Exercise 10.1 · Q1: Displacement of 40 km, 30° east of north, drawn to scale

Basics you need

Solution

1. Read the direction

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

   $$\begin{aligned} &30^\circ \text{ with ON}, \\ &90^\circ - 30^\circ = 60^\circ \text{ with OE}. \end{aligned}$$
   A compass direction of the form "$\theta$ east of north" always names the starting line (north) and then the side to turn towards (east). Reading it the other way round, as 30° north of east, would put the arrow 30° above the east line, which is a different direction: 60° east of north.
   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 unit) to represent 10 km. 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 we draw every length at the same reduced scale and state the scale; then the 4 cm arrow stands for 40 km. Any convenient scale is allowed (the book's own figure marks a scale of 10 km), 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. With a protractor on ON, mark 30° on the east side of ON 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 compass lines give the reference directions; the angle must be measured from ON (the starting line) and opened towards OE (the side named "east").
   The same construction, with south as the starting line and west as the turning side, gives Example 1 (30° west of south, Fig 10.4). 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° east of north, so it represents the given displacement.

   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 40 km, 30° west of south, 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 north line:

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

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

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

Every fact used, and where it comes from