Q. Let $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}},\ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ and a unit vector $\overrightarrow{\mathbf{c}}$ be coplanar. If $\overrightarrow{\mathbf{c}}$ is perpendicular to $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{c}}$ is equal to
Explanation
(a)
Required unit vector
$\ \overrightarrow{\mathbf{c}} = \frac{\overrightarrow{\mathbf{a}} \times \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right)}{\left| \overrightarrow{\mathbf{a}} \times \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \right|}\$
Now,
$\overrightarrow{\mathbf{a}} \times \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) = \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right)\overrightarrow{\mathbf{a}} - \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{a}} \right)\overrightarrow{\mathbf{b}}$
$= 3\left( 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) - 6(\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}})$
$= - 9\widehat{\dot{\mathbf{j}}} + 9\widehat{\mathbf{k}}$
$\therefore\ \overrightarrow{\mathbf{c}} = \frac{- 9\widehat{\dot{\mathbf{j}}} + 9\widehat{\mathbf{k}}}{\sqrt{9^{2} + 9^{2}}} = \pm \frac{1}{\sqrt{2}}( - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.