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

Work through these on your own. Sketch the vector before you quote a direction, and write the quadrant beside every angle. Check the length of any unit vector you find.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find the magnitude of $\mathbf{a} = 6\mathbf{i} + 8\mathbf{j}$ .
Find the magnitude of $\mathbf{b} = -7\mathbf{i} + 24\mathbf{j}$ .
Find the magnitude of $\mathbf{c} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}$ .
Find the direction of $\mathbf{a} = \mathbf{i} + \sqrt{3}\mathbf{j}$ as an angle from the positive $x$ -axis.
Find the direction of $\mathbf{b} = -\mathbf{i} - \mathbf{j}$ as an anticlockwise angle from the positive $x$ -axis, between $0^\circ$ and $360^\circ$ .
Find the direction of $\mathbf{c} = 3\mathbf{i} - 3\mathbf{j}$ as an anticlockwise angle from the positive $x$ -axis, between $0^\circ$ and $360^\circ$ .
Find the unit vector in the direction of $\mathbf{a} = 6\mathbf{i} + 8\mathbf{j}$ .
A force of $20$ N acts at $60^\circ$ above the horizontal. Find its horizontal and vertical components, exactly.
A cyclist rides $8$ km on a bearing of $030^\circ$ . Find the displacement from the start, with east as $\mathbf{i}$ and north as $\mathbf{j}$ , exactly.
A vector has magnitude $25$ and lies in the first quadrant. The angle $\theta$ it makes with the positive $x$ -axis satisfies $\tan\theta = \dfrac{7}{24}$ . Find its two components.
A drone is $3$ m east, $4$ m north and $12$ m above its base. Find its distance from the base.
A current has velocity $-4\mathbf{i} + 3\mathbf{j}$ m/s. Find its speed and the angle it makes with the positive $x$ -axis.

Answers and concise solutions

1. The magnitude

Answer: $|\mathbf{a}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ .

2. The magnitude

Answer: $|\mathbf{b}| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ . Squaring removes the negative sign.

3. The magnitude

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

4. The direction in degrees

Answer: Both components are positive, so the vector is in the first quadrant. $\tan\theta = \dfrac{\sqrt{3}}{1} = \sqrt{3}$ , so $\theta = 60^\circ$ .

5. The direction in degrees

Answer: Both components are negative, so the vector is in the third quadrant. Its angle to the negative $x$ -axis is $45^\circ$ , so its anticlockwise direction is $180^\circ + 45^\circ = 225^\circ$ .

6. The direction in degrees

Answer: The vector is in the fourth quadrant. Its angle below the positive $x$ -axis is $\tan^{-1}(1) = 45^\circ$ , so its anticlockwise direction is $360^\circ - 45^\circ = 315^\circ$ .

7. The unit vector

Answer: $|\mathbf{a}| = 10$ , so $\hat{\mathbf{a}} = \dfrac{1}{10}(6\mathbf{i} + 8\mathbf{j}) = \dfrac{3}{5}\mathbf{i} + \dfrac{4}{5}\mathbf{j}$ . Check: $\sqrt{(3/5)^2 + (4/5)^2} = 1$ .

8. The two components

Answer: $F_x = 20\cos 60^\circ = 20 \times \dfrac{1}{2} = 10$ N and $F_y = 20\sin 60^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ N. The force is $10\mathbf{i} + 10\sqrt{3}\mathbf{j}$ N.

9. The two components

Answer: A bearing is measured from north, so the east component uses sine and the north component uses cosine. East $= 8\sin 30^\circ = 4$ km and north $= 8\cos 30^\circ = 4\sqrt{3}$ km. The displacement is $4\mathbf{i} + 4\sqrt{3}\mathbf{j}$ km.

10. The two components

Answer: The components form a $7$ – $24$ – $25$ right-angled triangle, so $\cos\theta = \dfrac{24}{25}$ and $\sin\theta = \dfrac{7}{25}$ . The components are $25 \times \dfrac{24}{25} = 24$ and $25 \times \dfrac{7}{25} = 7$ .

11. The distance

Answer: The displacement is $3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$ m, so the distance is $\sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ m.

12. The speed and the direction

Answer: The speed is the magnitude: $\sqrt{(-4)^2 + 3^2} = 5$ m/s. The vector is in the second quadrant, so the angle from the positive $x$ -axis is $180^\circ - \tan^{-1}\left(\dfrac{3}{4}\right) = 180^\circ - 36.9^\circ = 143.1^\circ$ .