Find a vector of magnitude $6$, which is perpendicular to both the vectors $2\widehat {\rm{i}} - \widehat {\rm{j}} + 2\widehat {\rm{k}}$ and $4\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}}$.

As we know that any vector perpendicular to both the vectors $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$

is given by$\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}}$

$= \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$

and then we will find the vector with magnitude 6.

Let $\overrightarrow {\rm{a}} = 2\widehat {\rm{i}} - \widehat {\rm{j}} + 2\widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = 4\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}}$

So, any vector perpendicular to both the vectors $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$

is given by
$\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\2&{ - 1}&2\\4&{ - 1}&3\end{array}} \right|$

$= \widehat {\rm{i}}( - 3 + 2) - \widehat {\rm{j}}(6 - 8) + \widehat {\rm{k}}( - 2 + 4)$

$= - \widehat {\rm{i}} + 2\widehat {\rm{j}} + 2\widehat {\rm{k}} = \overrightarrow {\rm{r}}$

A vector of magnitude 6 in the direction of $\overrightarrow {\rm{r}}$

$= \frac{{\overrightarrow {\rm{r}} }}{{|\overrightarrow {\rm{r}} |}} \cdot 6 = \frac{{ - \widehat {\rm{i}} + 2\widehat {\rm{j}} + 2\widehat {\rm{k}}}}{{\sqrt {{1^2} + {2^2} + {2^2}} }} \cdot 6$

$= \frac{{ - 6}}{3}\widehat {\rm{i}} + \frac{{12}}{3}\widehat {\rm{j}} + \frac{{12}}{3}\widehat {\rm{k}}$

$= - 2\widehat {\rm{i}} + 4\widehat {\rm{j}} + 4\widehat {\rm{k}}$