Question

find the magnitude and direction angle (to the nearest tenth) for each vector. give the measure of the direction angle as an angle in 0,360°.
⟨-6√3,6⟩
a. 12; 150°
b. 24; 150°
c. 12; 330°
d. 24; 30°

Explanation:

Step1: Calculate the magnitude

For a vector \(\langle x,y
angle\), the magnitude \(r=\sqrt{x^{2}+y^{2}}\). Here \(x = - 6\sqrt{3}\), \(y = 6\).

$$
LATEXBLOCK0
$$

Step2: Calculate the direction angle

The formula for the direction angle \(\theta\) is \(\tan\theta=\frac{y}{x}\). Here \(\tan\theta=\frac{6}{-6\sqrt{3}}=-\frac{1}{\sqrt{3}}\).
Since \(x<0,y>0\), the vector is in the second - quadrant.
We know that \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), so \(\theta = 180^{\circ}-30^{\circ}=150^{\circ}\)

Answer:

A. \(12;150^{\circ}\)