Q. A unit vector in the plane of $\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ and $\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$ and perpendicular to 2$\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ is
Explanation
(d)
Let $\overrightarrow{\mathbf{r}}$ be a unit vector such that
$\overrightarrow{\mathbf{r}} = x\left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) + y\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right)$
$= (x + y)\widehat{\dot{\mathbf{i}}} + (2x + y)\widehat{\dot{\mathbf{j}}} + (x + 2y)\widehat{\mathbf{k}}$
Since, $\overrightarrow{\mathbf{r}} \bullet \left( 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 0$
$\Longrightarrow 2x + 2y + 2x + y + x + 2y = 0$
$\Longrightarrow y = - x$
$\therefore\ \overrightarrow{\mathbf{r}} = x\widehat{\dot{\mathbf{i}}} - x\widehat{\mathbf{k}} \Longrightarrow \overrightarrow{\mathbf{r}} = \frac{\widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}}}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.