Question

A force f=2i+3k act on a particle at r=.5j-2k .The torqe

Answer 1

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Here is Your Answer....!!!
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⭐CROSS PRODUCT gives the Torque

⭐tau ( T) = F × r

⭐we get ....

Answer 2

Let, the torque of the given force acting on a point be T.

So, vectorT = vector(r)×vector( F)

we know that torque is a vector quantity and we take the cross product of F and r in the formula .

we have ,

F = (2i^ - 3j^ + 4k)F

r = (3i^ + 2j^ +3k^)m

Now , we take the cross product of F and r

F × r = (2i^ - 3j^ + 4k) × (3i^ +2j^ +3k^)

= i^( -3×3-2×4) -j^ (2×3-3×4) +k^(2×2-3×-3)

= i^(-9-8) -j^(6-12) +k^(4+9)

= -17i^ + 6j^ + 13k^

Hence,

T = ( -17i^ + 8j^ + 13k^)Nm