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if `veca` and `vecb` non-zero ad non-collinear vectors, thenA. `veca xx vecb=[vecavecbhati]hati+[vecavecbhatj]hatj+[vecavecbhatk]hatk`B. `vecavecb=(veca.hati)+(vecahatj)(vecb.hatj)+(veca.hatk)(vecb.hatk)`C. if `vecu=hata-(hata.hatb)hatb` and `vecv=hata xx hatb`, then `|vecv|=|vecu|`D. if `vecc=vecaxx(vecaxxvecb),` then `vecc.veca=0` |
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Answer» Correct Answer - A::B::C::D (A) `bar(a)xxbar(b)=a_(1)hat(i)+a_(2)hat(j)+a_(3)hat(k)` Take dot with `hat(i)` `[bar(a)bar(b)hat(i)]=a_(1)` Similarly `a_(2)=[bar(a)bar(b)hat(j)]` `a_(3)=[bar(a)bar(b)hat(k)]` (B) take `bar(a)=a_(1)hat(i)+a_(2)hat(j)+a_(3)hat(k)` `bar(b)=b_(1)hat(i)+b_(2)hat(j)+b_(3)hat(k)` (C) `|bar(mu)|=sqrt(|hat(a)|^(2)+(hat(a).hat(b))^(2)|hat(b)|^(2)-2(hat(a).hat(b))(hat(a).hat(b))` `|bar(mu)|= sqrt(1+(hat(a).hat(b))^(2)-2(hat(a).hat(b))^(2))=|sintheta|` `barv=|sintheta|` (D) `bar(c).bar(a)=0` (obvious) |
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