Find the angle between the vectors $\mathbf{v}=2\mathbf{i}-\mathbf{j}+ \mathbf{k}$ and $\mathbf{w}=- \mathbf{i}+\mathbf{j}$.

Solution We find $\mathbf{v}\,{\cdot}\, \mathbf{w}$, $\Vert \mathbf{v} \Vert ,$ and $\Vert \mathbf{w}\Vert.$ $$\begin{eqnarray*} \mathbf{v}\,{\cdot}\, \mathbf{w} &=&(2) (-1) +(-1) (1) +(1) (0) =-2-1+0=-3 \\[4pt] \Vert \mathbf{v}\Vert &=&\sqrt{2^{2}+(-1) ^{2}+1^{2}}=\sqrt{6} \\[4pt] \Vert \mathbf{w}\Vert &=&\sqrt{(-1) ^{2}+1^{2}}=\sqrt{2} \end{eqnarray*}$$

Figure 36

Then if $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$, $$\begin{equation*} \cos \theta =\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert }=\dfrac{-3}{( \sqrt{6}) (\sqrt{2}) }= \dfrac{-3}{\sqrt{12}}=-\dfrac{\sqrt{3}}{2} \end{equation*}$$

Since $0\leq \theta \leq \pi$, the angle $\theta$ between $\mathbf{v}$ and $\mathbf{w}$ is $\dfrac{5\pi}{6}$ radians. See Figure 36.