Question

find the magnitude and positive direction angle of the vector $\langle - 4\sqrt{3},-4\
angle$.
the magnitude of the vector is 8.
(simplify your answer.)
the direction angle is $\square^{\circ}$.
(simplify your answer. use angle measures greater than or equal to 0 and less than 360.)

Explanation:

Step1: Calculate the reference angle

For a vector \(\langle x,y
angle=\langle - 4\sqrt{3},-4
angle\), we first find the reference angle \(\theta'\) using the formula \(\tan\theta'=\frac{\vert y\vert}{\vert x\vert}\).
Here, \(x = - 4\sqrt{3}\), \(y=-4\), so \(\tan\theta'=\frac{\vert - 4\vert}{\vert - 4\sqrt{3}\vert}=\frac{4}{4\sqrt{3}}=\frac{1}{\sqrt{3}}\).
Since \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), the reference angle \(\theta' = 30^{\circ}\).

Step2: Determine the direction angle

The vector \(\langle - 4\sqrt{3},-4
angle\) lies in the third - quadrant (because \(x<0\) and \(y < 0\)).
The formula for the direction angle \(\theta\) of a vector in the third - quadrant is \(\theta=180^{\circ}+\theta'\).
Substituting \(\theta' = 30^{\circ}\), we get \(\theta=180^{\circ}+30^{\circ}\).

Answer:

\(210\)