Question

1. an airplane flies in the path shown by the vector $overrightarrow{cd}$ on the graph below. what is the magnitude and direction of $overrightarrow{cd}$? (1 unit represents 1 mile)

c (-4,4)
d (0,1)
-6 -4 -2 -2 2 4
-4
6
4
2
-2
-4
72° south of west for 6 miles
42° south for 8 miles
26° north for 7 miles
53° east of south for 5 miles

Explanation:

Step1: Calculate the magnitude of the vector

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $x_1=-4,y_1 = 4,x_2=0,y_2 = 1$. So $d=\sqrt{(0 + 4)^2+(1 - 4)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.

Step2: Calculate the direction of the vector

The slope of the line passing through $C(-4,4)$ and $D(0,1)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{1 - 4}{0+4}=-\frac{3}{4}$. The angle $\theta$ that the vector makes with the positive - x - axis satisfies $\tan\theta=-\frac{3}{4}$. The reference angle $\alpha=\arctan(\frac{3}{4})\approx37^{\circ}$. Since the vector is in the fourth - quadrant, the angle measured from the positive y - axis (south - to - east direction) is $\theta = 90^{\circ}-\alpha=53^{\circ}$ east of south.

Answer:

$53^{\circ}$ east of south for 5 miles