Question

The resultant moment of three forces $\hat{ i }+2 \hat{ j }-3 \hat{ k }, 2 \hat{ i }+3 \hat{ j }+4 \hat{ k }$ and $-\hat{ i }-\hat{ j }+\hat{ k }$ acting on a particle at a point $P$ $(0,1,2)$ about the point $A (1,-2,0)$ is

• A. $6 \sqrt{2}$
• B. $\sqrt{140}$
• C. $\sqrt{21}$
• D. None

Answer 1

Answer: (B)

If $\vec{ F }$ be the resultant of the three given forces then
$\vec{ F }=(\hat{ i }+2 \hat{ j }-3 \hat{ k })+(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })+(-\hat{ i }-\hat{ j }+\hat{ k })=2 \hat{ i }+4 \hat{ j }+2 \hat{ k }$
If $O$ be the origin, then $\overrightarrow{ OP }= p . v$. of $P =\hat{ j }+2 \hat{ k }$
$\overrightarrow{ OA }= p . v .$ of $A =\hat{ i }-2 \hat{ j }$
$\therefore \overrightarrow{ AP }=\overrightarrow{ OP }-\overrightarrow{ OA }=(\hat{ j }+2 \hat{ k })-(\hat{ i }-2 \hat{ j })=-\hat{ i }+3 \hat{ j }+2 \hat{ k }$
$\therefore $ Vector moment of the given forces about
$A =$ vector moment of $\vec{ F }$ about $A =\overrightarrow{ AP } \times \overrightarrow{ F }$
$=(-\hat{ i }+3 \hat{ j }+2 \hat{ k }) \times(2 \hat{ i }+4 \hat{ j }+2 \hat{ k })=-2 \hat{ i }+6 \hat{ j }-10 \hat{ k }$
The magnitude of the moment $=\sqrt{4+36+100}=\sqrt{140}$