Q. A vector of magnitude 12 units perpendicular to the plane containing the vectors $4\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ and $3\widehat{\dot{\mathbf{i}}} + 8\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ is

A) $- 8\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}}$
B) $8\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}}$
C) $8\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}}$
D) $8\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} - 8\widehat{\mathbf{k}}$
Explanation

(c)
Let $\overrightarrow{\mathbf{a}} = 4\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
$and\ \overrightarrow{\mathbf{b}} = 3\widehat{\dot{\mathbf{i}}} + 8\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
$\therefore\ \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
4 & 6 & - 1 \\
3 & 8 & 1
\end{matrix} \right| = 14\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 14\widehat{\mathbf{k}}$
$\Longrightarrow \widehat{\mathbf{c}} = \frac{14\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 14\widehat{\mathbf{k}}}{\sqrt{14^{2} + 7^{2} + 14^{2}}} = \frac{14\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 14\widehat{\mathbf{k}}}{21}$
∴Required vector
$= 12 \bullet \frac{(14\widehat{\dot{\mathbf{i}}} - 7\widehat{\dot{\mathbf{j}}} + 14\widehat{\mathbf{k}})}{21} = 8\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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