Three forces 2i+3j−6k , 2i+3j−4k and i−j+k are acting on a particle (0,1,2). the magnitude of the moment of forces about the point (1,−2,0) is
A
4√26
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B
5√26
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C
5√23
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D
None of the above
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Solution
The correct option is D None of the above Radius vector= Final point − Initial point =−i−3j+2k Moment of forces = (2i+3j−6k)×(i−3j−2k)+(2i+3j−4k)×(i−3j−2k)+(i−j+k)×(i−3j−2k) =(−24i−2j−9k−18i−9k+5i+3j−2k=−37i+j−20k Magnitude is √(37)2+1+(−20)2=√1770