\boxed{(-2/7,3/7,6/7)}$ . Squared length: $(4+9+36)/49=1$ . Multiplier: $1/7>0$ .

6

$|\mathbf f|=\sqrt{1^2+(-1)^2+0^2}=\boxed{\sqrt2}$ .

\widehat{\mathbf f}=\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right) =\boxed{(\sqrt2/2,-\sqrt2/2,0)}.

Squared length: $1/2+1/2+0=1$ . Multiplier: $1/\sqrt2>0$ . Dividing by 2 would divide by the magnitude squared, not the magnitude.

7

$|\mathbf g|=\sqrt{2^2+0^2+(-3)^2}=\boxed{\sqrt{13}}$ .

$\widehat{\mathbf g}=\boxed{(2/\sqrt{13},0,-3/\sqrt{13})}$ . An equivalent form is $(2\sqrt{13}/13,0,-3\sqrt{13}/13)$ . Squared length: $4/13+0+9/13=1$ . Multiplier: $1/\sqrt{13}>0$ .

8

The components are $(-4,4,2)$ , so $|\mathbf h|=\sqrt{16+16+4}=\boxed{6}$ .

\widehat{\mathbf h}=-\frac46\mathbf i+\frac46\mathbf j+\frac26\mathbf k =\boxed{-\frac23\mathbf i+\frac23\mathbf j+\frac13\mathbf k}.

Squared length: $4/9+4/9+1/9=1$ . Multiplier: $1/6>0$ .

9

$|\mathbf v|=\sqrt{0+9+16}=5$ , so E is correct: divide every component by 5. Its squared length is $9/25+16/25=1$ , and its multiplier $1/5$ is positive.

A: 5 is the magnitude, a scalar; the question asks for a vector.
B: this is $\mathbf v/25$ . Its length is $5/25=1/5$ , not 1. It may fit division by magnitude squared.
C: this has length 1 but is $-\mathbf v/5$ , so it points in the opposite direction.
D: direction is correct, but the vector still has length 5; it has not been normalised.

These comparisons identify what each expression does. They do not prove why someone chose it.

10

(a) $|\mathbf d|=\sqrt{(-3)^2+6^2+2^2}=\sqrt{9+36+4}=7$ . Thus

\boxed{\widehat{\mathbf d}=(-3/7,6/7,2/7)}.

(b) Multiply the unit vector by 14:

14\widehat{\mathbf d}=14(-3/7,6/7,2/7)=\boxed{(-6,12,4)}.

(c) Its magnitude is $\sqrt{(-6)^2+12^2+4^2}=\sqrt{36+144+16}=\sqrt{196}=14$ . It equals $2\mathbf d$ ; the positive multiplier 2 confirms the direction. Before moving on: give both checks—length and direction—without relying on the answer looking familiar. This practice is not a controlled reassessment or proof of mastery.

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