Q. If the vectors $\overrightarrow{\mathbf{c}},\overrightarrow{\mathbf{a}} = x\widehat{\dot{\mathbf{i}}} + y\widehat{\dot{\mathbf{j}}} + z\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{j}}}$ are such that $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{c}}\ and\ \overrightarrow{\mathbf{b}}$ from a right handed system, then $\overrightarrow{\mathbf{c}}$ is
A) $z\widehat{\dot{\mathbf{i}}} - x\widehat{\mathbf{k}}$
B) $\overrightarrow{\mathbf{0}}$
C) $y\widehat{\dot{\mathbf{j}}}$
D) $- z\widehat{\dot{\mathbf{i}}} - x\widehat{\mathbf{k}}$
Explanation
(a)
Since, $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ from a right handed system
$\therefore\ \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}$
$= \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
0 & 1 & 0 \\
x & y & z
\end{matrix} \right| = z\widehat{\dot{\mathbf{i}}} - x\widehat{\mathbf{k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.