The moment of the force, →F=4ˆi+5ˆj−6ˆk at (2,0,–3) about the point (2,–2,–2) is given by
A
−8ˆi−4ˆj−7ˆk
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B
−4ˆi−ˆj−8ˆk
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C
−7ˆi−8ˆj−4ˆk
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D
−7ˆi−4ˆj−8ˆk
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Solution
The correct option is D−7ˆi−4ˆj−8ˆk →F=4ˆi+5ˆj−6ˆk→r=(2−2)ˆi+(0+2)ˆj+(−3+2)ˆk=2ˆj−1ˆk→r×→F=(2ˆj−1ˆk)×(4ˆi+5ˆj−6ˆk)=⎛⎜⎝ˆiˆjˆk02−145−6∣∣
∣
∣∣=−7ˆi−4ˆj−8ˆk