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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 |
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Answer» 190 `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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