Answer
$a.\qquad\sqrt{a_{1}^{2}+a_{2}^{2}},\qquad 2\sqrt{2}$
$ b.\qquad$w = $ < |w|\cos\theta,|w|\sin\theta > $
Work Step by Step
$a.$
The length of a vector w=$ < a_{1},a_{2} > $is $| $w $|=\sqrt{a_{1}^{2}+a_{2}^{2}}.$
In figure II, the x component of u, $a_{1}=2$, and the y component of u, $a_{2}=2$, so
$|u|=\sqrt{2^{2}+2^{2}}=\sqrt{2\cdot 2^{2}}=2\sqrt{2}$
$b.$
Sketching a vector w, we draw two perpendicular components,
which are the legs of a right triangle whose hypotenuse has length $|w|$.
The angle between the x-component and w is $\theta.$
The y-component is opposite this angle.
The cosine of $\theta $ is $ \displaystyle \frac{a_{1}}{|w|}$, the adjacent leg over the hypotenuse, so
$a_{1}=|w|\cos\theta.$
Similarly, observing the sine,
$ a_{2}=|w|\sin\theta$
w = $ < |w|\cos\theta,|w|\sin\theta > $
