Answer
a) $3 \vec{v}=\vec{u}$
b) $\frac{6}{\sqrt{26}} \hat{\imath}-\frac{8}{\sqrt{26}} \hat{\jmath}-\frac{2}{\sqrt{26}} \hat{k}=\vec{u}$
Work Step by Step
a) Let $\vec{v}=-3 \hat{\jmath}+2 \hat{\imath} .$ The vector $\vec{u}$ with the same direction of $\vec{v}$ and three times the length of $\vec{v}$ is
\[
\vec{u}=3 \vec{v}=(-3 \hat{\jmath}+2 \hat{\imath})3=-9 \hat{\jmath}+6 \hat{\imath}
\]
because $\vec{v} / / 3 \vec{v}$ and $\|3 \vec{v}\|=3\|\vec{v}\|$
$b)$
Let $\vec{v}=4 \hat{\jmath}+\hat{k}-3 \hat{\imath} .$ The unitary vector that is oppositely directed to $\vec{v}$ is
\[
\begin{aligned}
\hat{u} &=-\frac{1}{\|\vec{v}\|} \vec{v}=-\frac{1}{\sqrt{4^{2}+1^{2}+(-3)^{2}}}(-3 \hat{\imath}+4 \hat{\jmath}+\hat{k}) \\
&=-(-3 \hat{\imath}+4 \hat{\jmath}+\hat{k})\frac{1}{\sqrt{26}}
\end{aligned}
\]
So the vector with length 2 in the opposite direction of $\vec{v}$ is
\[
\vec{u}=2 \hat{u}=-(-3 \hat{\imath}+4 \hat{\jmath}+\hat{k}) \frac{2}{\sqrt{26}}
\]
