Answer
$4 \sqrt 2 i+4 \sqrt 2 j$
Work Step by Step
Let us consider that a vector $v$ is given by $v=pi+qj$ and makes an angle of $\alpha$ with the positive $x$-axis. The magnitude of the vector can be determined using the formula
$||v||=\sqrt{p^2+q^2} $
Then we find $p=||v||cos\alpha$ and $v=||v||sin\alpha$.
Here, we have: $p=||8|| \cos \ (45^\circ)=4 \sqrt 2$
and $v=||8|| \sin \ (45^\circ)=4 \sqrt 2$
Therefore $v=4 \sqrt 2 i+4 \sqrt 2 j$
