Vector magnitude, direction and unit vectors

A vector combines a magnitude, which tells you its size, with a direction. For example,

$$\mathbf a=\begin{pmatrix}3\\4\end{pmatrix}$$

means a displacement of $3$ units in the positive $x$ direction and $4$ units in the positive $y$ direction. Its magnitude is $5$, but the vector itself is not the number $5$: it also records where the displacement points.

This lesson develops three closely related skills:

1. finding the magnitude and direction of a vector;
2. producing a unit vector in a required direction;
3. constructing a vector when its magnitude and direction are known.

Prerequisites

You should be able to:

• use column vectors and $\mathbf i$, $\mathbf j$, $\mathbf k$ notation;
• apply Pythagoras’ theorem in two and three dimensions;
• use exact surds;
• solve right angled triangles using sine, cosine and tangent;
• interpret angles in all four quadrants.

Review vectors in two and three dimensions, vector arithmetic or trigonometric graphs if necessary.

Magnitude in two dimensions

For

$$\mathbf a=\begin{pmatrix}x\\y\end{pmatrix}=x\mathbf i+y\mathbf j,$$

the horizontal and vertical components are perpendicular. Pythagoras’ theorem therefore gives

$$\boxed{|\mathbf a|=\sqrt{x^2+y^2}}.$$

The notation $|\mathbf a|$ means the magnitude or length of $\mathbf a$. Some textbooks use $\lVert\mathbf a\rVert$ instead. Both quantities are scalars and are never negative.

Worked example 1: an exact magnitude

Find the magnitude of

$$\mathbf p=\begin{pmatrix}-5\\12\end{pmatrix}.$$

Substitute both components into the formula:

$$\begin{aligned} |\mathbf p| &=\sqrt{(-5)^2+12^2}\\ &=\sqrt{25+144}\\ &=\boxed{13}. \end{aligned}$$

The negative sign affects the direction, but not the length. Squaring makes its contribution positive.

Worked example 2: keep a surd exact

Let $\mathbf q=7\mathbf i-4\mathbf j$. Find $|\mathbf q|$.

The components are $7$ and $-4$, so

$$|\mathbf q|=\sqrt{7^2+(-4)^2}=\sqrt{65}.$$

Since $65$ has no square factor greater than $1$,

$$\boxed{|\mathbf q|=\sqrt{65}}.$$

Do not replace this by a rounded decimal unless the question asks for one. The exact answer remains useful in later algebra.

Magnitude is not the sum of the components

For $\begin{pmatrix}3\\4\end{pmatrix}$, the magnitude is $\sqrt{3^2+4^2}=5$, not $3+4=7$. The components describe perpendicular movements, so Pythagoras is required.

Self-check 1

Find the exact magnitude of $\mathbf a=-6\mathbf i-3\mathbf j$.

Answer

$$|\mathbf a|=\sqrt{(-6)^2+(-3)^2}=\sqrt{45}=\boxed{3\sqrt5}.$$

Magnitude in three dimensions

For

$$\mathbf a=\begin{pmatrix}x\\y\\z\end{pmatrix}=x\mathbf i+y\mathbf j+z\mathbf k,$$

Pythagoras can be applied twice. First combine the $x$ and $y$ components, then include the perpendicular $z$ component:

$$\boxed{|\mathbf a|=\sqrt{x^2+y^2+z^2}}.$$

Worked example 3: magnitude in three dimensions

Find the magnitude of

$$\mathbf v=2\mathbf i-3\mathbf j+6\mathbf k.$$ $$\begin{aligned} |\mathbf v| &=\sqrt{2^2+(-3)^2+6^2}\\ &=\sqrt{4+9+36}\\ &=\sqrt{49}\\ &=\boxed{7}. \end{aligned}$$

The same formula works for a displacement between two points after finding the displacement vector using finish minus start.

Worked example 4: distance between points in space

Points $A(1,-2,4)$ and $B(5,1,-2)$ are given. Find $AB$.

First find the vector from $A$ to $B$:

$$\overrightarrow{AB} =\begin{pmatrix}5-1\\1-(-2)\\-2-4\end{pmatrix} =\begin{pmatrix}4\\3\\-6\end{pmatrix}.$$

The distance is the magnitude of this vector:

$$AB=|\overrightarrow{AB}| =\sqrt{4^2+3^2+(-6)^2} =\sqrt{61}.$$

Hence

$$\boxed{AB=\sqrt{61}}.$$

Vector versus magnitude

$\overrightarrow{AB}$ is a vector, while $AB$ is a length. Thus $\overrightarrow{AB}=\begin{pmatrix}4\\3\\-6\end{pmatrix}$ but $AB=\sqrt{61}$. Keep this distinction in proofs and modelling questions.

How scaling changes magnitude

Multiplying a vector by a scalar $k$ multiplies its magnitude by $|k|$:

$$\boxed{|k\mathbf a|=|k|\,|\mathbf a|}.$$

The absolute value is essential. If $k<0$, the vector reverses direction, but its length is still positive.

For example, if $|\mathbf a|=6$, then

$$|3\mathbf a|=18, \qquad \left|-\frac12\mathbf a\right|=3.$$

Worked example 5: an unknown scalar

Suppose $\mathbf a=\begin{pmatrix}2\\-1\end{pmatrix}$ and $|k\mathbf a|=15$. Find the possible values of $k$.

First,

$$|\mathbf a|=\sqrt{2^2+(-1)^2}=\sqrt5.$$

Therefore

$$|k|\sqrt5=15.$$

Dividing by $\sqrt5$ gives

$$|k|=\frac{15}{\sqrt5}=3\sqrt5.$$

Hence

$$\boxed{k=3\sqrt5\quad\text{or}\quad k=-3\sqrt5}.$$

There are two answers because opposite vectors have the same magnitude.

Unit vectors

A unit vector has magnitude $1$. To turn any non-zero vector $\mathbf a$ into a unit vector pointing in the same direction, divide by its magnitude:

$$\boxed{\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}}.$$

The hat is a common notation for a unit vector, although notation varies. The reasoning is

$$\left|\frac{\mathbf a}{|\mathbf a|}\right| =\frac{|\mathbf a|}{|\mathbf a|}=1.$$

This process is called normalising the vector.

Worked example 6: unit vector in the same direction

Find a unit vector in the direction of

$$\mathbf d=\begin{pmatrix}-2\\5\end{pmatrix}.$$

First calculate its magnitude:

$$|\mathbf d|=\sqrt{(-2)^2+5^2}=\sqrt{29}.$$

Now divide every component by $\sqrt{29}$:

$$\boxed{ \widehat{\mathbf d} =\frac1{\sqrt{29}}\begin{pmatrix}-2\\5\end{pmatrix} =\begin{pmatrix}-2/\sqrt{29}\\5/\sqrt{29}\end{pmatrix}}.$$

A check confirms that

$$|\widehat{\mathbf d}| =\sqrt{\frac4{29}+\frac{25}{29}}=1.$$

Worked example 7: a specified magnitude and direction

Find the vector of magnitude $10$ in the same direction as

$$\mathbf a=3\mathbf i-4\mathbf j.$$

Since $|\mathbf a|=5$, the unit vector in its direction is

$$\frac{\mathbf a}{|\mathbf a|} =\frac35\mathbf i-\frac45\mathbf j.$$

Multiply the unit vector by the required magnitude:

$$10\left(\frac35\mathbf i-\frac45\mathbf j\right) =\boxed{6\mathbf i-8\mathbf j}.$$

Its magnitude is $\sqrt{6^2+(-8)^2}=10$, as required.

The zero vector cannot be normalised

The formula $\mathbf a/|\mathbf a|$ requires $|\mathbf a|\ne0$. Since $|\mathbf0|=0$, normalising $\mathbf0$ would involve division by zero. The zero vector has no defined direction.

Self-check 2

Find a vector of magnitude $12$ in the opposite direction to $\begin{pmatrix}4\\-3\end{pmatrix}$.

Answer

The given vector has magnitude $5$, so a unit vector in the opposite direction is

$$-\frac15\begin{pmatrix}4\\-3\end{pmatrix} =\begin{pmatrix}-4/5\\3/5\end{pmatrix}.$$

Multiplying by $12$ gives

$$\boxed{\begin{pmatrix}-48/5\\36/5\end{pmatrix}}.$$

Direction angles in two dimensions

The direction of a two dimensional vector is often given by an angle $\theta$ measured anticlockwise from the positive $x$ axis. For

$$\mathbf a=\begin{pmatrix}x\\y\end{pmatrix},$$

the component triangle suggests

$$\tan\theta=\frac yx.$$

However, $\tan^{-1}(y/x)$ alone does not identify the correct quadrant. A calculator may return an acute or negative principal value even when the vector points into the second or third quadrant.

A reliable method is:

1. inspect the signs of $x$ and $y$ to identify the quadrant;
2. find the reference angle $\alpha=\tan^{-1}(|y/x|)$;
3. convert $\alpha$ into the required direction angle.

Signs of $(x,y)$	Quadrant	Direction angle
$(+,+)$	first	$\theta=\alpha$
$(-,+)$	second	$\theta=180^\circ-\alpha$
$(-,-)$	third	$\theta=180^\circ+\alpha$
$(+,-)$	fourth	$\theta=360^\circ-\alpha$

If your calculator provides an $\operatorname{atan2}(y,x)$ function, it handles the quadrant using both components. Convert a negative result to an angle from $0^\circ$ to $360^\circ$ by adding $360^\circ$.

Worked example 8: direction in the second quadrant

Find the direction of $\mathbf a=\begin{pmatrix}-4\\7\end{pmatrix}$, measured anticlockwise from the positive $x$ axis.

The signs $(-,+)$ place the vector in the second quadrant. Its reference angle is

$$\alpha=\tan^{-1}\left(\frac74\right)\approx60.255^\circ.$$

Therefore

$$\theta=180^\circ-60.255^\circ=119.745^\circ.$$

To one decimal place, the direction is

$$\boxed{119.7^\circ}.$$

The rough size makes sense: the vector points up and left, so its direction must lie between $90^\circ$ and $180^\circ$.

Worked example 9: direction in the fourth quadrant

Find the direction of $5\mathbf i-12\mathbf j$.

The vector lies in the fourth quadrant. The reference angle is

$$\alpha=\tan^{-1}\left(\frac{12}{5}\right)\approx67.380^\circ.$$

Hence

$$\theta=360^\circ-67.380^\circ =\boxed{292.6^\circ}$$

to one decimal place.

Do not lose the component signs

Writing only $\tan^{-1}(|y/x|)$ gives a reference angle, not necessarily the direction angle. Always identify the quadrant from the original signed components.

Self-check 3

Find the direction of $\begin{pmatrix}-3\\-3\sqrt3\end{pmatrix}$, measured anticlockwise from the positive $x$ axis.

Answer

Both components are negative, so the vector is in the third quadrant. The reference angle is

$$\alpha=\tan^{-1}\left(\frac{3\sqrt3}{3}\right) =\tan^{-1}(\sqrt3)=60^\circ.$$

Thus

$$\boxed{\theta=180^\circ+60^\circ=240^\circ}.$$

Forming a vector from magnitude and direction

Suppose a vector has magnitude $r$ and direction angle $\theta$, measured anticlockwise from the positive $x$ axis. Resolving it horizontally and vertically gives

$$\boxed{ \mathbf a =\begin{pmatrix}r\cos\theta\\r\sin\theta\end{pmatrix}}$$

or, equivalently,

$$\boxed{\mathbf a=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j}.$$

The vector $\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}$ has magnitude $1$ because

$$\sqrt{\cos^2\theta+\sin^2\theta}=1.$$

Multiplying it by $r$ gives the required magnitude.

Worked example 10: magnitude and direction to components

A force has magnitude $20\ \mathrm N$ and acts at $35^\circ$ above the positive horizontal. Write the force as a column vector.

The horizontal component is adjacent to the angle and the vertical component is opposite:

$$\mathbf F =\begin{pmatrix}20\cos35^\circ\\20\sin35^\circ\end{pmatrix}\mathrm N.$$

Therefore

$$\boxed{ \mathbf F\approx \begin{pmatrix}16.4\\11.5\end{pmatrix}\mathrm N}$$

to $3$ significant figures. Both components are positive, as expected from the stated direction.

Worked example 11: exact components

Find the vector of magnitude $8$ with direction $150^\circ$.

$$\begin{aligned} \mathbf a &=\begin{pmatrix}8\cos150^\circ\\8\sin150^\circ\end{pmatrix}\\ &=\begin{pmatrix}8(-\sqrt3/2)\\8(1/2)\end{pmatrix}\\ &=\boxed{\begin{pmatrix}-4\sqrt3\\4\end{pmatrix}}. \end{aligned}$$

The signs provide a useful check: $150^\circ$ lies in the second quadrant, so the $x$ component should be negative and the $y$ component positive.

Bearings and compass directions

A bearing is measured clockwise from north and written with three figures. This differs from the usual mathematical direction angle, which is measured anticlockwise from the positive $x$ axis.

If $\mathbf i$ points east and $\mathbf j$ points north, a vector of magnitude $r$ on a bearing $\beta$ has components

$$\boxed{ \mathbf a =r\sin\beta\,\mathbf i+r\cos\beta\,\mathbf j}.$$

Sine gives the east component because the bearing angle is measured from the north axis. This is the reverse of the usual $r\cos\theta$, $r\sin\theta$ order.

Worked example 12: velocity on a bearing

An aircraft flies at $240\ \mathrm{km\,h^{-1}}$ on a bearing of $120^\circ$. Write its velocity in east and north components.

$$\begin{aligned} \mathbf v &=240\sin120^\circ\,\mathbf i +240\cos120^\circ\,\mathbf j\\ &=240\left(\frac{\sqrt3}{2}\right)\mathbf i +240\left(-\frac12\right)\mathbf j\\ &=\boxed{120\sqrt3\,\mathbf i-120\mathbf j\ \mathrm{km\,h^{-1}}}. \end{aligned}$$

A bearing of $120^\circ$ points south east, so positive east and negative north components are sensible.

Worked example 13: components to a bearing

A velocity is

$$\mathbf v=-5\mathbf i+12\mathbf j\ \mathrm{m\,s^{-1}},$$

where $\mathbf i$ points east and $\mathbf j$ points north. Find its speed and bearing.

The speed is

$$|\mathbf v|=\sqrt{(-5)^2+12^2}=13\ \mathrm{m\,s^{-1}}.$$

The vector points north west. Its angle west of north is

$$\alpha=\tan^{-1}\left(\frac5{12}\right)\approx22.620^\circ.$$

Bearings increase clockwise, so the bearing is

$$360^\circ-22.620^\circ=337.380^\circ.$$

Thus the velocity has speed $\boxed{13\ \mathrm{m\,s^{-1}}}$ on a bearing of $\boxed{337^\circ}$ to the nearest degree.

Self-check 4

A displacement of $30$ km is made on a bearing of $210^\circ$. Find its east and north components exactly.

Answer

$$\begin{aligned} \mathbf d &=30\sin210^\circ\,\mathbf i+30\cos210^\circ\,\mathbf j\\ &=30\left(-\frac12\right)\mathbf i +30\left(-\frac{\sqrt3}{2}\right)\mathbf j\\ &=\boxed{-15\mathbf i-15\sqrt3\mathbf j\ \mathrm{km}}. \end{aligned}$$

The bearing points south west, so both components are negative.

Solving equations involving magnitudes

An equation such as $|\mathbf a|=c$ becomes an ordinary scalar equation after squaring. Remember that squaring can lead to two possible values of an unknown component.

Worked example 14: find an unknown component

The vector

$$\mathbf p=\begin{pmatrix}k\\4\end{pmatrix}$$

has magnitude $\sqrt{41}$. Find $k$.

Use the magnitude formula:

$$\sqrt{k^2+4^2}=\sqrt{41}.$$

Squaring both sides gives

$$k^2+16=41,$$

so

$$k^2=25.$$

Therefore

$$\boxed{k=5\quad\text{or}\quad k=-5}.$$

Both vectors have the same length but point in different directions.

Worked example 15: use a direction condition

The vector $\mathbf a=\begin{pmatrix}k\\-8\end{pmatrix}$ has magnitude $10$ and points into the fourth quadrant. Find $k$.

The magnitude condition gives

$$k^2+(-8)^2=10^2,$$

so

$$k^2=36$$

and initially $k=\pm6$. In the fourth quadrant the horizontal component is positive and the vertical component negative. Therefore

$$\boxed{k=6}.$$

The direction information selects one of the two algebraic possibilities.

Common misconceptions

• A vector and its magnitude are different types of quantity. $\mathbf a$ is a vector, while $|\mathbf a|$ is a non-negative scalar.
• Negative components do not make a negative magnitude. Every component is squared in the magnitude formula.
• Dividing by a magnitude produces length $1$. Multiplying by the magnitude usually does not.
• Opposite direction requires a negative sign. The unit vector opposite to $\mathbf a$ is $-\mathbf a/|\mathbf a|$.
• An inverse tangent gives only a reference angle unless the quadrant is handled. Check the signs of both components.
• Bearings start from north, not east. For east and north components use $r\sin\beta$ and $r\cos\beta$ respectively.
• Exact values should remain exact. Keep $\sqrt{13}$ or $4\sqrt3$ unless a decimal accuracy is requested.

Mixed self-check

1. Find the magnitude of $\mathbf a=\begin{pmatrix}1\\-2\\2\end{pmatrix}$.
2. Find a unit vector in the direction from $P(2,-1)$ to $Q(-4,7)$.
3. A vector has magnitude $14$ and direction $225^\circ$. Find its exact components.
4. A vector has components $8\mathbf i+15\mathbf j$. Find its magnitude and its direction anticlockwise from the positive $x$ axis.

Answers

$$|\mathbf a|=\sqrt{1^2+(-2)^2+2^2}=\boxed{3}.$$

2. First,

$$\overrightarrow{PQ}=\begin{pmatrix}-4-2\\7-(-1)\end{pmatrix} =\begin{pmatrix}-6\\8\end{pmatrix}, \qquad |\overrightarrow{PQ}|=10.$$

Hence the required unit vector is

$$\boxed{\frac1{10}\begin{pmatrix}-6\\8\end{pmatrix} =\begin{pmatrix}-3/5\\4/5\end{pmatrix}}.$$

$$\begin{aligned} \mathbf a &=\begin{pmatrix}14\cos225^\circ\\14\sin225^\circ\end{pmatrix}\\ &=\boxed{\begin{pmatrix}-7\sqrt2\\-7\sqrt2\end{pmatrix}}. \end{aligned}$$

4. The magnitude is

$$\sqrt{8^2+15^2}=\boxed{17}.$$

The vector is in the first quadrant, so

$$\theta=\tan^{-1}\left(\frac{15}{8}\right) \approx\boxed{61.9^\circ}.$$

Summary

For $\mathbf a=\begin{pmatrix}x\\y\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}x\\y\\z\end{pmatrix}$,

$$|\mathbf a|=\sqrt{x^2+y^2}, \qquad |\mathbf b|=\sqrt{x^2+y^2+z^2}.$$

For any non-zero vector $\mathbf a$,

$$\text{unit vector in its direction}=\frac{\mathbf a}{|\mathbf a|}.$$

A two dimensional vector of magnitude $r$ and standard direction angle $\theta$ is

$$\begin{pmatrix}r\cos\theta\\r\sin\theta\end{pmatrix}.$$

When finding a direction, identify the quadrant before interpreting an inverse tangent. When working with bearings, remember that the angle is measured clockwise from north.

Next, use these ideas to calculate position vectors and distance and then apply vectors in vector modelling. For physical applications involving velocity, see two dimensional motion with vectors.