Answer
$\frac{\pi}{2}$ rad = $90^\circ$
Work Step by Step
$\mathbf{u}=3\mathbf{i}+2\mathbf{j}+\mathbf{k}$ and $\mathbf{v}=2\mathbf{i}-3\mathbf{j}$
$\mathbf{u}\cdot\mathbf{v}=3\times2+2\times(-3)+1\times0=6-6+0=0$
$\|\mathbf{u}\|=\sqrt{3^2+2^2+1^2}=\sqrt{14}$ and $\|\mathbf{v}\|=\sqrt{2^2+(-3)^2}=\sqrt{13}$
${\theta}=\cos^{-1}\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}=\cos^{-1}\frac{0}{\sqrt{14}\sqrt{13}}=\cos^{-1}{0}=\frac{\pi}{2}$ rad = $90^\circ$
