Answer
$\left(\dfrac{30\sqrt{13}}{13},-\dfrac{20\sqrt{13}}{13}\right)$
Work Step by Step
We have to find a vector $w$ so that
$|w|=10$
with the direction $v=(3,-2)$.
First determine the unit vector of $v$: $u=\dfrac{\overrightarrow{v}}{|v|}=\dfrac{(3,-2)}{\sqrt{3^2+(-2)^2}}=\dfrac{(3,-2)}{\sqrt{13}}=\left(\dfrac{3}{\sqrt{13}},-\dfrac{2}{\sqrt{13}}\right)=\left(\dfrac{3\sqrt{13}}{13},-\dfrac{2\sqrt{13}}{13}\right)$
A vector $w$ of length 10 and parallel to $v$ is:
$w=10u=10\left(\dfrac{3\sqrt{13}}{13},-\dfrac{2\sqrt{13}}{13}\right)=\left(\dfrac{30\sqrt{13}}{13},-\dfrac{20\sqrt{13}}{13}\right)$
