If $\theta$ is the angle between the vector $2\mathbf{i}-2\mathbf{j}+4\mathbf{k}\text{ and }3\mathbf{i}+\mathbf{j}+2\mathbf{k}$ then sin$\theta$ is

Let $\mathbf{a}=2\mathbf{i}-2\mathbf{j}+4\mathbf{k},\mathbf{b}=3\mathbf{i}+\mathbf{j}+2\mathbf{k}$
$\begin{array}{l}\mathbf{a}\times \mathbf{b}=\left|\begin{array}{rrr}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& -2& 4\\ 3& 1& 2\end{array}\right|=\mathbf{i}(-4-4)-\mathbf{j}(4-12)+\mathbf{k}(2+6)=-8\mathbf{i}+8\mathbf{j}+8\mathbf{k}\\ |\mathbf{a}\times \mathbf{b}|=\sqrt{64+64+64}=8\sqrt{3},\left|\mathbf{a}\right|=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6},\left|\mathbf{b}\right|=\sqrt{9+1+4}=\sqrt{14}\\ \sin \theta =\frac{|\mathbf{a}\times \mathbf{b}|}{\left|\mathbf{a}\right|\left|\mathbf{b}\right|}=\frac{8\sqrt{3}}{\left(2\sqrt{6}\right)\left(\sqrt{14}\right)}=\frac{2}{\sqrt{7}}\end{array}$