Find the torque of a force F = 7i + 3j+5k about the origin the force acts on a particle whose position vector is r= i+ j +k

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Solution

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$F o r c e (\vec{F}) = (7 i + 3 j + 5 k) a n d P o i n t (\vec{r}) = (i + j + k) . S o, t o r q u e (τ) = \vec{r} \times \vec{F} \Rightarrow τ = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 5 \\ 1 & 1 & 1 \end{matrix}| \Rightarrow τ = (3 - 5) \hat{i} - (7 + 5) \hat{j} + (7 - 3) \hat{k} = (- 2 \hat{i} - 12 \hat{j} + 4 \hat{k}) a n d t h e m a g n i t u d e o f t h e t o r q u e b e = \sqrt{2^{2} + 12^{2} + 4^{2}} = \sqrt{164} = 2 \sqrt{41}$

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