Answer
Magnitude: $8$
Direction angle: $120^{\circ}$
Work Step by Step
The magnitude $|u|$ of a vector $u=\left\langle a,b\right\rangle$ is given by $|u|=\sqrt {a^2+b^2}$.
For the given vector
$u=\left\langle -4,4\sqrt 3\right\rangle \Rightarrow ~ a=-4, b=4\sqrt 3.$
Therefore,
$|u|=|\left\langle -4,4\sqrt 3\right\rangle|=\sqrt {(-4)^2+(4\sqrt 3)^2}=\sqrt {16+4^2(\sqrt {3})^2}=\sqrt{16+16\times3}=\sqrt{64}=8$
Hence the magnitude of the vector $\left\langle -4,4\sqrt 3\right\rangle$ is $8$.
The direction angle $\theta$ of the vector $u=\left\langle a,b\right\rangle$ is given by
$\theta=\tan^{-1}\left(\frac{b}{a}\right)$
Therefore, for $u=\left\langle -4,4\sqrt 3\right\rangle $ we have
$\theta=\tan^{-1}\left(\frac{-4}{4\sqrt3}\right)=-\tan^{-1}\left(\frac{1}{\sqrt3}\right)=-60^{\circ}$
Adding $180^{\circ}$ will give direction angle
$\theta=-60^{\circ}+180^{\circ}=120^{\circ}$ with positive $x$-axis.
Hence, the magnitude of $\left\langle -4,4\sqrt 3\right\rangle$ is $8$ and the direction angle is $120^{\circ}$.
