Ext Maths · 7 Transformations and vectors

7.3 Magnitude of a vector

Before you start

Column vectors, and adding, subtracting and multiplying them by a scalar (7.2).
Pythagoras' theorem: in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Squaring negative numbers: ${(- 3)}^{2} = 9$ .
Rounding to 3 significant figures.

Syllabus objectives covered (1)

Calculate the magnitude of a vector \binom{x}{y} as √(x² + y²).

Part 1 of 4

Magnitude is the length of the arrow

The magnitude of a vector is its length. It tells you how far the movement goes and ignores which way it goes.
Magnitude is written with modulus signs, straight lines either side: $| 𝐚 |$ is the magnitude of $𝐚$ , and $| \vec{A B} |$ is the magnitude of $\vec{A B}$ . Read $| 𝐚 |$ as 'the magnitude of a'.
Here is where the formula comes from. The vector $(\binom{x}{y})$ goes $x$ across and $y$ up. Those two movements are the two shorter sides of a right-angled triangle, and the vector itself is the hypotenuse.
The components are the two short sides and the vector is the hypotenuse: $| (\binom{- 3}{4}) | = \sqrt{9 + 16} = 5$ . The 3 left still counts as a length of 3.
So Pythagoras gives the magnitude directly:
$| (\binom{x}{y}) | = \sqrt{x^{2} + y^{2}}$
This formula is not on the formula list printed in the paper, so you must remember it. If you forget it, sketch the right-angled triangle and use Pythagoras, which is exactly what the formula is.
For example, $\vec{P Q} = (\binom{- 3}{4})$ has magnitude $\sqrt{{(- 3)}^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$ .
If $A$ and $B$ are points, $| \vec{A B} |$ is simply the length of the line $A B$ , so this is also how you find the distance between two points from their vectors.

Part 2 of 4

Negative components and your calculator

A magnitude is a length, so it can never be negative. Squaring removes every minus sign: ${(- 3)}^{2} = 9$ , not $- 9$ .
The calculator trap: on most scientific calculators, typing $- 3^{2}$ without brackets gives $- 9$ , because the calculator squares the 3 first and then applies the minus. Type ${(- 3)}^{2}$ with the brackets.
Because the sign disappears anyway, you can square the size of each number and ignore its sign: $| (\binom{- 5}{- 12}) | = \sqrt{25 + 144} = \sqrt{169} = 13$ .
The same idea explains why direction never changes magnitude: $| \vec{B A} | = | \vec{A B} |$ and $| - 𝐚 | = | 𝐚 |$ . Reversing a vector flips its signs, and the squares are unchanged.

Part 3 of 4

Exact answers and 3 significant figures

When $x^{2} + y^{2}$ is a perfect square, the magnitude is a whole number and you give it exactly: $| (\binom{5}{- 12}) | = \sqrt{169} = 13$ .
Usually it is not a perfect square. $| (\binom{2}{- 4}) | = \sqrt{4 + 16} = \sqrt{20}$ , which is not a whole number.
On Paper 4 the rule is to give non-exact answers correct to 3 significant figures unless the question says otherwise. $\sqrt{20} = 4.4721 \dots$ , so write $4.47$ .
Write down the unrounded value (at least four figures) before you round. If you round too early and then use that value again, the final answer can drift out of range.
If a question asks for the answer as an exact value or in surd form, leave the square root in: $\sqrt{20}$ , which simplifies to $2 \sqrt{5}$ (surds are 1.18).

Part 4 of 4

Magnitude of a combined vector, and working backwards

To find something like $| 𝐚 + 2 𝐛 |$ , first work out $𝐚 + 2 𝐛$ as a single column vector (7.2), then find the magnitude of that column. The order is: combine, then measure.
The tempting shortcut $| 𝐚 + 𝐛 | = | 𝐚 | + | 𝐛 |$ is almost always wrong. With $𝐚 = (\binom{3}{0})$ and $𝐛 = (\binom{0}{4})$ , $| 𝐚 + 𝐛 | = | (\binom{3}{4}) | = 5$ , but $| 𝐚 | + | 𝐛 | = 3 + 4 = 7$ .
That is geometry, not a coincidence: going straight from start to finish is shorter than going along two sides of a triangle. The two only agree when $𝐚$ and $𝐛$ point in exactly the same direction.
Multiplying by a scalar does scale the magnitude: $| 3 𝐚 | = 3 | 𝐚 |$ , and $| - 2 𝐚 | = 2 | 𝐚 |$ , because a magnitude is never negative.
Sometimes you are given the magnitude and a component is missing. Square both sides to remove the root, then solve. A squared unknown has two square roots, so expect a positive and a negative answer.
$| (\binom{k}{12}) | = 13 \Rightarrow k^{2} + 144 = 169 \Rightarrow k^{2} = 25 \Rightarrow k = 5 or k = - 5$

Worked examples

Example 1. $𝐩 = (\binom{- 7}{24})$ . Calculate $| 𝐩 |$ .

Show the working

Write the formula with the components substituted, keeping the bracket round the negative.
$| 𝐩 | = \sqrt{{(- 7)}^{2} + 24^{2}}$
Square each component. The minus sign disappears.
$= \sqrt{49 + 576} = \sqrt{625}$
625 is a perfect square ( $25^{2}$ ), so the answer is exact and needs no rounding.

Answer: $| 𝐩 | = 25$

Example 2. $𝐚 = (\binom{3}{- 1})$ and $𝐛 = (\binom{- 2}{4})$ . Calculate $| 𝐚 - 2 𝐛 |$ , giving your answer correct to 3 significant figures.

Show the working

Combine first. Multiply $𝐛$ by 2.
$2 𝐛 = (\binom{- 4}{8})$
Subtract, with brackets round the negatives.
$𝐚 - 2 𝐛 = (\binom{3 - (- 4)}{- 1 - 8}) = (\binom{7}{- 9})$
Now find the magnitude of that single vector.
$| (\binom{7}{- 9}) | = \sqrt{49 + 81} = \sqrt{130}$
$\sqrt{130} = 11.4017 \dots$ Write this down, then round to 3 significant figures.

Answer: $| 𝐚 - 2 𝐛 | = 11.4$ (3 s.f.)

Example 3. $\vec{O P} = (\binom{k}{- 6})$ and $| \vec{O P} | = 10$ . Find the possible values of $k$ .

Show the working

Write the magnitude using the formula and set it equal to 10.
$\sqrt{k^{2} + {(- 6)}^{2}} = 10$
Square both sides to remove the square root.
$k^{2} + 36 = 100$
Solve for $k^{2}$ .
$k^{2} = 64$
Take the square root, and remember both signs: $8^{2} = 64$ and ${(- 8)}^{2} = 64$ . Both vectors $(\binom{8}{- 6})$ and $(\binom{- 8}{- 6})$ have length 10.

Answer: $k = 8$ or $k = - 8$ .

Where people go wrong

Mistake: Squaring a negative component to get a negative number, e.g. ${(- 3)}^{2} = - 9$ .

Instead: ${(- 3)}^{2} = 9$ . Use brackets on the calculator.

A negative times a negative is positive. Without brackets, the calculator squares 3 and then makes it negative.

Mistake: Writing $| 𝐚 + 𝐛 | = | 𝐚 | + | 𝐛 |$ .

Instead: Work out $𝐚 + 𝐛$ as one column vector first, then find its magnitude.

The straight route is shorter than two sides of a triangle, so the two are usually different.

Mistake: Forgetting to square: $| (\binom{3}{4}) | = \sqrt{3 + 4}$ .

Instead: $\sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$ .

The formula is Pythagoras, which works with the squares of the sides, not the sides.

Mistake: Giving only $k = 8$ when $k^{2} = 64$ .

Instead: $k = 8$ or $k = - 8$ .

Both a vector pointing right and one pointing left can have the same length. The question saying 'values' is the hint.

Check yourself

Calculate $| (\binom{5}{- 12}) |$ .
Answer
$\sqrt{25 + 144} = \sqrt{169} = 13$ .
Calculate $| (\binom{- 4}{7}) |$ , correct to 3 significant figures.
Answer
$\sqrt{16 + 49} = \sqrt{65} = 8.062 \dots = 8.06$ .
$A$ is $(1, - 2)$ and $B$ is $(7, 6)$ . Find $| \vec{A B} |$ .
Answer
$\vec{A B} = (\binom{6}{8})$ , so $| \vec{A B} | = \sqrt{36 + 64} = 10$ .
$𝐚 = (\binom{1}{3})$ and $𝐛 = (\binom{2}{- 5})$ . Is $| 𝐚 + 𝐛 |$ equal to $| 𝐚 | + | 𝐛 |$ ? Work both out to 3 significant figures.
Answer
No. $𝐚 + 𝐛 = (\binom{3}{- 2})$ , so $| 𝐚 + 𝐛 | = \sqrt{13} = 3.61$ . But $| 𝐚 | + | 𝐛 | = \sqrt{10} + \sqrt{29} = 8.55$ .
You type $- 3^{2}$ into your calculator and get $- 9$ . What has gone wrong, and what should you type?
Answer
The calculator squared 3 and then applied the minus sign. Type ${(- 3)}^{2}$ , which gives 9.
$| (\binom{t}{t}) | = \sqrt{50}$ . Find the possible values of $t$ .
Answer
$t^{2} + t^{2} = 50$ , so $2 t^{2} = 50$ , $t^{2} = 25$ , and $t = 5$ or $t = - 5$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. $𝐮 = (\binom{6}{- 8})$ and $𝐯 = (\binom{- 5}{2})$ .

(a) Calculate $| 𝐮 |$ . 2 marks No calculator

Mark scheme

M1 $\sqrt{6^{2} + {(- 8)}^{2}}$ oe
A1 10

(b) Calculate $| 𝐮 + 2 𝐯 |$ , giving your answer correct to 3 significant figures. 3 marks

Mark scheme

B1 $(\binom{- 4}{- 4})$ seen
M1 $\sqrt{{(- 4)}^{2} + {(- 4)}^{2}}$ oe, FT their vector
A1 5.66 awrt (5.656…)

Question 2. $\vec{A B} = (\binom{2 k}{k})$ , where $k$ is a constant, and $| \vec{A B} | = \sqrt{80}$ .

(a) Find the two possible values of $k$ . 3 marks No calculator

Mark scheme

M1 ${(2 k)}^{2} + k^{2} = 80$ oe
M1 $5 k^{2} = 80$ or $k^{2} = 16$
A1 $k = 4$ and $k = - 4$

(b) Explain why both values of $k$ give the same magnitude. 1 mark No calculator

Mark scheme

B1 the components are squared, so their signs do not affect the magnitude oe

Question 3. A drone leaves its base and flies with displacement $(\binom{7}{- 3})$ , then with displacement $(\binom{- 2}{9})$ . The components are in kilometres.

(a) Calculate the distance of the drone from its base after both flights. 3 marks

Mark scheme

M1 $(\binom{5}{6})$ seen
M1 $\sqrt{5^{2} + 6^{2}}$ , FT their vector
A1 7.81 awrt (km)

(b) Calculate the total distance flown by the drone. 2 marks

Mark scheme

M1 $\sqrt{7^{2} + {(- 3)}^{2}} + \sqrt{{(- 2)}^{2} + 9^{2}}$ oe
A1 16.8 awrt (km)

How this is examined

Magnitude of a combination like $| 𝐮 + 2 𝐯 |$ : A 2- or 3-mark 'Calculate' that follows a part where you combined vectors, often asking for 3 significant figures The single combined column vector written down, then $\sqrt{x^{2} + y^{2}}$ with that vector's numbers substituted, then the value to 3 s.f. with the unrounded value seen first.
Magnitude of a single given vector: A 1- or 2-mark 'Calculate' or 'Work out' such as $| \vec{A B} |$ $\sqrt{{(top)}^{2} + {(bottom)}^{2}}$ written with the numbers in, brackets round negatives, then the value. If the sum is a perfect square, give the exact whole number.
A missing component from a given magnitude: A 2- or 3-mark 'Find' with an unknown inside the column vector The equation with the root removed, e.g. $k^{2} + 36 = 100$ , then both values, e.g. $k = 8$ or $k = - 8$ .

Words to know

Magnitude: The length of a vector, ignoring its direction. $| (\binom{x}{y}) | = \sqrt{x^{2} + y^{2}}$ .
Modulus signs: The straight lines in $| 𝐚 |$ , meaning 'the magnitude of'.
Component: One of the two numbers in a column vector: the movement across (top) or up (bottom).
Exact value: An answer left unrounded, such as $13$ or $\sqrt{20}$ , rather than a decimal approximation.

All Ext Maths lessons Next: 7.4 Vector geometry