Answer
$|v|=1,$
$\theta=225^{o}$
Work Step by Step
The magnitude of v = $< a_{1},a_{2} > $is $| $v $|=\sqrt{a_{1}^{2}+a_{2}^{2}}.$
The direction is given by$\qquad \displaystyle \theta=\tan^{-1}(\frac{a_{2}}{a_{1}})$
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$|v|=\sqrt{(-\frac{\sqrt{2}}{2})^{2}+(-\frac{\sqrt{2}}{2})^{2}}=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}=1$
The vector v points to the 3rd quadrant, as both components are negative.
Our calculator will give us a 1st quadrant value... so we will add or subtract $180^{o}$.
$\displaystyle \tan^{-1}(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}})=\tan^{-1}1=45^{0}$
$ ($we add $180^{o}$ to find the angle in the 3rd quadrant)
$\theta=45^{o}+180^{o}=225^{o}$
