1.

Vectors a and b are inclined at an angle theta=120^(@). If |a|=1, |b|=2, then [(a+3b)xx(3a+b)]^(2) is equal to

Answer»

190
275
300
192

Solution :Given ,`|a|=1,|b|=2`
`THEREFORE [(a+3b)xx(3a+b)]^2=[0+axxb+9bxxa+0]^2`
`=[-8axxb]^2=64[|a|^2|b|^2sin^2theta]`
`=64[1xx4xxsin^2 120^@]`
`=64xx4xx(3)/(4)=192`


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