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 $\left[0,360^{\circ}\right)$.$$\langle- 4,4 \sqrt{3}\rangle$$
Step 1
The magnitude of a vector $\langle x, y \rangle$ is given by $\sqrt{x^2 + y^2}$. So, for the vector $\langle -4, 4\sqrt{3} \rangle$, the magnitude is $\sqrt{(-4)^2 + (4\sqrt{3})^2}$. Show more…
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