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

Options

A. `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`

Answer

A, B, C, D

Explanation

(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)