Maths · 8 Transformations and vectors

8.3 Magnitude of a vector

Before you start

Column vectors such as $(\begin{matrix} x \\ y \end{matrix})$ , and the notation $\vec{A B}$ and $𝐚$ (8.2).
Writing $\vec{A B}$ from the coordinates of $A$ and $B$ , adding and subtracting vectors, and multiplying a vector by a number (8.2).
Pythagoras' theorem (7.1).
Simplifying a surd, such as $\sqrt{45} = 3 \sqrt{5}$ (1.17).

Syllabus objectives covered (1)

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

Part 1 of 5

Magnitude means length

A vector carries two pieces of information: a direction and a size. In 8.2 you used the column vector $(\begin{matrix} x \\ y \end{matrix})$ to describe a movement: $x$ across (positive to the right) and $y$ up (positive upwards).
The magnitude of a vector is its size: the length of the arrow, measured in a straight line from its tail to its head. It ignores the direction completely.
The syllabus writes magnitude with modulus signs. $| 𝐚 |$ is the magnitude of $𝐚$ , and $| \vec{A B} |$ is the magnitude of $\vec{A B}$ . Read $| 𝐚 |$ as "the magnitude of a".
The answer is a single number, not a column vector. It is a length, so it is never negative. It is zero only for the zero vector $(\begin{matrix} 0 \\ 0 \end{matrix})$ .
A vector along a grid line is easy: $(\begin{matrix} 5 \\ 0 \end{matrix})$ is 5 units long, and $(\begin{matrix} 0 \\ - 3 \end{matrix})$ is 3 units long. It points down, but its length is still 3, not $- 3$ . The new idea in this lesson is the vector that goes across and up at the same time.

Part 2 of 5

Pythagoras gives you the formula

Draw $(\begin{matrix} 3 \\ 4 \end{matrix})$ on a grid. From the tail, go 3 across, then 4 up. Those two moves are at right angles to each other.
graph diagram — not drawn yet
A square grid with an arrow labelled a from the origin to the point (3, 4); a dashed horizontal segment from (0, 0) to (3, 0) labelled 3, a dashed vertical segment from (3, 0) to (3, 4) labelled 4, and a right-angle mark at (3, 0).
The two components are the shorter sides of a right-angled triangle, so the arrow is the hypotenuse.
So the move across, the move up and the arrow make a right-angled triangle. The arrow is the hypotenuse, because it is opposite the right angle.
Pythagoras gives the length of the hypotenuse.
$| (\begin{matrix} 3 \\ 4 \end{matrix}) | = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
The same triangle exists for every column vector. That gives the formula in the objective:
$| (\begin{matrix} x \\ y \end{matrix}) | = \sqrt{x^{2} + y^{2}}$
Do three operations, always in this order: square each component, add the squares, then take the square root of the total. If you stop before the square root, you get $25$ for this vector. That is the square of the length, not the length.
This formula is not in the List of formulas printed on Papers 2 and 4, so you must know it. You do not need to learn it as a separate fact: it is Pythagoras, and you can rebuild it from a sketch of the triangle.
Use a size check on every answer. The magnitude is at least as big as the larger component (ignore its sign), and at most the two component sizes added together. For $(\begin{matrix} 3 \\ 4 \end{matrix})$ : $4 \leq 5 \leq 7$ , so $5$ is possible. An answer of $25$ or $7$ fails the check.

Part 3 of 5

Negative components

A negative component means the move is to the left or down. The length of that move is still positive. $(\begin{matrix} - 5 \\ 12 \end{matrix})$ goes 5 left and 12 up, so its triangle has shorter sides 5 and 12.
In the formula, squaring deals with the sign for you, because the square of a negative number is positive: ${(- 5)}^{2} = 25$ .
$| (\begin{matrix} - 5 \\ 12 \end{matrix}) | = \sqrt{{(- 5)}^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$
The trap is on the calculator. Without brackets, $- 5^{2}$ means "the negative of $5^{2}$ ", which is $- 25$ . The calculator does exactly what you typed, and you get $\sqrt{- 25 + 144} = \sqrt{119} = 10.9$ . That looks believable, but it is wrong.
There are two safe ways. Put brackets round every negative component, ${(- 5)}^{2}$ . Or ignore the sign before you start, because ${(- 5)}^{2}$ and $5^{2}$ are equal.
This also means that a vector and its negative have the same magnitude: $| - 𝐚 | = | 𝐚 |$ . Reversing an arrow does not change its length. So $| \vec{B A} | = | \vec{A B} |$ .
The reverse is not true. Equal magnitudes do not mean equal vectors. $(\begin{matrix} 3 \\ 4 \end{matrix})$ , $(\begin{matrix} 4 \\ - 3 \end{matrix})$ and $(\begin{matrix} 0 \\ - 5 \end{matrix})$ all have magnitude 5, but they point in different directions.

Part 4 of 5

Find the vector first, then its magnitude

Sometimes the vector is not given as a column. It can be $\vec{A B}$ between two points, or a combination such as $2 𝐩 - 𝐪$ . The rule is the same for both: first write the vector as one column vector, then use $\sqrt{x^{2} + y^{2}}$ on that column.
From two points. $\vec{A B}$ is the movement from $A$ to $B$ , so its components are "end minus start". For $A (1, 2)$ and $B (7, 10)$ :
$\vec{A B} = (\begin{matrix} 7 - 1 \\ 10 - 2 \end{matrix}) = (\begin{matrix} 6 \\ 8 \end{matrix}), | \vec{A B} | = \sqrt{36 + 64} = 10$
This magnitude is the length of the line segment $A B$ : the straight-line distance between the two points. So a question about how far apart two points are is a magnitude question.
From a combination. Work out the combination component by component, as in 8.2, until you have one column. Only then square and add.
Do not add magnitudes instead. Take $𝐚 = (\begin{matrix} 3 \\ 4 \end{matrix})$ and $𝐛 = (\begin{matrix} 6 \\ - 8 \end{matrix})$ . Then $| 𝐚 | = 5$ and $| 𝐛 | = 10$ , but $𝐚 + 𝐛 = (\begin{matrix} 9 \\ - 4 \end{matrix})$ , so $| 𝐚 + 𝐛 | = \sqrt{81 + 16} = \sqrt{97} = 9.85$ , not $15$ .
graph diagram — not drawn yet
On a square grid, arrow a from (0, 0) to (3, 4), arrow b from (3, 4) to (9, -4), and arrow a + b from (0, 0) to (9, -4), so the three arrows form a triangle; sides labelled |a| = 5, |b| = 10 and |a + b| = 9.85.
The direct arrow a + b is shorter than the route along a and then along b.
The reason: when you draw $𝐚$ and $𝐛$ nose to tail, $𝐚$ , $𝐛$ and $𝐚 + 𝐛$ make a triangle. One side of a triangle is always shorter than the other two sides added together. The two results are equal only when $𝐚$ and $𝐛$ point the same way, so that there is no triangle.
One shortcut is safe: multiplying a vector by a number $k$ multiplies its length by the size of $k$ . Each component is multiplied by $k$ , so each square is multiplied by $k^{2}$ , and the square root of $k^{2}$ is the size of $k$ . With $𝐚 = (\begin{matrix} 3 \\ 4 \end{matrix})$ :
$| 2 𝐚 | = | (\begin{matrix} 6 \\ 8 \end{matrix}) | = 10 = 2 \times 5, | - 3 𝐚 | = | (\begin{matrix} - 9 \\ - 12 \end{matrix}) | = 15 = 3 \times 5$
A negative $k$ reverses the direction, but the length is still multiplied by a positive number: $3$ , not $- 3$ .

Part 5 of 5

Exact answers, rounding, and working backwards

Many column vectors do not give a whole-number magnitude. What you write then depends on the paper and on the question.
With a calculator: give a non-exact answer to 3 significant figures. Keep the full value on the calculator until the last line.
$| (\begin{matrix} 4 \\ - 6 \end{matrix}) | = \sqrt{16 + 36} = \sqrt{52} = 7.2111 \dots = 7.21$
Without a calculator, or when the question asks for an exact answer: leave the square root in. If the question asks for the simplest form, take out the largest square factor (1.17).
$\sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13}$
To compare two magnitudes, compare the exact values. $\sqrt{74}$ and $\sqrt{74}$ are certainly equal. Two rounded decimals that match do not prove that two lengths are equal.
Working backwards. Sometimes the magnitude is given and one component is a letter. Write the formula as an equation and solve it. If $𝐯 = (\begin{matrix} k \\ 8 \end{matrix})$ and $| 𝐯 | = 10$ :
$k^{2} + 8^{2} = 10^{2} \Rightarrow k^{2} = 100 - 64 = 36 \Rightarrow k = 6 or k = - 6$
Square the magnitude too. The equation is $k^{2} + 64 = 100$ , not $k^{2} + 64 = 10$ . Squaring both sides of $\sqrt{k^{2} + 64} = 10$ is what removes the square root.
Keep both signs. $(\begin{matrix} 6 \\ 8 \end{matrix})$ and $(\begin{matrix} - 6 \\ 8 \end{matrix})$ both have magnitude 10, so both values of $k$ are correct unless the question tells you that $k$ is positive.

Worked examples

Example 1. $𝐚 = (\begin{matrix} - 7 \\ 24 \end{matrix})$ . Calculate $| 𝐚 |$ without a calculator.

Show the working

Write the formula with the components in. Put brackets round the negative component.
$| 𝐚 | = \sqrt{{(- 7)}^{2} + 24^{2}}$
Square each component. ${(- 7)}^{2} = 49$ , which is positive because a negative times a negative is positive. $24^{2} = 576$ .
Add the squares, then take the square root. $49 + 576 = 625$ , and $625 = 25^{2}$ .
$| 𝐚 | = \sqrt{625} = 25$
Size check: $25$ is bigger than $24$ , the larger component, and smaller than $7 + 24 = 31$ . The answer is possible.

Answer: $| 𝐚 | = 25$

Example 2. $A$ is the point $(- 3, 5)$ and $B$ is the point $(4, 1)$ . Calculate $| \vec{A B} |$ .

Show the working

Find the column vector first. $\vec{A B}$ goes from $A$ to $B$ , so subtract the coordinates of $A$ from the coordinates of $B$ .
$\vec{A B} = (\begin{matrix} 4 - (- 3) \\ 1 - 5 \end{matrix}) = (\begin{matrix} 7 \\ - 4 \end{matrix})$
Check the signs with a picture in your head. From $x = - 3$ to $x = 4$ is 7 to the right. From $y = 5$ to $y = 1$ is 4 down. Right and down give $7$ and $- 4$ .
Now find the magnitude of that column vector.
$| \vec{A B} | = \sqrt{7^{2} + {(- 4)}^{2}} = \sqrt{49 + 16} = \sqrt{65}$
$\sqrt{65}$ is not a whole number. With a calculator, $\sqrt{65} = 8.0622 \dots$ , which is $8.06$ to 3 significant figures.

Answer: $| \vec{A B} | = \sqrt{65} = 8.06$ (3 s.f.)

Example 3. $𝐩 = (\begin{matrix} 3 \\ - 1 \end{matrix})$ and $𝐪 = (\begin{matrix} - 1 \\ 4 \end{matrix})$ . Calculate $| 2 𝐩 - 𝐪 |$ .

Show the working

Make one column vector first. Multiply $𝐩$ by 2, which doubles each component.
$2 𝐩 = (\begin{matrix} 6 \\ - 2 \end{matrix})$
Subtract $𝐪$ component by component. Take care with the double negative in the top line.
$2 𝐩 - 𝐪 = (\begin{matrix} 6 - (- 1) \\ - 2 - 4 \end{matrix}) = (\begin{matrix} 7 \\ - 6 \end{matrix})$
Now find the magnitude of that one column vector.
$| 2 𝐩 - 𝐪 | = \sqrt{7^{2} + {(- 6)}^{2}} = \sqrt{49 + 36} = \sqrt{85} = 9.2195 \dots$
Compare the wrong shortcut: $2 | 𝐩 | - | 𝐪 | = 2 \sqrt{10} - \sqrt{17} = 2.20$ . That is a different number. This is why you combine the vectors before you find any magnitude.

Answer: $| 2 𝐩 - 𝐪 | = \sqrt{85} = 9.22$ (3 s.f.)

Example 4. $𝐯 = (\begin{matrix} k \\ - 5 \end{matrix})$ and $| 𝐯 | = 13$ . Find the possible values of $k$ .

Show the working

Write the magnitude formula as an equation, with the given magnitude on the right.
$\sqrt{k^{2} + {(- 5)}^{2}} = 13$
Square both sides to remove the square root. This is why $13$ becomes $169$ .
$k^{2} + 25 = 169$
Subtract 25 from both sides.
$k^{2} = 144$
Take the square root and keep both signs, because $12^{2}$ and ${(- 12)}^{2}$ are both $144$ .
$k = 12 or k = - 12$
Check one value: $\sqrt{12^{2} + {(- 5)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13$ . Correct.

Answer: $k = 12$ or $k = - 12$

Where people go wrong

Mistake: Adding the components: $| (\begin{matrix} 3 \\ 4 \end{matrix}) | = 3 + 4 = 7$ .

Instead: Square each component, add the squares, then take the square root: $\sqrt{9 + 16} = 5$ .

The magnitude is the hypotenuse of the triangle that the components make. The route across and then up is longer than the straight route along the arrow.

Mistake: Typing $- 5^{2}$ on the calculator for the square of $- 5$ , and getting $- 25$ .

Instead: Type ${(- 5)}^{2}$ with brackets, or square $5$ instead, because the sign makes no difference after squaring.

Without brackets, $- 5^{2}$ means the negative of $5^{2}$ . The result, for example $\sqrt{- 25 + 144} = 10.9$ instead of $13$ , looks believable, so it is easy to miss.

Mistake: Working out $| 𝐚 + 𝐛 |$ as $| 𝐚 | + | 𝐛 |$ .

Instead: Add the vectors first to get one column vector, then find the magnitude of that column vector.

$𝐚$ , $𝐛$ and $𝐚 + 𝐛$ make a triangle, and one side of a triangle is shorter than the other two added together.

Mistake: Stopping at $x^{2} + y^{2}$ , and giving $25$ as the magnitude of $(\begin{matrix} 3 \\ 4 \end{matrix})$ .

Instead: Finish with the square root: $\sqrt{25} = 5$ .

$x^{2} + y^{2}$ is the square of the length. The size check finds the mistake: a magnitude can never be more than the two component sizes added together, and $25 > 3 + 4$ .

Mistake: Giving only $k = 12$ as the solution of $k^{2} = 144$ .

Instead: Give $k = 12$ and $k = - 12$ , unless the question says that $k$ is positive.

A component can be negative, and ${(- 12)}^{2} = 144$ too. Both vectors have the same magnitude.

Check yourself

Calculate $| (\begin{matrix} 8 \\ - 15 \end{matrix}) |$ .
Answer
$\sqrt{64 + 225} = \sqrt{289} = 17$
Calculate the magnitude of $(\begin{matrix} 2 \\ - 5 \end{matrix})$ . Give the exact value, and then the value to 3 significant figures.
Answer
$\sqrt{4 + 25} = \sqrt{29} = 5.39$ (3 s.f.)
Find the magnitude of $(\begin{matrix} - 6 \\ 3 \end{matrix})$ as a surd in its simplest form.
Answer
$\sqrt{36 + 9} = \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$
$A$ is the point $(1, 2)$ and $B$ is the point $(- 11, 7)$ . Calculate $| \vec{A B} |$ .
Answer
$\vec{A B} = (\begin{matrix} - 12 \\ 5 \end{matrix})$ , so $| \vec{A B} | = \sqrt{144 + 25} = \sqrt{169} = 13$
A drone flies from its launch point along the vector $(\begin{matrix} 120 \\ - 50 \end{matrix})$ , in metres. Calculate the straight-line distance from the launch point to where the drone stops.
Answer
$\sqrt{120^{2} + {(- 50)}^{2}} = \sqrt{14400 + 2500} = \sqrt{16900} = 130$ m
$𝐚 = (\begin{matrix} 5 \\ 0 \end{matrix})$ and $𝐛 = (\begin{matrix} 0 \\ 12 \end{matrix})$ . Calculate $| 𝐚 + 𝐛 |$ , and compare it with $| 𝐚 | + | 𝐛 |$ .
Answer
$𝐚 + 𝐛 = (\begin{matrix} 5 \\ 12 \end{matrix})$ , so $| 𝐚 + 𝐛 | = 13$ . But $| 𝐚 | + | 𝐛 | = 5 + 12 = 17$ . The two are different, so you cannot add magnitudes.
$(\begin{matrix} k \\ 20 \end{matrix})$ has magnitude $29$ . Find the possible values of $k$ .
Answer
$k^{2} + 400 = 841$ , so $k^{2} = 441$ and $k = 21$ or $k = - 21$ .
Explain why $| - 𝐚 | = | 𝐚 |$ for any vector $𝐚$ .
Answer
$- 𝐚$ has the same components as $𝐚$ with their signs changed. Squaring removes the signs, so $x^{2} + y^{2}$ is the same, and so is its square root. In a picture, $- 𝐚$ is the same arrow pointing the other way, so its length is the same.

Exam questions

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

Question 1. $𝐚 = (\begin{matrix} 9 \\ - 12 \end{matrix})$ and $𝐛 = (\begin{matrix} - 5 \\ 8 \end{matrix})$ .

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

Mark scheme

M1 √(9² + (−12)²) oe, or 81 + 144 seen
A1 15 cao

(b) Calculate $| 𝐚 + 2 𝐛 |$ . Give your answer in exact form. 3 marks No calculator

Mark scheme

B1 $𝐚 + 2 𝐛 = (\begin{matrix} - 1 \\ 4 \end{matrix})$ soi
M1 √((−1)² + 4²) oe; FT their column vector
A1 √17 cao

Question 2. The points $P (- 4, 1)$ , $Q (3, 6)$ and $R (3, - 4)$ are the corners of triangle $P Q R$ .

(a) Write $\vec{P Q}$ as a column vector. 1 mark

Mark scheme

B1 $(\begin{matrix} 7 \\ 5 \end{matrix})$

(b) Calculate $| \vec{P Q} |$ . 2 marks

Mark scheme

M1 √(7² + 5²) oe; FT their (a)
A1 8.60 awrt, or √74

(c) Show that triangle $P Q R$ is isosceles. 3 marks

Mark scheme

M1 $\vec{P R} = (\begin{matrix} 7 \\ - 5 \end{matrix})$ or $\vec{Q R} = (\begin{matrix} 0 \\ - 10 \end{matrix})$ seen
M1 |PR| = √(7² + (−5)²) = √74 oe
A1 states PQ = PR (both √74 oe), so two sides are equal and the triangle is isosceles [QR = 10]. Any other correct complete method scores full marks, e.g. the midpoint of QR is (3, 1), level with P, and QR is vertical, so P is on the perpendicular bisector of QR

Question 3. $𝐮 = (\begin{matrix} k \\ 7 \end{matrix})$ , where $k$ is a constant, and $| 𝐮 | = 25$ .

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

Mark scheme

M1 k² + 7² = 25² oe
M1 dep: k² = 625 − 49 or k² = 576
A1 24 and −24, both required; ±24 oe

(b) Write down $| - 3 𝐮 |$ . 1 mark No calculator

Mark scheme

B1 75

How this is examined

The magnitude of a vector you must build first: $\vec{A B}$ from two points, or a combination such as $𝐚 + 2 𝐛$ : A Calculate part of 2 or 3 marks, on its own or after a part that asks you to Write the column vector The correct single column vector first, e.g. $\vec{A B} = (\begin{matrix} 7 \\ - 4 \end{matrix})$ from "end minus start"; then the square root of the sum of the squares of its components, $\sqrt{7^{2} + {(- 4)}^{2}}$ ; then the value, $\sqrt{65} = 8.06$ . Where a scheme allows FT, a correct method on a wrong column vector can still earn the method mark.
Using magnitudes to show that two sides are equal, or that a triangle is isosceles: A Show (that) of about 3 marks on a triangle given by the coordinates of its corners Each column vector written, each magnitude worked out, best in exact form, e.g. $| \vec{P Q} | = \sqrt{74}$ and $| \vec{P R} | = \sqrt{74}$ , then a final sentence that states the result: two sides are equal, so the triangle is isosceles. The result written without the working does not show it.
Working back from a given magnitude to a missing component: A Find of about 3 marks: a column vector with a letter in it, and its magnitude given The equation $k^{2} + 7^{2} = 25^{2}$ written down; $k^{2} = 576$ ; then both answers, $k = 24$ and $k = - 24$ , unless the question says that $k$ is positive.
The magnitude of a column vector, straight from $\sqrt{x^{2} + y^{2}}$ : A 2-mark Calculate, with or without a calculator The formula with the numbers in and brackets round any negative, e.g. $\sqrt{9^{2} + {(- 12)}^{2}}$ ; then the value, $15$ . With a calculator, a non-exact answer goes to 3 significant figures.
Exact magnitudes when the square root does not come out whole: The final line of a non-calculator magnitude part, or a part that asks for the answer in exact form or as a surd The square root left in, e.g. $\sqrt{17}$ , and simplified when the question asks for the simplest form, e.g. $\sqrt{52} = 2 \sqrt{13}$ . A rounded decimal is not an exact answer.

Words to know

Vector: A quantity with both a size and a direction, written in this topic as a column vector.
Component: One of the two numbers in a column vector: the top one is the move across, the bottom one is the move up.
Magnitude: The length of a vector. For $(\begin{matrix} x \\ y \end{matrix})$ it is $\sqrt{x^{2} + y^{2}}$ . It is never negative.
Modulus signs: The vertical bars in $| 𝐚 |$ , meaning "the magnitude of".
Scalar: A quantity with a size but no direction. A magnitude is a scalar.
Exact form: An answer with the square root left in, such as $2 \sqrt{13}$ , instead of a rounded decimal.

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