1.11.3 Magnitude and direction of a vector - Revision Guide

What you'll learn

How to calculate the magnitude of a vector from its components.
How to find a vector's direction, including choosing the correct quadrant.
How to convert between component form and magnitude/direction form.
How to apply these techniques to two-dimensional and three-dimensional vectors.

Prerequisites: vectors and components

A vector is a quantity with both a size and a direction. For example, a displacement of 5 m east is a vector because it tells you how far to travel and in which direction.

In two dimensions, a vector can be written in component form as

a=(xy).\mathbf{a}= \begin{pmatrix} x\\ y \end{pmatrix}.

The number $x$ is the horizontal component and the number $y$ is the vertical component.

A positive horizontal component means movement to the right.
A negative horizontal component means movement to the left.
A positive vertical component means movement upwards.
A negative vertical component means movement downwards.

For example,

(3−4)\begin{pmatrix} 3\\ -4 \end{pmatrix}

represents movement 3 units to the right and 4 units downwards.

Vectors can also be written using the unit vectors $i\mathbf{i}$ and $j\mathbf{j}$ :

(xy)=xi+yj,\begin{pmatrix} x\\ y \end{pmatrix} =x\mathbf{i}+y\mathbf{j},

where $i\mathbf{i}$ points one unit in the positive horizontal direction and $j\mathbf{j}$ points one unit in the positive vertical direction.

Definition

Component form

The component form of a vector lists its signed movement in each coordinate direction. In two dimensions, $(xy)\begin{pmatrix}x\\y\end{pmatrix}$ means $x$ units horizontally and $y$ units vertically.

Magnitude of a vector

The magnitude of a vector is its length or size. The magnitude of $a\mathbf{a}$ is written as $∣a∣|\mathbf{a}|$ .

The horizontal and vertical components form the perpendicular sides of a right-angled triangle. Therefore, Pythagoras' theorem gives

∣a∣=x2+y2.|\mathbf{a}|=\sqrt{x^2+y^2}.

A vector with horizontal and vertical components forming a right-angled triangle

Key Idea

Magnitude formula

For $a=(xy)\mathbf{a}=\begin{pmatrix}x\\y\end{pmatrix}$ ,

∣a∣=x2+y2.|\mathbf{a}|=\sqrt{x^2+y^2}.

The signs of the components do not affect the magnitude because the components are squared.

Example

Finding the magnitude of a vector

Find the magnitude of

a=(−512).\mathbf{a}= \begin{pmatrix} -5\\ 12 \end{pmatrix}.

Substitute the horizontal and vertical components into the magnitude formula:
$∣a∣=(−5)2+122.|\mathbf{a}|=\sqrt{(-5)^2+12^2}.$
Square each component:
$∣a∣=25+144=169.|\mathbf{a}|=\sqrt{25+144}=\sqrt{169}.$
Take the positive square root, since a length cannot be negative:
$∣a∣=13.|\mathbf{a}|=13.$

Common Mistake

Forgetting brackets around negative components

Write $5)^2$ , not $5^2$ . The first expression equals 25, whereas the second is interpreted as $5^2)=-25$ .

Magnitude in three dimensions

A three-dimensional vector has three components:

a=(xyz)=xi+yj+zk.\mathbf{a}= \begin{pmatrix} x\\ y\\ z \end{pmatrix} =x\mathbf{i}+y\mathbf{j}+z\mathbf{k}.

Its magnitude is found by extending Pythagoras' theorem:

∣a∣=x2+y2+z2.|\mathbf{a}|=\sqrt{x^2+y^2+z^2}.

Example

Finding a three-dimensional magnitude

Find the magnitude of $a=2i−3j+6k\mathbf{a}=2\mathbf{i}-3\mathbf{j}+6\mathbf{k}$ .

Identify the three components as $x = 2$ , $y = - 3$ and $z = 6$ .
Substitute them into the three-dimensional magnitude formula:
$∣a∣=22+(−3)2+62.|\mathbf{a}|=\sqrt{2^2+(-3)^2+6^2}.$
Simplify:
$∣a∣=4+9+36=49=7.|\mathbf{a}|=\sqrt{4+9+36}=\sqrt{49}=7.$

Direction of a two-dimensional vector

The direction angle $θ\theta$ is usually measured anticlockwise from the positive horizontal axis, unless the question states a different convention.

For a vector in the first quadrant,

tan⁡θ=yx,\tan\theta=\frac{y}{x},

θ=tan⁡−1(yx).\theta=\tan^{-1}\left(\frac{y}{x}\right).

Here, $tan^{-1}$ means the inverse tangent function on your calculator, not the reciprocal of tangent.

Example

Finding a direction in the first quadrant

Find the direction of

a=(47),\mathbf{a}= \begin{pmatrix} 4\\ 7 \end{pmatrix},

giving your answer to 1 decimal place.

Since both components are positive, the vector lies in the first quadrant, so the calculator's inverse tangent result will give the required angle directly.
Use the ratio of the vertical component to the horizontal component:
$θ=tan⁡−1(74).\theta=\tan^{-1}\left(\frac{7}{4}\right).$
Evaluate and round only at the end:
$θ≈60.3∘.\theta\approx60.3^\circ.$

Choosing the correct quadrant

Using $tan^{-1}(y/x)$ alone can produce an angle in the wrong quadrant because tangent has a period of $180∘180^\circ$ .

The signs of the components tell you where the vector lies:

$x > 0$ , $y > 0$ : first quadrant.
$x < 0$ , $y > 0$ : second quadrant.
$x < 0$ , $y < 0$ : third quadrant.
$x > 0$ , $y < 0$ : fourth quadrant.

A reliable method is to find the positive reference angle

α=tan⁡−1(∣yx∣)\alpha=\tan^{-1}\left(\left|\frac{y}{x}\right|\right)

and then use the quadrant to obtain $θ\theta$ :

first quadrant: $θ=α\theta=\alpha$ ;
second quadrant: $θ=180∘−α\theta=180^\circ-\alpha$ ;
third quadrant: $θ=180∘+α\theta=180^\circ+\alpha$ ;
fourth quadrant: $θ=360∘−α\theta=360^\circ-\alpha$ .

Example

Finding a direction in the second quadrant

Find the direction of

b=(−34),\mathbf{b}= \begin{pmatrix} -3\\ 4 \end{pmatrix},

measured anticlockwise from the positive horizontal axis.

The horizontal component is negative and the vertical component is positive, so the vector lies in the second quadrant.
Find the reference angle:
$α=tan⁡−1(43)≈53.1∘.\alpha=\tan^{-1}\left(\frac{4}{3}\right)\approx53.1^\circ.$
Subtract the reference angle from $180∘180^\circ$ :
$θ=180∘−53.1∘=126.9∘.\theta=180^\circ-53.1^\circ=126.9^\circ.$

Tip

Use your calculator's coordinate-aware angle function

Some calculators provide a function such as $Arg⁡(x+yi)\operatorname{Arg}(x+yi)$ or $atan2⁡(y,x)\operatorname{atan2}(y,x)$ that uses both components and identifies the quadrant. Check whether it returns an angle between $−180∘-180^\circ$ and $180∘180^\circ$ ; if it gives a negative angle, add $360∘360^\circ$ when an angle from $0∘0^\circ$ to $360∘360^\circ$ is required.

Common Mistake

Vectors on the axes

The formula $tan^{-1}(y/x)$ cannot be used when $x = 0$ because division by zero is undefined. Read the direction directly: an upward vector has direction $90∘90^\circ$ , while a downward vector has direction $270∘270^\circ$ .

From magnitude and direction to components

Suppose a vector has magnitude $r$ and direction $θ\theta$ , measured anticlockwise from the positive horizontal axis.

From right-angled triangle trigonometry,

cos⁡θ=xrandsin⁡θ=yr.\cos\theta=\frac{x}{r} \qquad\text{and}\qquad \sin\theta=\frac{y}{r}.

Therefore,

x=rcos⁡θandy=rsin⁡θ.x=r\cos\theta \qquad\text{and}\qquad y=r\sin\theta.

The vector is

a=(rcos⁡θrsin⁡θ).\mathbf{a}= \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix}.

This is called its magnitude/direction form.

Key Idea

Converting to components

A vector of magnitude $r$ at an angle $θ\theta$ to the positive horizontal axis has component form

a=(rcos⁡θrsin⁡θ).\mathbf{a}= \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix}.

Cosine gives the horizontal component; sine gives the vertical component.

Example

Converting magnitude and direction to components

A vector has magnitude 10 and direction $150∘150^\circ$ . Find its exact component form.

Use the component formula:
$a=(10cos⁡150∘10sin⁡150∘).\mathbf{a}= \begin{pmatrix} 10\cos150^\circ\\ 10\sin150^\circ \end{pmatrix}.$
Use the exact trigonometric values $cos⁡150∘=−32\cos150^\circ=-\frac{\sqrt3}{2}$ and $sin⁡150∘=12\sin150^\circ=\frac12$ :
$a=(10(−32)10(12)).\mathbf{a}= \begin{pmatrix} 10\left(-\frac{\sqrt3}{2}\right)\\ 10\left(\frac12\right) \end{pmatrix}.$
Simplify:
$a=(−535).\mathbf{a}= \begin{pmatrix} -5\sqrt3\\ 5 \end{pmatrix}.$
The signs are sensible because a direction of $150∘150^\circ$ lies in the second quadrant.

Common Mistake

Swapping sine and cosine

When the angle is measured from the horizontal axis, the horizontal component is $rcos⁡θr\cos\theta$ and the vertical component is $rsin⁡θr\sin\theta$ . If the angle is measured from the vertical, this relationship changes, so draw a triangle and identify the adjacent side.

The zero vector

The zero vector has every component equal to zero:

0=(00).\mathbf{0}= \begin{pmatrix} 0\\ 0 \end{pmatrix}.

Its magnitude is zero, but it has no defined direction. There is no arrow of positive length from which an angle can be measured.

Exam technique

In the exam

For component form, calculate the magnitude using Pythagoras and use the component signs to identify the quadrant before finding the direction.
Check the angle convention carefully: it may be anticlockwise from the positive horizontal axis, an angle from another axis, or a three-figure bearing.
When converting back to components, use $x=rcos⁡θx=r\cos\theta$ and $y=rsin⁡θy=r\sin\theta$ , then check that their signs agree with the stated direction.
Keep exact surds where possible and delay decimal rounding until the final answer.

Self review

Check yourself

What are the magnitude and direction of $(−8−6)\begin{pmatrix}-8\\-6\end{pmatrix}$ ?
A vector has magnitude 12 and direction $210∘210^\circ$ . What is its exact component form?
Why does $tan^{-1}(y/x)$ need a quadrant check, and what happens when $x = 0$ ?

What you'll learn

How to calculate the magnitude of a vector from its components.
How to find a vector's direction, including choosing the correct quadrant.
How to convert between component form and magnitude/direction form.
How to apply these techniques to two-dimensional and three-dimensional vectors.

Prerequisites: vectors and components

A vector is a quantity with both a size and a direction. For example, a displacement of 5 m east is a vector because it tells you how far to travel and in which direction.

In two dimensions, a vector can be written in component form as

a=(xy).\mathbf{a}= \begin{pmatrix} x\\ y \end{pmatrix}.

The number $x$ is the horizontal component and the number $y$ is the vertical component.

A positive horizontal component means movement to the right.
A negative horizontal component means movement to the left.
A positive vertical component means movement upwards.
A negative vertical component means movement downwards.

For example,

(3−4)\begin{pmatrix} 3\\ -4 \end{pmatrix}

represents movement 3 units to the right and 4 units downwards.

Vectors can also be written using the unit vectors $i\mathbf{i}$ and $j\mathbf{j}$ :

(xy)=xi+yj,\begin{pmatrix} x\\ y \end{pmatrix} =x\mathbf{i}+y\mathbf{j},

where $i\mathbf{i}$ points one unit in the positive horizontal direction and $j\mathbf{j}$ points one unit in the positive vertical direction.

Definition

Component form

Magnitude of a vector

The magnitude of a vector is its length or size. The magnitude of $a\mathbf{a}$ is written as $∣a∣|\mathbf{a}|$ .

The horizontal and vertical components form the perpendicular sides of a right-angled triangle. Therefore, Pythagoras' theorem gives

∣a∣=x2+y2.|\mathbf{a}|=\sqrt{x^2+y^2}.

A vector with horizontal and vertical components forming a right-angled triangle

Key Idea

Magnitude formula

For $a=(xy)\mathbf{a}=\begin{pmatrix}x\\y\end{pmatrix}$ ,

∣a∣=x2+y2.|\mathbf{a}|=\sqrt{x^2+y^2}.

The signs of the components do not affect the magnitude because the components are squared.

Example

Finding the magnitude of a vector

Find the magnitude of

a=(−512).\mathbf{a}= \begin{pmatrix} -5\\ 12 \end{pmatrix}.

Substitute the horizontal and vertical components into the magnitude formula:
$∣a∣=(−5)2+122.|\mathbf{a}|=\sqrt{(-5)^2+12^2}.$
Square each component:
$∣a∣=25+144=169.|\mathbf{a}|=\sqrt{25+144}=\sqrt{169}.$
Take the positive square root, since a length cannot be negative:
$∣a∣=13.|\mathbf{a}|=13.$

Common Mistake

Forgetting brackets around negative components

Write $5)^2$ , not $5^2$ . The first expression equals 25, whereas the second is interpreted as $5^2)=-25$ .

Magnitude in three dimensions

A three-dimensional vector has three components:

a=(xyz)=xi+yj+zk.\mathbf{a}= \begin{pmatrix} x\\ y\\ z \end{pmatrix} =x\mathbf{i}+y\mathbf{j}+z\mathbf{k}.

Its magnitude is found by extending Pythagoras' theorem:

∣a∣=x2+y2+z2.|\mathbf{a}|=\sqrt{x^2+y^2+z^2}.

Example

Finding a three-dimensional magnitude

Find the magnitude of $a=2i−3j+6k\mathbf{a}=2\mathbf{i}-3\mathbf{j}+6\mathbf{k}$ .

Identify the three components as $x = 2$ , $y = - 3$ and $z = 6$ .
Substitute them into the three-dimensional magnitude formula:
$∣a∣=22+(−3)2+62.|\mathbf{a}|=\sqrt{2^2+(-3)^2+6^2}.$
Simplify:
$∣a∣=4+9+36=49=7.|\mathbf{a}|=\sqrt{4+9+36}=\sqrt{49}=7.$

Direction of a two-dimensional vector

The direction angle $θ\theta$ is usually measured anticlockwise from the positive horizontal axis, unless the question states a different convention.

For a vector in the first quadrant,

tan⁡θ=yx,\tan\theta=\frac{y}{x},

θ=tan⁡−1(yx).\theta=\tan^{-1}\left(\frac{y}{x}\right).

Here, $tan^{-1}$ means the inverse tangent function on your calculator, not the reciprocal of tangent.

Example

Finding a direction in the first quadrant

Find the direction of

a=(47),\mathbf{a}= \begin{pmatrix} 4\\ 7 \end{pmatrix},

giving your answer to 1 decimal place.

Since both components are positive, the vector lies in the first quadrant, so the calculator's inverse tangent result will give the required angle directly.
Use the ratio of the vertical component to the horizontal component:
$θ=tan⁡−1(74).\theta=\tan^{-1}\left(\frac{7}{4}\right).$
Evaluate and round only at the end:
$θ≈60.3∘.\theta\approx60.3^\circ.$

Choosing the correct quadrant

Using $tan^{-1}(y/x)$ alone can produce an angle in the wrong quadrant because tangent has a period of $180∘180^\circ$ .

The signs of the components tell you where the vector lies:

$x > 0$ , $y > 0$ : first quadrant.
$x < 0$ , $y > 0$ : second quadrant.
$x < 0$ , $y < 0$ : third quadrant.
$x > 0$ , $y < 0$ : fourth quadrant.

A reliable method is to find the positive reference angle

α=tan⁡−1(∣yx∣)\alpha=\tan^{-1}\left(\left|\frac{y}{x}\right|\right)

and then use the quadrant to obtain $θ\theta$ :

first quadrant: $θ=α\theta=\alpha$ ;
second quadrant: $θ=180∘−α\theta=180^\circ-\alpha$ ;
third quadrant: $θ=180∘+α\theta=180^\circ+\alpha$ ;
fourth quadrant: $θ=360∘−α\theta=360^\circ-\alpha$ .

Example

Finding a direction in the second quadrant

Find the direction of

b=(−34),\mathbf{b}= \begin{pmatrix} -3\\ 4 \end{pmatrix},

measured anticlockwise from the positive horizontal axis.

The horizontal component is negative and the vertical component is positive, so the vector lies in the second quadrant.
Find the reference angle:
$α=tan⁡−1(43)≈53.1∘.\alpha=\tan^{-1}\left(\frac{4}{3}\right)\approx53.1^\circ.$
Subtract the reference angle from $180∘180^\circ$ :
$θ=180∘−53.1∘=126.9∘.\theta=180^\circ-53.1^\circ=126.9^\circ.$

Tip

Use your calculator's coordinate-aware angle function

Common Mistake

Vectors on the axes

From magnitude and direction to components

Suppose a vector has magnitude $r$ and direction $θ\theta$ , measured anticlockwise from the positive horizontal axis.

From right-angled triangle trigonometry,

cos⁡θ=xrandsin⁡θ=yr.\cos\theta=\frac{x}{r} \qquad\text{and}\qquad \sin\theta=\frac{y}{r}.

Therefore,

x=rcos⁡θandy=rsin⁡θ.x=r\cos\theta \qquad\text{and}\qquad y=r\sin\theta.

The vector is

a=(rcos⁡θrsin⁡θ).\mathbf{a}= \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix}.

This is called its magnitude/direction form.

Key Idea

Converting to components

A vector of magnitude $r$ at an angle $θ\theta$ to the positive horizontal axis has component form

a=(rcos⁡θrsin⁡θ).\mathbf{a}= \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix}.

Cosine gives the horizontal component; sine gives the vertical component.

Example

Converting magnitude and direction to components

A vector has magnitude 10 and direction $150∘150^\circ$ . Find its exact component form.

Use the component formula:
$a=(10cos⁡150∘10sin⁡150∘).\mathbf{a}= \begin{pmatrix} 10\cos150^\circ\\ 10\sin150^\circ \end{pmatrix}.$
Use the exact trigonometric values $cos⁡150∘=−32\cos150^\circ=-\frac{\sqrt3}{2}$ and $sin⁡150∘=12\sin150^\circ=\frac12$ :
$a=(10(−32)10(12)).\mathbf{a}= \begin{pmatrix} 10\left(-\frac{\sqrt3}{2}\right)\\ 10\left(\frac12\right) \end{pmatrix}.$
Simplify:
$a=(−535).\mathbf{a}= \begin{pmatrix} -5\sqrt3\\ 5 \end{pmatrix}.$
The signs are sensible because a direction of $150∘150^\circ$ lies in the second quadrant.

Common Mistake

Swapping sine and cosine

The zero vector

The zero vector has every component equal to zero:

0=(00).\mathbf{0}= \begin{pmatrix} 0\\ 0 \end{pmatrix}.

Its magnitude is zero, but it has no defined direction. There is no arrow of positive length from which an angle can be measured.

Exam technique

In the exam

For component form, calculate the magnitude using Pythagoras and use the component signs to identify the quadrant before finding the direction.
Check the angle convention carefully: it may be anticlockwise from the positive horizontal axis, an angle from another axis, or a three-figure bearing.
When converting back to components, use $x=rcos⁡θx=r\cos\theta$ and $y=rsin⁡θy=r\sin\theta$ , then check that their signs agree with the stated direction.
Keep exact surds where possible and delay decimal rounding until the final answer.

Self review

Check yourself

What are the magnitude and direction of $(−8−6)\begin{pmatrix}-8\\-6\end{pmatrix}$ ?
A vector has magnitude 12 and direction $210∘210^\circ$ . What is its exact component form?
Why does $tan^{-1}(y/x)$ need a quadrant check, and what happens when $x = 0$ ?

1.13.2 Magnitude and direction of a vector

What you'll learn

Prerequisites: vectors and components

Magnitude of a vector

Magnitude in three dimensions

Direction of a two-dimensional vector

Choosing the correct quadrant

From magnitude and direction to components

The zero vector

Review 1.11.3 Magnitude and direction of a vector by teaching Genie

1.13.2 Magnitude and direction of a vector

What you'll learn

Prerequisites: vectors and components

Magnitude of a vector

Magnitude in three dimensions

Direction of a two-dimensional vector

Choosing the correct quadrant

From magnitude and direction to components

The zero vector

Review 1.11.3 Magnitude and direction of a vector by teaching Genie

1.13.2 Magnitude and direction of a vector

What you'll learn

Prerequisites: vectors and components

Magnitude of a vector

Magnitude in three dimensions

Direction of a two-dimensional vector

Choosing the correct quadrant

From magnitude and direction to components

The zero vector

Review 1.11.3 Magnitude and direction of a vector by teaching Genie

1.11.3 Magnitude and direction of a vector Revision Guide

1.13.2 Magnitude and direction of a vector

What you'll learn

Prerequisites: vectors and components

Magnitude of a vector

Magnitude in three dimensions

Direction of a two-dimensional vector

Choosing the correct quadrant

From magnitude and direction to components

The zero vector

Review 1.11.3 Magnitude and direction of a vector by teaching Genie

1.11.3 Magnitude and direction of a vector Revision Guide