The Magnitude of a Vector

What you'll learn

How to read a column vector as horizontal and vertical movement.
How to find the magnitude, or length, of a vector using Pythagoras.
How to combine vectors in triangles, parallelograms and trapezia before finding a magnitude.
How to give exact answers using square roots and simplified surds.

Vectors as movements

A vector describes a movement with both size and direction. In geometry, $AB⃗\vec{AB}$ means “the vector from A to B”. The order of the letters matters.

Definition

Column vector and components

A column vector is a vector written in the form

(xy)\begin{pmatrix} x \\ y \end{pmatrix}

where the top number, $x$ , is the horizontal component and the bottom number, $y$ , is the vertical component.

For example,

AB⃗=(5−2)\vec{AB}=\begin{pmatrix} 5 \\ -2 \end{pmatrix}

means “move 5 units to the right and 2 units down” from A to B.

If you reverse a vector, you change the signs of both components:

BA⃗=−AB⃗\vec{BA}=-\vec{AB}

So if $AB⃗=(5−2)\vec{AB}=\begin{pmatrix} 5 \\ -2 \end{pmatrix}$ , then

BA⃗=(−52)\vec{BA}=\begin{pmatrix} -5 \\ 2 \end{pmatrix}

Example

Using the order of letters

Suppose

AB⃗=(4−3)CB⃗=(−26)\vec{AB}=\begin{pmatrix} 4 \\ -3 \end{pmatrix} \quad \vec{CB}=\begin{pmatrix} -2 \\ 6 \end{pmatrix}

Find $AC⃗\vec{AC}$ .

You want to travel from A to C. A useful route is A to B, then B to C:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
You are given $CB⃗\vec{CB}$ , but you need $BC⃗\vec{BC}$ . Reverse the vector by changing both signs:
$BC⃗=−CB⃗=(2−6)\vec{BC}=-\vec{CB} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$
Add the components:
$AC⃗=(4−3)+(2−6)=(6−9)\vec{AC} = \begin{pmatrix} 4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -6 \end{pmatrix} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}$

Common Mistake

Mixing up $C B$ and $B C$

$CB⃗\vec{CB}$ and $BC⃗\vec{BC}$ have the same length, but they point in opposite directions. Always change both signs when reversing a vector.

Magnitude means length

Definition

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v⃗\vec{v}$ is written as $∣v⃗∣|\vec{v}|$ .

For a column vector $v⃗=(xy)\vec{v}=\begin{pmatrix} x \\ y \end{pmatrix}$ , the horizontal and vertical components form a right-angled triangle. So we use Pythagoras:

∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}

A vector split into horizontal and vertical components, showing magnitude as the hypotenuse

Key Idea

The magnitude formula

To find the length of $(xy)\begin{pmatrix} x \\ y \end{pmatrix}$ , square both components, add them, then square root.

Example

Finding the magnitude of a column vector

Find the magnitude of

v⃗=(−815)\vec{v}=\begin{pmatrix} -8 \\ 15 \end{pmatrix}

Identify the two components:
$\quad y=15$
Substitute into the magnitude formula:
$∣v⃗∣=(−8)2+152|\vec{v}|=\sqrt{(-8)^2+15^2}$
Square the components and add:
$∣v⃗∣=64+225=289|\vec{v}|=\sqrt{64+225}=\sqrt{289}$
Square root the answer:
$∣v⃗∣=17|\vec{v}|=17$

Common Mistake

Forgetting that negatives square to positives

If a component is negative, keep it in brackets when squaring. For example, $8)^2=64$ , not -64.

Exact magnitudes and surds

Sometimes the square root is not a whole number. In that case, you may be asked for the exact magnitude.

Definition

Exact answer and surd

An exact answer has not been rounded. A surd is a square root that is left in exact form, such as $29\sqrt{29}$ .

If possible, simplify the square root by taking out a square factor:

116=4×29=229\sqrt{116}=\sqrt{4 \times 29}=2\sqrt{29}

Example

Giving an exact magnitude

Find the exact magnitude of

p⃗=(10−4)\vec{p}=\begin{pmatrix} 10 \\ -4 \end{pmatrix}

Use the magnitude formula:
$∣p⃗∣=102+(−4)2|\vec{p}|=\sqrt{10^2+(-4)^2}$
Simplify inside the square root:
$∣p⃗∣=100+16=116|\vec{p}|=\sqrt{100+16}=\sqrt{116}$
Take out the square factor 4:
$116=4×29=229\sqrt{116}=\sqrt{4 \times 29}=2\sqrt{29}$

Tip

Spot square factors

When simplifying a square root, look for factors like 4, 9, 16, 25, 36, 49 and 64.

Find the vector first, then find its magnitude

Many exam questions do not give you the vector you need directly. You must build it from other vectors first.

A helpful rule is:

AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}

Rearranging gives:

BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}

Triangle vector diagram showing AB plus BC equals AC

Key Idea

Two-stage method

For magnitude questions in geometry, first find the correct column vector, then apply $∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}$ .

Example

Finding a hidden vector before its magnitude

Given

AB⃗=(5−7)CB⃗=(−14)\vec{AB}=\begin{pmatrix} 5 \\ -7 \end{pmatrix} \quad \vec{CB}=\begin{pmatrix} -1 \\ 4 \end{pmatrix}

find the magnitude of $AC⃗\vec{AC}$ .

Use the route from A to B to C:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
Reverse $CB⃗\vec{CB}$ to get $BC⃗\vec{BC}$ :
$BC⃗=−CB⃗=(1−4)\vec{BC}=-\vec{CB} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$
Add the vectors:
$AC⃗=(5−7)+(1−4)=(6−11)\vec{AC} = \begin{pmatrix} 5 \\ -7 \end{pmatrix} + \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 6 \\ -11 \end{pmatrix}$
Find the magnitude:
$∣AC⃗∣=62+(−11)2=36+121=157|\vec{AC}|=\sqrt{6^2+(-11)^2} = \sqrt{36+121} = \sqrt{157}$

Parallelograms and diagonals

Definition

Parallelogram

A parallelogram is a quadrilateral with both pairs of opposite sides parallel and equal in length.

In a parallelogram $A B C D$ , the diagonal $AC⃗\vec{AC}$ can be made by going along $AB⃗\vec{AB}$ and then $BC⃗\vec{BC}$ :

AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}

So if you know $AB⃗\vec{AB}$ and $AC⃗\vec{AC}$ , you can find:

BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}

Example

Magnitude in a parallelogram

$A B C D$ is a parallelogram.

AB⃗=(32)AC⃗=(11−4)\vec{AB}=\begin{pmatrix} 3 \\ 2 \end{pmatrix} \quad \vec{AC}=\begin{pmatrix} 11 \\ -4 \end{pmatrix}

Find the magnitude of $BC⃗\vec{BC}$ .

Use the diagonal relationship:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
Rearrange to find $BC⃗\vec{BC}$ :
$BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}$
Subtract the column vectors component by component:
$BC⃗=(11−4)−(32)=(8−6)\vec{BC} = \begin{pmatrix} 11 \\ -4 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$
Find the magnitude:
$∣BC⃗∣=82+(−6)2=64+36=100=10|\vec{BC}|=\sqrt{8^2+(-6)^2} = \sqrt{64+36} = \sqrt{100} = 10$

Tip

Subtracting column vectors

Subtract top from top, and bottom from bottom. Keep the order exactly as written.

Trapezia and scalar multiples

Definition

Trapezium and scalar multiple

A trapezium is a quadrilateral with at least one pair of parallel sides. A scalar multiple means a vector has been multiplied by a number, such as $3AB⃗3\vec{AB}$ .

If you are told $DC⃗=3AB⃗\vec{DC}=3\vec{AB}$ , then $DC⃗\vec{DC}$ points in the same direction as $AB⃗\vec{AB}$ and is three times as long.

To solve these questions, build the missing side carefully using vector paths.

Example

Exact magnitude in a trapezium

$A B C D$ is a trapezium and

DC⃗=3AB⃗\vec{DC}=3\vec{AB}

Also,

DA⃗=(2−1)DB⃗=(53)\vec{DA}=\begin{pmatrix} 2 \\ -1 \end{pmatrix} \quad \vec{DB}=\begin{pmatrix} 5 \\ 3 \end{pmatrix}

Find the exact magnitude of $BC⃗\vec{BC}$ .

First find $AB⃗\vec{AB}$ . Going from D to A, then A to B, gives D to B:
$DA⃗+AB⃗=DB⃗\vec{DA}+\vec{AB}=\vec{DB}$
Rearrange:
$AB⃗=DB⃗−DA⃗\vec{AB}=\vec{DB}-\vec{DA}$
Substitute and subtract:
$AB⃗=(53)−(2−1)=(34)\vec{AB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
Use $DC⃗=3AB⃗\vec{DC}=3\vec{AB}$ :
$DC⃗=3(34)=(912)\vec{DC} = 3\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}$
To get from B to C, go from B to D, then D to C:
$BC⃗=BD⃗+DC⃗\vec{BC}=\vec{BD}+\vec{DC}$
Since $BD⃗=−DB⃗\vec{BD}=-\vec{DB}$ , substitute both vectors:
$BC⃗=(−5−3)+(912)=(49)\vec{BC} = \begin{pmatrix} -5 \\ -3 \end{pmatrix} + \begin{pmatrix} 9 \\ 12 \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix}$
Find the exact magnitude:
$∣BC⃗∣=42+92=16+81=97|\vec{BC}|=\sqrt{4^2+9^2} = \sqrt{16+81} = \sqrt{97}$

Common Mistake

Do not assume the diagram is to scale

In vector geometry, the written vector information is more reliable than how the shape looks. Use the vector equations, not visual guessing.

Exam technique

In the exam

Write a vector path for the vector you need, such as $AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$ .
Reverse vectors carefully: $BA⃗=−AB⃗\vec{BA}=-\vec{AB}$ , so both components change sign.
Only find the magnitude after you have the final column vector, then use $∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}$ .

Self review

Check yourself

If $PQ⃗=(−38)\vec{PQ}=\begin{pmatrix} -3 \\ 8 \end{pmatrix}$ , what is $QP⃗\vec{QP}$ ?
Why does a negative component not make the magnitude negative?
In a parallelogram, how can you find $BC⃗\vec{BC}$ if you know $AB⃗\vec{AB}$ and $AC⃗\vec{AC}$ ?

The Magnitude of a Vector

What you'll learn

How to read a column vector as horizontal and vertical movement.
How to find the magnitude, or length, of a vector using Pythagoras.
How to combine vectors in triangles, parallelograms and trapezia before finding a magnitude.
How to give exact answers using square roots and simplified surds.

Vectors as movements

A vector describes a movement with both size and direction. In geometry, $AB⃗\vec{AB}$ means “the vector from A to B”. The order of the letters matters.

Definition

Column vector and components

A column vector is a vector written in the form

(xy)\begin{pmatrix} x \\ y \end{pmatrix}

where the top number, $x$ , is the horizontal component and the bottom number, $y$ , is the vertical component.

For example,

AB⃗=(5−2)\vec{AB}=\begin{pmatrix} 5 \\ -2 \end{pmatrix}

means “move 5 units to the right and 2 units down” from A to B.

If you reverse a vector, you change the signs of both components:

BA⃗=−AB⃗\vec{BA}=-\vec{AB}

So if $AB⃗=(5−2)\vec{AB}=\begin{pmatrix} 5 \\ -2 \end{pmatrix}$ , then

BA⃗=(−52)\vec{BA}=\begin{pmatrix} -5 \\ 2 \end{pmatrix}

Example

Using the order of letters

Suppose

AB⃗=(4−3)CB⃗=(−26)\vec{AB}=\begin{pmatrix} 4 \\ -3 \end{pmatrix} \quad \vec{CB}=\begin{pmatrix} -2 \\ 6 \end{pmatrix}

Find $AC⃗\vec{AC}$ .

You want to travel from A to C. A useful route is A to B, then B to C:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
You are given $CB⃗\vec{CB}$ , but you need $BC⃗\vec{BC}$ . Reverse the vector by changing both signs:
$BC⃗=−CB⃗=(2−6)\vec{BC}=-\vec{CB} = \begin{pmatrix} 2 \\ -6 \end{pmatrix}$
Add the components:
$AC⃗=(4−3)+(2−6)=(6−9)\vec{AC} = \begin{pmatrix} 4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -6 \end{pmatrix} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}$

Common Mistake

Mixing up $C B$ and $B C$

$CB⃗\vec{CB}$ and $BC⃗\vec{BC}$ have the same length, but they point in opposite directions. Always change both signs when reversing a vector.

Magnitude means length

Definition

Magnitude of a vector

The magnitude of a vector is its length. The magnitude of $v⃗\vec{v}$ is written as $∣v⃗∣|\vec{v}|$ .

For a column vector $v⃗=(xy)\vec{v}=\begin{pmatrix} x \\ y \end{pmatrix}$ , the horizontal and vertical components form a right-angled triangle. So we use Pythagoras:

∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}

A vector split into horizontal and vertical components, showing magnitude as the hypotenuse

Key Idea

The magnitude formula

To find the length of $(xy)\begin{pmatrix} x \\ y \end{pmatrix}$ , square both components, add them, then square root.

Example

Finding the magnitude of a column vector

Find the magnitude of

v⃗=(−815)\vec{v}=\begin{pmatrix} -8 \\ 15 \end{pmatrix}

Identify the two components:
$\quad y=15$
Substitute into the magnitude formula:
$∣v⃗∣=(−8)2+152|\vec{v}|=\sqrt{(-8)^2+15^2}$
Square the components and add:
$∣v⃗∣=64+225=289|\vec{v}|=\sqrt{64+225}=\sqrt{289}$
Square root the answer:
$∣v⃗∣=17|\vec{v}|=17$

Common Mistake

Forgetting that negatives square to positives

If a component is negative, keep it in brackets when squaring. For example, $8)^2=64$ , not -64.

Exact magnitudes and surds

Sometimes the square root is not a whole number. In that case, you may be asked for the exact magnitude.

Definition

Exact answer and surd

An exact answer has not been rounded. A surd is a square root that is left in exact form, such as $29\sqrt{29}$ .

If possible, simplify the square root by taking out a square factor:

116=4×29=229\sqrt{116}=\sqrt{4 \times 29}=2\sqrt{29}

Example

Giving an exact magnitude

Find the exact magnitude of

p⃗=(10−4)\vec{p}=\begin{pmatrix} 10 \\ -4 \end{pmatrix}

Use the magnitude formula:
$∣p⃗∣=102+(−4)2|\vec{p}|=\sqrt{10^2+(-4)^2}$
Simplify inside the square root:
$∣p⃗∣=100+16=116|\vec{p}|=\sqrt{100+16}=\sqrt{116}$
Take out the square factor 4:
$116=4×29=229\sqrt{116}=\sqrt{4 \times 29}=2\sqrt{29}$

Tip

Spot square factors

When simplifying a square root, look for factors like 4, 9, 16, 25, 36, 49 and 64.

Find the vector first, then find its magnitude

Many exam questions do not give you the vector you need directly. You must build it from other vectors first.

A helpful rule is:

AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}

Rearranging gives:

BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}

Triangle vector diagram showing AB plus BC equals AC

Key Idea

Two-stage method

For magnitude questions in geometry, first find the correct column vector, then apply $∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}$ .

Example

Finding a hidden vector before its magnitude

Given

AB⃗=(5−7)CB⃗=(−14)\vec{AB}=\begin{pmatrix} 5 \\ -7 \end{pmatrix} \quad \vec{CB}=\begin{pmatrix} -1 \\ 4 \end{pmatrix}

find the magnitude of $AC⃗\vec{AC}$ .

Use the route from A to B to C:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
Reverse $CB⃗\vec{CB}$ to get $BC⃗\vec{BC}$ :
$BC⃗=−CB⃗=(1−4)\vec{BC}=-\vec{CB} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$
Add the vectors:
$AC⃗=(5−7)+(1−4)=(6−11)\vec{AC} = \begin{pmatrix} 5 \\ -7 \end{pmatrix} + \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 6 \\ -11 \end{pmatrix}$
Find the magnitude:
$∣AC⃗∣=62+(−11)2=36+121=157|\vec{AC}|=\sqrt{6^2+(-11)^2} = \sqrt{36+121} = \sqrt{157}$

Parallelograms and diagonals

Definition

Parallelogram

A parallelogram is a quadrilateral with both pairs of opposite sides parallel and equal in length.

In a parallelogram $A B C D$ , the diagonal $AC⃗\vec{AC}$ can be made by going along $AB⃗\vec{AB}$ and then $BC⃗\vec{BC}$ :

AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}

So if you know $AB⃗\vec{AB}$ and $AC⃗\vec{AC}$ , you can find:

BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}

Example

Magnitude in a parallelogram

$A B C D$ is a parallelogram.

AB⃗=(32)AC⃗=(11−4)\vec{AB}=\begin{pmatrix} 3 \\ 2 \end{pmatrix} \quad \vec{AC}=\begin{pmatrix} 11 \\ -4 \end{pmatrix}

Find the magnitude of $BC⃗\vec{BC}$ .

Use the diagonal relationship:
$AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$
Rearrange to find $BC⃗\vec{BC}$ :
$BC⃗=AC⃗−AB⃗\vec{BC}=\vec{AC}-\vec{AB}$
Subtract the column vectors component by component:
$BC⃗=(11−4)−(32)=(8−6)\vec{BC} = \begin{pmatrix} 11 \\ -4 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$
Find the magnitude:
$∣BC⃗∣=82+(−6)2=64+36=100=10|\vec{BC}|=\sqrt{8^2+(-6)^2} = \sqrt{64+36} = \sqrt{100} = 10$

Tip

Subtracting column vectors

Subtract top from top, and bottom from bottom. Keep the order exactly as written.

Trapezia and scalar multiples

Definition

Trapezium and scalar multiple

A trapezium is a quadrilateral with at least one pair of parallel sides. A scalar multiple means a vector has been multiplied by a number, such as $3AB⃗3\vec{AB}$ .

If you are told $DC⃗=3AB⃗\vec{DC}=3\vec{AB}$ , then $DC⃗\vec{DC}$ points in the same direction as $AB⃗\vec{AB}$ and is three times as long.

To solve these questions, build the missing side carefully using vector paths.

Example

Exact magnitude in a trapezium

$A B C D$ is a trapezium and

DC⃗=3AB⃗\vec{DC}=3\vec{AB}

Also,

DA⃗=(2−1)DB⃗=(53)\vec{DA}=\begin{pmatrix} 2 \\ -1 \end{pmatrix} \quad \vec{DB}=\begin{pmatrix} 5 \\ 3 \end{pmatrix}

Find the exact magnitude of $BC⃗\vec{BC}$ .

First find $AB⃗\vec{AB}$ . Going from D to A, then A to B, gives D to B:
$DA⃗+AB⃗=DB⃗\vec{DA}+\vec{AB}=\vec{DB}$
Rearrange:
$AB⃗=DB⃗−DA⃗\vec{AB}=\vec{DB}-\vec{DA}$
Substitute and subtract:
$AB⃗=(53)−(2−1)=(34)\vec{AB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
Use $DC⃗=3AB⃗\vec{DC}=3\vec{AB}$ :
$DC⃗=3(34)=(912)\vec{DC} = 3\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}$
To get from B to C, go from B to D, then D to C:
$BC⃗=BD⃗+DC⃗\vec{BC}=\vec{BD}+\vec{DC}$
Since $BD⃗=−DB⃗\vec{BD}=-\vec{DB}$ , substitute both vectors:
$BC⃗=(−5−3)+(912)=(49)\vec{BC} = \begin{pmatrix} -5 \\ -3 \end{pmatrix} + \begin{pmatrix} 9 \\ 12 \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix}$
Find the exact magnitude:
$∣BC⃗∣=42+92=16+81=97|\vec{BC}|=\sqrt{4^2+9^2} = \sqrt{16+81} = \sqrt{97}$

Common Mistake

Do not assume the diagram is to scale

In vector geometry, the written vector information is more reliable than how the shape looks. Use the vector equations, not visual guessing.

Exam technique

In the exam

Write a vector path for the vector you need, such as $AC⃗=AB⃗+BC⃗\vec{AC}=\vec{AB}+\vec{BC}$ .
Reverse vectors carefully: $BA⃗=−AB⃗\vec{BA}=-\vec{AB}$ , so both components change sign.
Only find the magnitude after you have the final column vector, then use $∣v⃗∣=x2+y2|\vec{v}|=\sqrt{x^2+y^2}$ .

Self review

Check yourself

If $PQ⃗=(−38)\vec{PQ}=\begin{pmatrix} -3 \\ 8 \end{pmatrix}$ , what is $QP⃗\vec{QP}$ ?
Why does a negative component not make the magnitude negative?
In a parallelogram, how can you find $BC⃗\vec{BC}$ if you know $AB⃗\vec{AB}$ and $AC⃗\vec{AC}$ ?

The Magnitude of a Vector

What you'll learn

Vectors as movements

Magnitude means length

Exact magnitudes and surds

Find the vector first, then find its magnitude

Parallelograms and diagonals

Trapezia and scalar multiples

Recap questions

You've reached the end

The Magnitude of a Vector

What you'll learn

Vectors as movements

Magnitude means length

Exact magnitudes and surds

Find the vector first, then find its magnitude

Parallelograms and diagonals

Trapezia and scalar multiples

Recap questions

You've reached the end

The Magnitude of a Vector

What you'll learn

Vectors as movements

Magnitude means length

Exact magnitudes and surds

Find the vector first, then find its magnitude

Parallelograms and diagonals

Trapezia and scalar multiples

Recap questions

You've reached the end

The Magnitude of a Vector Revision Guide

The Magnitude of a Vector

What you'll learn

Vectors as movements

Magnitude means length

Exact magnitudes and surds

Find the vector first, then find its magnitude

Parallelograms and diagonals

Trapezia and scalar multiples

Recap questions

You've reached the end

The Magnitude of a Vector Revision Guide