## Precalculus (10th Edition)

$21i-2j-5k$
We know that for two vectors $k=a+bi$ and $l=c+di$, $k+l=((a+c)+(b+d)i$. Also, if $m$ is a constant $m\cdot k=m(a+bi)=ma+mbi.$ \begin{align*} 4v-3w=4(3i+j-2k)-3(-3i+2j-k)=12i+4j-8k+9i-6j+3k=i(12+9)+j(4-6)+k(-8+3)=21i-2j-5k\end{align*}