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# Engineering

 Force \vec{F}=2\hat{i}+2\hat{j}+k is acting a point \vec{o}=i+3j+2k . The torque acting about point  \vec{r}=2\hat{i}+\hat{j}+2\hat{k} is 

Answers (1)
A Akshay

Torque is basically the cross product of the position vector r ( about a point ) and the force F acting at that particular point.

Here the position vector r,

\text { position at point }(1,3,2) \text { with respect to point }(2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})

\begin{array}{l} =(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})-(2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\ =-1 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\ \text { Force } \mathrm{F}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+1 \hat{\mathrm{k}} \end{array}

\begin{aligned} &\text { so, torque }(\mathrm{T})=\mathrm{r} \times \mathrm{f}\\ &=\left[\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ -1 & 2 & 0 \\ 2 & 2 & 1 \end{array}\right]\\ &=\mathrm{i}(2-0)-\mathrm{j}(-1-0)+\mathrm{k}(-2-4) \end{aligned}

\begin{aligned} &=2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\\ &\text { Hence, the answer is } 2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}} -6 \hat{\mathrm{k}} \end{aligned}

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