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Physics Question on System of Particles & Rotational Motion

The torque of force F⃗=(2i^−3j^+4k^)\vec{F}=(2\hat{i}-3\hat{j}+4\hat{k}) newton acting at a point r⃗=(3i^+2j^+3k^)\vec{r}=(3\hat{i}+2\hat{j}+3\hat{k}) metre about origin is:

A

6i⃗−6j⃗+12k⃗N−m6\vec{i}-6\vec{j}+12\vec{k}N-m

B

−6i⃗+6j⃗−12k⃗N−m-6\vec{i}+6\vec{j}-12\vec{k}N-m

C

17i⃗−6j⃗−13k⃗N−m17\vec{i}-6\vec{j}-13\vec{k}N-m

D

−17i⃗+6j⃗+13k⃗N−m-17\vec{i}+6\vec{j}+13\vec{k}N-m

Answer

17i⃗−6j⃗−13k⃗N−m17\vec{i}-6\vec{j}-13\vec{k}N-m

Explanation

Solution

Here :F⃗=2i^−3j^+4kN: \vec{ F }=2 \hat{ i }-3 \hat{ j }+ 4 \mathbf { k } N Position vector of a point r⃗=3i^+2j^+3k^m\vec{ r }=3 \hat{ i }+2 \hat{ j }+3 \hat{ k } m The torque acting at a point about the origin is given by τ⃗=r⃗×F⃗=(3i^+2j^+3k^)×(2i^−3j^+4k^)\vec{\tau} =\vec{ r } \times \vec{ F }=( 3 \hat{ i }+2 \hat{ j }+3 \hat{ k }) \times(2 \hat{ i }-3 \hat{ j }+4 \hat{ k }) =∣i^j^k^ 323 2−34∣=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\\ 3 & 2 & 3 \\\ 2 & -3 & 4\end{vmatrix} =i^[8−(−9)]−j^(12−6)+k^(−9−4)=\hat{ i }[8-(-9)]-\hat{ j }(12-6)+\hat{ k }(-9-4) =17i^−6j^−13k^ N−m=17 \hat{ i }-6 \hat{ j }-13 \hat{ k } \,N - m