Q. A unit vector perpendicular to the plane of $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} - 6\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}} = 4\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ is
A) $\frac{4\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}}{\sqrt{26}}$
B) $\frac{2\widehat{\dot{\mathbf{i}}} - 6\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}}{7}$
C) $\frac{3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}}{7}$
D) $\frac{2\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}}}{7}$
Explanation
(c)
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
2 & - 6 & - 3 \\
4 & 3 & - 1
\end{matrix} \right| = 15\widehat{\dot{\mathbf{i}}} - 10\widehat{\dot{\mathbf{j}}} + 30\widehat{\mathbf{k}}$
$and\ \left| \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right| = \sqrt{15^{2} + ( - 10)^{2} + (30)^{2}} = 35$
$\therefore Required\ vector = \frac{3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}}{7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.