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Question: The torque of the force \(\vec { r } = ( 3 \hat { i } + 2 \hat { j } + 3 \hat { k } )\)m about the o...

The torque of the force r⃗=(3i^+2j^+3k^)\vec { r } = ( 3 \hat { i } + 2 \hat { j } + 3 \hat { k } )m about the origin be

A

6i^−6j^+12k^6 \hat { i } - 6 \hat { j } + 12 \hat { k }

B

17i^−6j^−13k^17 \hat { i } - 6 \hat { j } - 13 \hat { k }

C

−6i^+6j^−12k^- 6 \hat { i } + 6 \hat { j } - 12 \hat { k }

D

−17i^+6j^+13k^- 17 \hat { i } + 6 \hat { j } + 13 \hat { k }

Answer

17i^−6j^−13k^17 \hat { i } - 6 \hat { j } - 13 \hat { k }

Explanation

Solution

τ⃗=r⃗×F⃗\vec { \tau } = \vec { r } \times \vec { F }

=∣i^j^k^3232−34∣= \left| \begin{array} { r r r } \hat { i } & \hat { j } & \hat { k } \\ 3 & 2 & 3 \\ 2 & - 3 & 4 \end{array} \right| =[(2×4)−(3×−3)]i^+[(2×3)−(3×4)]j^+[(3×−3)−(2×2)]k^= [ ( 2 \times 4 ) - ( 3 \times - 3 ) ] \hat { i } + [ ( 2 \times 3 ) - ( 3 \times 4 ) ] \hat { j } + [ ( 3 \times - 3 ) - ( 2 \times 2 ) ] \hat { k } =17i^−6j^−13k^= 17 \hat { i } - 6 \hat { j } - 13 \hat { k }