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if `veca` and `vecb` non-zero ad non-collinear vectors, thenA. `vecaxxvecb=[vecavecbhati]hati+[vecavecbhatj]hatj+[vecavecbhatk]hatk`B. veca.vecb=(veca.hati)(vecb.hati)+(veca.hatj)(vecb.hatj)+(veca.hatk)(vecb.hatk)`C. if `vecu=hata-(hata.hatb)hatb` and `vecv=hataxxhatb` then `|vecv|=|vecu|`D. if `vecc=vecaxx(vecaxxvecb)`, then `vecc.veca=0` |
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Answer» Correct Answer - A::B::C::D (A). `vecaxxvecb=a_(1)hati+a-(2)hatj+a_(3)hatk` Take dot with `hati` `[overline(a)overline(b)hati]=a_(1)` similarly `a_(2)=[overline(a)overline(b)hatj]` `a_(2)=[overline(a)overline(b)hatk]` (B) take `overline(a)=a_(1)hati+a_(2)hatj+a-(3)hatk` `overline(b)=b_(1)hati+b_(2)hatj+b_(3)hatk` (C) `|overline(u)|=sqrt(|hata|^(2)+(hata.hatb)^(2)|hat(b)|^(2)-2(hata.hatb)(hat(a).hat(b)))` `|overline(u)|=sqrt(1+(hata.hatb)^(2)-2(hata.hatb)^(2))=|sintheta|` `vecv=|sintheta|` (D). `vecc.veca=0` (obvious) |
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