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

1. $\cos 60^\circ = \dfrac{1}{2}$ and $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$.
2. $\tan 45^\circ = 1$, so $\tan^{-1}(1) = 45^\circ$.
3. $\sqrt{25 + 144} = \sqrt{169} = 13$.

If question 1 was hard, revise the exact trigonometric values before you continue.

The magnitude is the hypotenuse of a right-angled triangle whose two shorter sides are the components.

Square each component, add the two squares, then take the square root.

The vector has length $13$.

The negative sign belongs to a direction, not to a length. Squaring removes it.

Use $(-8)^2 = 64$ and $6^2 = 36$.

The magnitude is $10$.

In three dimensions there are three squares to add before the square root.

Work out $2^2 + (-3)^2 + 6^2$.

The magnitude is $7$.

Both components are equal, so the vector lies on a familiar line through the origin.

$\tan\theta = \dfrac{1}{1} = 1$.

Both components are positive, so the vector is in the first quadrant. $\tan^{-1}(1) = 45^\circ$, so the direction is $45^\circ$ from the positive $x$-axis.

The $x$-component is negative and the $y$-component is positive. Sketch it to choose the quadrant.

The inverse tangent gives $45^\circ$ to the negative $x$-axis. Subtract it from $180^\circ$.

The vector is in the second quadrant. Its angle to the negative $x$-axis is $\tan^{-1}(1) = 45^\circ$, so its direction from the positive $x$-axis is

The $x$-component is $0$, so the formula for $\tan\theta$ divides by zero. Draw the vector instead.

The vector points straight down, which is $90^\circ$ clockwise from the positive $x$-axis.

The vector points straight downwards. As an anticlockwise angle from the positive $x$-axis, that is $270^\circ$. It can also be written as $-90^\circ$. Both name the same direction.

The horizontal component uses cosine and the vertical component uses sine.

Work out $25\cos 40^\circ$ and $25\sin 40^\circ$.

The force is $19.2\mathbf{i} + 16.1\mathbf{j}$ N.

A bearing is measured clockwise from north. The east component uses sine and the north component uses cosine.

Work out $15\sin 60^\circ$ and $15\cos 60^\circ$ with exact values.

The displacement is $\dfrac{15\sqrt{3}}{2}\mathbf{i} + \dfrac{15}{2}\mathbf{j}$ km. Both components are positive, so the walker finishes north-east of the start.

A unit vector has magnitude $1$. Divide the vector by its own magnitude.

Question 1 gives $|\mathbf{a}| = 13$. Divide each component by $13$.

Check: $\sqrt{(5/13)^2 + (12/13)^2} = \sqrt{25/169 + 144/169} = 1$, so the unit vector has magnitude $1$.