Answer
Magnitude: $17$
Direction angle: $331.93^{\circ}$
Work Step by Step
The magnitude $|u|$ of vector $u=\left\langle a,b\right\rangle$ is given by $|u|=\sqrt {a^2+b^2}$.
For the given vector
$u=\left\langle 15,-8\right\rangle \Rightarrow ~ a=15, b=-8.$
Therefore,
$|u|=|\left\langle 15,-8\right\rangle|=\sqrt {(15)^2+(-8)^2}=\sqrt {225+64}=\sqrt {289}=17$
Hence the magnitude of the vector $\left\langle 15,-8\right\rangle$ is $17$.
The direction angle $\theta$ of vector $u=\left\langle a,b\right\rangle$ is given by
$\theta=\tan^{-1}\left(\frac{b}{a}\right)$
Therefore, for $u=\left\langle 15,-8\right\rangle $ we have
$\theta=\tan^{-1}\left(\frac{-8}{15}\right)=-28.07^{\circ}$
Adding $360°$ yields the direction angle $\theta=-28.07^{\circ}+360^{\circ}=331.93^{\circ}$.
