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Q.

A force F→=4i^+3j^+4k^ is applied on intersection point of x=2 plane and x-axis. The magnitude of torque of this force about a point (2, 3, 4) is________ (Round off to the Nearest Integer)

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answer is 20.

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Detailed Solution

τ→=r→×F→ As, r→=(2i^+3j^+4k^)−2i^=3j^+4k^⇒τ→=i^j^k^034434 ⇒τ→=i^(12−12)−j^(0−16)+k^(0−12)⇒τ→=16j^−12k^ ⇒|τ→|=(16)2+(−12)2=256+144=20

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