## Precalculus (10th Edition)

$\sqrt{5}$.
We know that if $v=ai+bj+ck$ and $w=di+ej+fk$, then $v+w=(a+d)i+(b+e)j+(c+f)k$ and that if $z$ is a constant, then $zv=(za)i+(zb)j+(zc)k$. Also, the magnitude of a vector $v=ai+bj+ck$ is: $||v||=\sqrt{a^2+b^2+c^2}$. Hence here: $||v+w||=||(3i-5j+2k)+(-2i+3j-2k)||=||i-2j||=\sqrt{1^2+(-2)^2}=\sqrt{1+4}=\sqrt{5}$.