Magnitude of a Vector

Find the length of a vector with Pythagoras, combine vectors first when needed, and give exact surd answers.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1 · 32 real past-paper questions below

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What Magnitude Means

The magnitude of a vector is its length. You write it with straight lines round it, like $∣ a ∣$ or $∣ A B ∣$ .

Formula

(x y) = x^{2} + y^{2}

The top number $x$ goes across and the bottom number $y$ goes up or down. They are the two short sides of a right-angled triangle, and the vector is the long side.

Key idea

A length is never negative. Squaring makes both signs disappear.

Finding a Magnitude

Square each number, add, then take the square root.

Worked example

1
$a = (- 5 12)$ — The top number is 5 across (to the left) and the bottom is 12 up. They are the two short sides of a right-angled triangle.
2
$(- 5)^{2} = 25$ — Square the top number. Keep the brackets, because a negative times a negative is positive.
3
$1 2^{2} = 144$ — Square the bottom number.
4
$25 + 144 = 169$ — Add the two squares. This is the long side squared.
5
$∣ a ∣ = 169$ — Take the square root to get back to a length.
6
$∣ a ∣ = 13$ — A length is positive, so there is only one answer.

Common mistake

Writing −25 for the square of −5. A negative times a negative is positive.

Combine First, Then Find the Length

The 3-mark question gives two vectors and asks for the length of a third. Find the third vector first, then use Pythagoras.

Worked example

1
$A B = (7 - 2), C B = (- 1 4)$ — You are given AB and CB but asked for the length of AC. Find the vector AC first.
2
$A C = A B + B C$ — Walk from A to B, then from B to C.
3
$B C = - C B = (1 - 4)$ — You are given CB, which goes the other way. Reverse it by changing both signs.
4
$A C = (7 - 2) + (1 - 4)$ — Put in the two columns.
5
$A C = (7 + 1 - 2 + (- 4)) = (8 - 6)$ — Add the tops, and add the bottoms.
6
$∣ A C ∣^{2} = 8^{2} + (- 6)^{2} = 64 + 36$ — Now use Pythagoras on the new vector. Square each number, and the negative becomes positive.
7
$∣ A C ∣ = 100 = 10$ — Add to get 100, then take the square root.

Common mistake

Adding the two lengths. The lengths of AB and BC add to about 11.4, not 10. Combine the vectors first.

Length Between Two Points

A very common question gives two points and asks how far apart they are. Find the vector between them, then its length.

Worked example

1
$P (- 1, 5), Q (4, - 3)$ — The distance between two points is the length of the vector that joins them.
2
$P Q = q - p$ — Use end minus start to find the vector from P to Q.
3
$P Q = (4 - 3) - (- 1 5)$ — Put in the two columns, with brackets round the negative number.
4
$4 - (- 1) = 5, - 3 - 5 = - 8$ — Subtract the tops, then the bottoms. Taking away −1 adds 1.
5
$∣ P Q ∣^{2} = 5^{2} + (- 8)^{2} = 25 + 64$ — Square each number. The −8 squared is positive.
6
$∣ P Q ∣ = 89 = 9.43$ — Add to get 89, take the square root, and round to 3 significant figures.

Key idea

Q minus P or P minus Q both work here. The signs all flip, but squaring removes them, so the length is the same.

The Length of a Multiple

Some questions give a multiple of a vector, like $E F = - 2 A B$ , and ask for its length. The length of the multiple is the length of $A B$ times the size of the number.

Formula

∣ k v ∣ = ∣ k ∣ \times ∣ v ∣

Here $k$ is the number you multiply by, and $∣ k ∣$ is that number with any minus sign dropped. $∣ v ∣$ is the length of the vector.

Worked example

1
$A B = (- 8 - 15), E F = - 2 A B$ — You want |EF|. Find the length of AB first, because it is quick.
2
$∣ A B ∣^{2} = (- 8)^{2} + (- 15)^{2} = 64 + 225$ — Square each number. Both negatives become positive.
3
$∣ A B ∣ = 289 = 17$ — Add to get 289, then take the square root.
4
$∣ E F ∣ = ∣ - 2 ∣ \times 17$ — Use the rule for a multiple.
5
$∣ E F ∣ = 2 \times 17 = 34$ — A length is never negative, so the −2 becomes 2.
6
$(1630) : 1 6^{2} + 3 0^{2} = 1156 = 3 4^{2}$ — Check it: −2AB is this column, and its length is 34.

Common mistake

Answering −34. The minus sign in the multiplier only reverses the direction, and a length is never negative.

Exact Answers in a Shape

The 5-mark question says exact: leave the square root in. You usually build the vector from a shape first.

Worked example

1
$D A = (- 1 2), D B = (34), D C = 3 A B$ — These are the facts. AB is not given, so build it from DA and DB.
2
$A B = A D + D B$ — Walk from A to D, then from D to B.
3
$A D = - D A = (1 - 2)$ — DA goes the other way, so change both signs.
4
$A B = (1 - 2) + (34) = (42)$ — Add the tops, then the bottoms.
5
$D C = 3 A B = 3 (42)$ — The question says DC is 3 times AB.
6
$D C = (126)$ — Multiply each number by 3.
7
$B C = B D + D C$ — You want BC. Walk from B to D, then from D to C.
8
$B D = - D B = (- 3 - 4)$ — DB goes the other way, so change both signs.
9
$B C = (- 3 - 4) + (126) = (92)$ — Add the tops, then the bottoms.
10
$∣ B C ∣ = 9^{2} + 2^{2} = 81 + 4$ — Square each number and add.
11
$∣ B C ∣ = 85$ — The question says exact, so leave the root. 85 = 5 × 17 has no square number in it, so this cannot be simplified.

You may also be asked for the answer as $k n$ . Find the length, then take out the biggest square number.

Worked example

1
$v = (2 - 4)$ — Find |v| exactly.
2
$∣ v ∣ = 2^{2} + (- 4)^{2} = 4 + 16$ — Square each number and add. The negative becomes positive.
3
$∣ v ∣ = 20$ — Add, and stop before the root.
4
$20 = 4 \times 5$ — Write 20 as a square number times another number. 4 is the biggest square that goes into 20.
5
$4 \times 5 = 4 \times 5$ — The root of a product is the product of the roots.
6
$∣ v ∣ = 25$ — √4 is 2. This is the exact answer, with nothing left to take out of the 5.

Common mistake

Giving 9.22 when the question says exact. A decimal loses the mark.

Finding a Missing Number From the Length

Sometimes the length is given and a number inside the vector is missing. Write the Pythagoras equation and solve it.

Worked example

1
$(k 8) = 10$ — The length is given and the top number is missing.
2
$k^{2} + 8^{2} = 1 0^{2}$ — Square each number on the left and the length on the right. Squaring both sides gets rid of the root.
3
$k^{2} + 64 = 100$ — 8² = 64 and 10² = 100.
4
$k^{2} = 36$ — Take 64 from both sides.
5
$k = 6 or - 6$ — Take the square root. There are two answers, because 6² and (−6)² are both 36.

The missing number can also sit under a square root. The method is the same.

Worked example

1
$(5 y) = 7$ — Now the missing number is inside a square root.
2
$5^{2} + (y)^{2} = 7^{2}$ — Square each number and the length. Squaring a square root gives the number back, so (√y)² is just y.
3
$25 + y = 49$ — 5² = 25 and 7² = 49.
4
$y = 24$ — Take 25 from both sides. Check: 25 + 24 = 49 and √49 = 7.

Key idea

Both answers make an arrow 10 long. One leans right, the other leans left.

Which Method? Read the Question

Look for the word magnitude, length or distance.

You see	You do
Magnitude of one vector	square, add, square root
Two vectors, length of a third	find the third vector first
Length of 3AB or −2AB	length of AB times the number, ignoring any minus sign
Exact	leave the square root
Answer as k√n	take out the biggest square number
Length given, number missing	solve the squared equation
Distance between two points	vector between them, then length
Vector between two points	Position Vectors

Key idea

Real distances work the same way: find the vector in km, then its length in km.

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