Q. The moment about the point $M( - 2,\ 4, - 6)$ of the force represented in magnitude and position $\overrightarrow{\mathbf{AB}}$ where the points $A$ and $B$ have the coordinates (1, 2, $- 3)$ and $(3, - 4,2)$ respectively, is

A) $8\widehat{\mathbf{i}} - 9\widehat{\mathbf{j}} - 14\widehat{\mathbf{k}}$
B) $2\widehat{\mathbf{i}} - 6\widehat{\mathbf{j}} + 5\widehat{\mathbf{k}}$
C) $- 3\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} - 3\widehat{\mathbf{k}}$
D) $- 5\widehat{\mathbf{i}} - 8\widehat{\mathbf{j}} - 8\widehat{\mathbf{k}}$
Explanation

(a)
Force $\overrightarrow{\mathbf{F}}\mathbf{=}\overrightarrow{\mathbf{AB}} = (3 - 1)\widehat{\mathbf{i}} + ( - 4 - 2)\widehat{\mathbf{j}} + (2 + 3)\widehat{\mathbf{k}}$
$= 2\widehat{\mathbf{i}} - 6\widehat{\mathbf{j}} + 5\widehat{\mathbf{k}}$
Moment of force $\overrightarrow{\mathbf{F}}$ with respect to $M = \overrightarrow{\mathbf{MA}}\mathbf{\times}\overrightarrow{\mathbf{F}}$
$\because\overrightarrow{\mathbf{MA}} = (1 + 2)\widehat{\mathbf{i}} + (2 - 4)\widehat{\mathbf{j}} + ( - 3 + 6)\widehat{\mathbf{k}} = 3\widehat{\mathbf{i}} - 2\widehat{\mathbf{j}} + 3\widehat{\mathbf{k}}$
Now, $\overrightarrow{\mathbf{MA}}\mathbf{\times}\overrightarrow{\mathbf{F}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
3 & - 2 & 3 \\
2 & - 6 & 5
\end{matrix} \right|$
$= \widehat{\mathbf{i}}( - 10 + 18) + \widehat{\mathbf{j}}(6 - 15) + \widehat{\mathbf{k}}( - 18 + 4)$
$= 8\widehat{\mathbf{i}} - 9\widehat{\mathbf{j}} - 14\widehat{\mathbf{k}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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

Want more practice?
Practice Vector algebra MCQs