If A=(1,1,1) and C=(0,1,−1) are given vectors and B is a vector satisfying A×B=C and A⋅B=3, then 9|B|2 is equal to
A
33
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B
22
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C
44
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D
55
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Solution
The correct option is D 33 Let B=(x,y,z), so that x+y+z=3, Also j−k=C=∣∣
∣∣ijk111xyz∣∣
∣∣ =(z−y)i+(x−z)j+(y−x)k Hence z=y,x−z=1 and x−y=1. That is, z=y and x=1+y, Therefore, 1+y+y+y=3, Thus y=2/3,z=2/3 and x=1+2/3=5/3 So 9|B|2=25+4+4=33