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A point of application of a force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is moved from ${\vec{r}}_{0} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque about a point $\vec{r} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is:

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a

42i^+30j^−6k^

b

Zero

c

42i^+30j^+6k^

d

42i^−30j^+6k^

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detailed solution

Correct option is A

Torque acting about a point P [ Position vector of point P is P→ (say)] due to force F→ is given byτ→=P→×F→P→=r→0−r→=(i^+2j^+3k^)−(3i^−2j^−3k^)=−2i^+4j^+6k^τ→=(−2i^+4j^+6k^)×(4i^−5j^+3k^)=i^(12+30)−j^(−6−24)+k^(10−16)=42i^+30j^−6k^

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Let $\vec{F}$ be the force acting on a particle having position vector $\vec{r}$ and $\vec{τ}$ be the torque of this force about the origin. Then

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