Find the sine of the angle between the vectors $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}$ and $\overrightarrow {\rm{b}} = 2\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$.

As we know, if $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ are in their component form, then $\cos \theta$

$= \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{\sqrt {a_1^2 + a_2^2 + a_3^2} \sqrt {b_1^2 + b_2^2 + b_3^2} }}$.

After getting $\cos \theta$, we shall find the sine of the angle.

Here, ${a_1} = 3,{a_2} = 1,{a_3} = 2$ and ${b_1} = 2,{b_2} = - 2,{b_3} = 4$

As we know. $\cos \theta = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{\sqrt {a_1^2 + a_2^2 + a_3^2} \sqrt {b_1^2 + b_2^2 + b_3^2} }}$

$= \frac{{3 \times 2 + 1 \times ( - 2) + 2 \times 4}}{{\sqrt {{3^2} + {1^2} + {2^2}} \sqrt {{2^2} + {{( - 2)}^2} + {4^2}} }}$

$= \frac{{6 - 2 + 8}}{{\sqrt {14} \sqrt {24} }} = \frac{{12}}{{2\sqrt {14} \sqrt 6 }} = \frac{6}{{\sqrt {84} }} = \frac{6}{{2\sqrt {21} }} = \frac{3}{{\sqrt {21} }}$

$\therefore$ $\sin \theta = \sqrt {1 - {{\cos }^2}\theta }$

$= \sqrt {1 - \frac{9}{{21}}} = \sqrt {\frac{{12}}{{21}}} = \frac{{2\sqrt 3 }}{{\sqrt 3 \sqrt 7 }} = \frac{2}{{\sqrt 7 }}$