The force acting on a particle is \( \vec{F} = \hat{i} + 2\hat{j} + 3\hat{k} \text{ N} \). Find the torque (in N m) of this force about origin if position vector of the particle is \( \vec{r} = 7\hat{i} + 3\hat{j} + 5\hat{k} \text{ m} \).
\( \hat{i} + 16\hat{j} - 11\hat{k} \)
\( -\hat{i} - 16\hat{j} + 11\hat{k} \)
\( \hat{i} + 16\hat{j} + 11\hat{k} \)
\( -\hat{i} + 9\hat{j} + 11\hat{k} \)
Solution:
Torque \( \vec{\tau} \) is given by \( \vec{r} \times \vec{F} \). Evaluating the cross product: \( \vec{\tau} = \det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 3 & 5 \ 1 & 2 & 3 \end{pmatrix} = -\hat{i} - 16\hat{j} + 11\hat{k} \text{ N m} \).
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