Magnitude and direction angle?

Question

What is the magnitude and direction angle of $v=4i+4j$? I have no idea how to start this problem, so any hints/formulas/tips and tricks would be useful, thank you!

Answer 1

We have: $||\vec{v}|| = \sqrt{x^2+y^2}$ if $\vec{v} = x\vec{i} + y\vec{j}$,

and the angle $\theta $ that $\vec{v}$ makes with the $x$-axis is:

$\theta = \tan^{-1}\left(\dfrac{y}{x}\right)$. So with $x = y = 4$, we have:

$||\vec{v}|| = \sqrt{4^2+4^2} = 4\sqrt{2}$, and $\theta = \tan^{-1} \left(\dfrac{4}{4}\right) = \tan^{-1} (1) = 45^\circ$.