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If `veca and vecb` are two vectors and angle between them is `theta` , thenA. `|vecaxxvecb|^(2)+ (veca.vecb)^(2)= |veca|^(2)|vecb|^(2)`B. `|vecaxxvecb|^(2)+ (veca.vecb)^(2), if theta= pi//4`C. `veca xx vecb = (veca. Vecb) hatn` ( where `hatn` is a normal unit vector ) `if theta f= pi//4`D. `(veca xx vecb ) . (veca + vecb) =0`

Answer» Correct Answer - a,b,c,d
`vecaxx vecb= |veca||vecb| sin theta hatn `
`or |vecaxx vecb|=|veca||vecb|sintheta`
`or sin theta (|vecaxxvecb|)/(|veca||vecb|)`
`veca. vecb= |veca||vecb|cos theta`
`Rightarrow cos theta= (|veca.vecb|)/(|veca||vecb|)`
From (i) and (ii) .
`sin^(2)theta + cos ^(2) theta=1`
`if theta= pi//4 "then" sintheta=costheta= 1//sqrt2.` Therefore,
`|vecaxxvecb|= (|veca||vecb|)/sqrt2 and veca.vecb= (|veca||vecb|)/sqrt2`
`|vecaxxvecb|= veca.vecb`
`vecaxxvecb= |veca||vecb|sinthetahatn = (|veca||vecb|)/sqrt2hatn`
`(veca.vecb)hatn`


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