Question

A constant force $F=3i+2j–4k$ is applied at the point $A(1,–2,2)$. Find the vector moment of $F$ about the point $B(2,–1,3)$.

• A. $\left(6i–7j+k\right)$
• B. $\left(-2i+6j-10k\right)$
• C. $\left(-4i-2j-5k\right)$
• D. $Noneoftheabove$

Answer 1

The correct option is A

$\left(6i–7j+k\right)$

Step 1. Given data:

$A(1,-2,2)$ and$B(2,-1,3)$

Force, $\overrightarrow{F}=\left(3i+2j–4k\right)$

Step 2. Finding resultant of point $\mathit{A}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{B}$.

Resultant, $\overrightarrow{R}=\overrightarrow{BA}$

$\overrightarrow{R}=\overrightarrow{OA}-\overrightarrow{OB}$

$\overrightarrow{R}=\left(i-2j+2k\right)-\left(2i-j+3k\right)$

$\overrightarrow{R}=\left(-i-j-k\right)$

Step 3. Calculating vector moment of $\mathit{F}$ about the point $\mathit{B}$,

$\overrightarrow{\tau }=\overrightarrow{R}x\overrightarrow{F}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -1& -1\\ 3& 2& -4\end{array}\right|$

$=\left(6i-7j+k\right)$

Hence, correct option is (A).