Answer
$ \lt \sqrt {32}, \sqrt {32} \gt $
Work Step by Step
Since $v$ has the same direction as $u$ and since it is a multiple of a unit vector $n$, we can write it as: $\overrightarrow{v} = < 3 n, 3n>$
To find the unit vector, we must first find the magnitude:
$\sqrt {((3n)^{2} + (3n)^{2}} = 8$
or, $ 9\ n^2+9 \ n^2 = 64$
This implies that $n^2=\dfrac{32}{9} \implies n=\dfrac{\sqrt {32}}{3}$
Now, $\overrightarrow{v} = < 3 (\dfrac{\sqrt {32}}{3}), 3 (\dfrac{\sqrt {32}}{3})> = \lt \sqrt {32}, \sqrt {32} \gt $
