Find a unit vector perpendicular to each of the vectors ( a + b )… - Vidyadip

Question

Find a unit vector perpendicular to each of the vectors $\left( {\vec a + \vec b} \right)$ and $\left( {\vec a - \vec b} \right)$, where $\vec a = \hat i + \hat j + \hat k, \ \vec b = \hat i + 2\hat j + 3\hat k$.

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Answer

$\vec a+\vec b=(\vec i+\vec j+\vec k)+(\vec i+2\vec j+3\vec k)=2\vec i+3\vec j+4\vec k$
$\vec a-\vec b=(\vec i+\vec j+\vec k)-(\vec i+2\vec j+3 \vec k=-\vec j-2\vec k$
A vector which is perpendicular to both $(\vec a + \vec b)$ are $(\vec a - \vec b)$ is given by
$(\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&4 \\ 0&{ - 1}&{ - 2} \end{array}} \right|$
$ = - 2\hat i + 4\hat j - 2\hat k$
Let $\vec c = - 2\hat i + 4\hat j - 2\hat k$
$\left| {\vec c} \right| = \sqrt {4 + 16 + 4} $
$ = \sqrt {24} $
$ = 2\sqrt 6 $
Required unit vector is
$\frac{{\vec c}}{{\left| {\vec c} \right|}} = - \frac{1}{{\sqrt 6 }}\hat i + \frac{2}{{\sqrt 6 }}\hat j - \frac{1}{{\sqrt 6 }}\hat k$

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