`vecp, vecq, vec` be vectors such that `vecq.vecr=0` and `vecp.vecq!=0`. Let `alpha` is real constant such that `vecx.vecp=alpha, vecx xx vecq=vecr`, then `vecx-lamda_(1)vecq=lamda_(2)(vecp xx vecr)` where

Options

A. `lamda_(1)=(alpha)/(vecp.vecq)`
B. `lamda_(2)=1/(vecp.vecq)`
C. `lamda_(2)=1/(vecr.vecq)`
D. `lamda_(1)=(alpha)/(vecr.vecq)`

Answer

A, B

Explanation

`vecpxx(vecx xx vecq)=(vecp.vecq)-(vecp.vecx)vecq=vecpxxvecr`
`vecx=((vecp.vecx)vecq)/(vecp.vecq)+((vecpxxvecr))/(vecp.vecq)=(alpha)/(vecp.vecq) vecq+((vecp+vecr))/(vecp.vecq)`