Answer
$\frac{1}{2}<2,-2,1>$
Work Step by Step
$\|\mathbf{v}\|=\frac{3}{2}$ and $\mathbf{u}=<2,-2,1>$
$\|\mathbf{u}\|=\sqrt{2^2+2^2+1^2}=3$
Unit vector in direction of $\mathbf{u}$: $\frac{1}{3}<2,-2,1>$
Vector in direction of $\mathbf{u}$ with magnitude $\frac{3}{2}$: $\frac{3}{2}\times\frac{1}{3}<2,-2,1>$
$=\frac{1}{2}<2,-2,1>$
