A force F = (4i - 5j +3k )N is acting at a point having a position vector r1=(i+2j+3k) .The torgue acting about a point having a position vector r2 = (3i-2j-3k)is

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Solution

Dear Student ,
$t o r q u e = f o r c e \times d i s p l a c e m e n t f r o m t h e a x i s o f r o t a t i o n s o t o r q u e = (4 \vec{i} - 5 \vec{j} + 3 \hat{k}) \times ((\vec{i} + 2 \vec{j} + 3 \hat{k}) - (3 \vec{i} - 2 \vec{j} - 3 \hat{k})) = (4 \vec{i} - 5 \vec{j} + 3 \hat{k}) \times (- 2 \vec{i} + 4 \vec{j} + 6 \hat{k}) = ((- 30 - 12) \vec{i} - (24 - 6) \vec{j} + (16 + 10) \hat{k}) = (- 42 \vec{i} + 30 \vec{j} + 26 \hat{k}) N m$
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