10.2 Magnitude and direction; component and magnitude/direction form — Learn

Prerequisites: Pythagoras' theorem; the trigonometric ratios $\sin$ , $\cos$ and $\tan$ ; finding an angle from a tangent; bearings from GCSE.

By the end of this section you can: find the magnitude of a vector in two and three dimensions; find the direction of a two-dimensional vector as an angle or a bearing; convert a vector from component form to magnitude/direction form and back; write a unit vector in the direction of a given vector.

1. The magnitude of a vector

The magnitude of a vector is its length. The magnitude of $\mathbf{a}$ is written $|\mathbf{a}|$ . The magnitude is a scalar, so it is never negative.

The components of a vector are the two sides of a right-angled triangle. The vector itself is the hypotenuse. Pythagoras' theorem gives the length.

For a two-dimensional vector,

\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}, \qquad |\mathbf{a}| = \sqrt{a_1^2 + a_2^2}.

For a three-dimensional vector, the same idea extends by one more square:

\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}, \qquad |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

Example 1. Find the magnitude of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ .

|\mathbf{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

The vector has length $5$ .

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

Square every component, add, then take the square root:

|\mathbf{b}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.

The sign of a component disappears when it is squared. A vector pointing down or backwards is not "negative" in length.

2. The direction of a two-dimensional vector

The direction of a vector is the angle it makes with a fixed line. Two conventions are common.

The angle measured anticlockwise from the positive $x$ -axis.
A bearing, measured clockwise from north.

At A level you must read which convention a question uses. If the question says "from the positive $x$ -axis", use the first. If it gives a direction such as "on a bearing of $120$ °", use the second.

For a vector $a_1\mathbf{i} + a_2\mathbf{j}$ with $a_1 \neq 0$ , the angle $\theta$ from the positive $x$ -axis satisfies

\tan\theta = \frac{a_2}{a_1}.

The inverse tangent gives only one angle. Use the signs of $a_1$ and $a_2$ to place the vector in the correct quadrant, then adjust the angle.

Example 3. Find the direction of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ , as an angle from the positive $x$ -axis.

Both components are positive, so the vector is in the first quadrant. Then

\theta = \tan^{-1}\left(\frac{4}{3}\right) = 53.1^\circ \text{ to } 1 \text{ d.p.}

Example 4. Find the direction of $\mathbf{c} = -3\mathbf{i} + 4\mathbf{j}$ from the positive $x$ -axis.

The $x$ -component is negative and the $y$ -component is positive, so the vector is in the second quadrant. The inverse tangent gives the angle to the negative $x$ -axis:

\tan^{-1}\left(\frac{4}{3}\right) = 53.1^\circ.

Subtract this from $180^\circ$ :

\theta = 180^\circ - 53.1^\circ = 126.9^\circ.

The angle is $126.9^\circ$ anticlockwise from the positive $x$ -axis.

3. From magnitude and direction to components

If a vector has magnitude $r$ and makes an angle $\theta$ with the positive $x$ -axis, its components are

a_1 = r\cos\theta, \qquad a_2 = r\sin\theta.

Example 5. A force of $10$ N acts at $30^\circ$ above the horizontal. Write the force in component form.

F_x = 10\cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}, \qquad F_y = 10\sin 30^\circ = 10 \times \frac{1}{2} = 5.

The force is $5\sqrt{3}\mathbf{i} + 5\mathbf{j}$ newtons, or about $8.66\mathbf{i} + 5\mathbf{j}$ N.

For a bearing, the angle is measured clockwise from north. The east component uses sine and the north component uses cosine:

a_{\text{east}} = r\sin\theta, \qquad a_{\text{north}} = r\cos\theta.

Example 6. A ship sails $20$ km on a bearing of $120^\circ$ . Find its displacement from the start, with east as $\mathbf{i}$ and north as $\mathbf{j}$ .

\text{east} = 20\sin 120^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}, \qquad \text{north} = 20\cos 120^\circ = 20 \times \left(-\frac{1}{2}\right) = -10.

The displacement is $10\sqrt{3}\mathbf{i} - 10\mathbf{j}$ km. The negative north component means the ship finishes $10$ km south of its starting line.

4. Unit vectors

A unit vector has magnitude $1$ . It records a direction only. The unit vector in the direction of $\mathbf{a}$ is

\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}.

Example 7. Find the unit vector in the direction of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ .

From Example 1, $|\mathbf{a}| = 5$ , so

\hat{\mathbf{a}} = \frac{1}{5}(3\mathbf{i} + 4\mathbf{j}) = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}.

Check the length: $\sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = 1$ . The unit vector has magnitude $1$ , as it must.

Watch out. Three slips are common on this page.

Adding magnitudes. The magnitude of a sum is not the sum of the magnitudes. Two vectors of length $3$ and $4$ can have a sum of any length from $1$ to $7$ , depending on their directions. Find the sum first, then its magnitude.

The blind inverse tangent. $\tan^{-1}(4/3)$ is always $53.1^\circ$ , but the vector $(-3, 4)$ points into the second quadrant, so its direction is $126.9^\circ$ . Always place the vector with the signs of its components before you quote an angle.

Mixing the two conventions. An angle of $30^\circ$ from the $x$ -axis and a bearing of $30^\circ$ point in different directions. A bearing of $30^\circ$ is $60^\circ$ from the positive $x$ -axis. Re-check: sketch the vector in the correct quadrant, then ask whether the angle you found matches the sketch.

You should now be able to

find the magnitude of a vector in two or three dimensions;
find the direction of a two-dimensional vector as an angle or a bearing;
keep the correct quadrant when you use an inverse tangent;
convert from magnitude and direction to components, including with bearings;
write and check a unit vector.