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