Question

A force $\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}$ acts at a point (2, - 3, 1). Then magnitude of torque of this force about point (0, 0, 2) will be

• A. 6
• B. $3\sqrt{5}$
• C. $6\sqrt{5}$
• D. None of these

Answer 1

The correct option is C. $6\sqrt{5}$.

Given,

Force $\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}$

Position vector $\overrightarrow{r}=(2\hat{i}-3\hat{j}+\hat{k})-\left(2\hat{k}\right)=2\hat{i}-3\hat{j}-\hat{k}$

We know that,

Torque $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$⟹\overrightarrow{\tau }=(2\hat{i}-3\hat{j}-\hat{k})\times (2\hat{i}+3\hat{j}-\hat{k})=6\hat{i}+12\hat{k}$

So, magnitude of torque $|\overrightarrow{\tau }|=\sqrt{6^2+{12}^2}=6\sqrt{5}N-m$.