Example 1: Displacement of 40 km, 30° west of south, drawn to scale

Basics you need

Solution

1. Read the direction

   "30° west of south" means: start on the line pointing south, then turn through 30° towards west. The required direction makes

   $$\begin{aligned} &30^\circ \text{ with OS}, \\ &90^\circ - 30^\circ = 60^\circ \text{ with OW}. \end{aligned}$$
   A compass direction of the form "$\theta$ west of south" names the starting line (south) and then the side to turn towards (west). Read the other way round, as 30° south of west, the arrow would be only 30° below the west line, which is a different direction: 60° west of south.
   Directions on a map are measured from one of the four main lines N, E, S, W. Basics: "30° east of north": start, turn, go

2. Choose a scale

   Take 1 cm (one scale length) to represent 10 km, as the scale bar of the book's Fig 10.4 does. Then the arrow must have length

   $$\frac{40\ \text{km}}{10\ \text{km per cm}} = 4\ \text{cm}.$$
   The magnitude of a vector is the length of its directed line segment (Section 10.2). A 40 km arrow cannot fit on paper, so every length is drawn at the same reduced scale and the scale is stated; then the 4 cm arrow stands for 40 km. Any convenient scale is allowed, as long as it is written down.
   The distance between the initial and terminal points of a vector is its magnitude, written $|\overrightarrow{OP}|$ (Section 10.2). Lesson: a vector is a length with a direction

3. Draw the compass lines and the angle

   Draw a vertical line NS and a horizontal line EW, perpendicular to each other and meeting at O. Mark north at the top and east on the right, so south is at the bottom and west on the left. With a protractor on OS, mark 30° on the west side of OS and draw a line from O through that mark.

   O will be the initial point of the displacement, the place where the movement starts. The angle must be measured from OS (the starting line) and opened towards OW (the side named "west"), so the new line runs down and to the left.
   The same construction from north towards east gives Exercise 10.1 Q1 (40 km, 30° east of north). Lesson: Examples 1–3

4. Draw the arrow OP

   On the line just drawn, mark P with $OP = 4$ cm, and put the arrowhead at P.

   The directed line segment $\overrightarrow{OP}$ has initial point O, magnitude 40 km (4 cm at the scale) and direction 30° west of south, so it represents the given displacement. This is the book's Fig 10.4.

   The arrowhead is what makes the segment a vector: OP with the arrow at P means "from O to P". The same segment with the arrow at O would be $\overrightarrow{PO} = -\overrightarrow{OP}$⁠, the displacement of 40 km, 30° east of north, which is the opposite direction.
   A directed line segment has magnitude as well as direction, so it is a vector (Fig 10.1(iii) and Definition 1, Section 10.2). Lesson: a vector is a length with a direction

5. Check where P lands

   Drop perpendiculars from P to the two compass lines. In the right triangle with hypotenuse OP = 40 km and the angle 30° at O next to the south line:

   $$\begin{aligned} \text{south part} &= 40\cos 30^\circ = 20\sqrt3\ \text{km}, \\ \text{west part} &= 40\sin 30^\circ = 20\ \text{km}. \end{aligned}$$

   So P is 20 km west and $20\sqrt3 \approx 34.64$ km south of O. At the scale 1 cm = 10 km that is 2 cm to the left and about 3.46 cm down, which is how P can also be plotted with a ruler alone.

   The side next to the 30° angle lies along OS, so it is $40\cos 30^\circ$⁠; the side opposite the angle is parallel to OW, so it is $40\sin 30^\circ$⁠. The west part is smaller than the south part, as it should be: the arrow is closer to south (30°) than to west (60°).
   $\cos 30^\circ = \frac{\sqrt3}{2}$ and $\sin 30^\circ = \frac12$⁠. Basics: cos θ is a shadow, sin θ is a height

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