if `veca` and `vecb` non-zero ad non-collinear vectors, then

Options

A. `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`

Answer

A, B, C, D

Explanation

(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)