Prove that ` vec R+([ vec Rdot( vecbetaxx( vecbetaxx vecalpha))] vecalpha)/(| vecalphaxx vecbeta|^2)+([ vec Rdot( vecalphaxx( vecalphaxx vecbeta))] vecbeta)/(| vecalphaxx vecbeta|^2)=([ vec R vecalpha vecbeta]( vecalphaxx vecbeta))/(| vecalphaxx vecbeta|^2)`

Explanation

`vecalpha , vecbeta and vecalpha xx vecbeta` are three non-coplanar vectors. Any vector `vecR` can be respresented as a linear combination of these vectors. Thus ,
`vecR=k_(1)vecalpha+k_(2)vecbeta+k_(3)(vecalphaxxvecbeta)`
Take dot product of (i) with `(vecalpha xx vec beta)` . we have
`vecR.(vecalphaxxvecbeta)=k_(3)(vecalpha xxvecbeta)=k_(3)|vecalphaxxvecbeta|^(2)`
`k_(3)=(vecR.(vecalphaxxvecbeta))/(|vecalphaxxvecbeta|^(2))=([vecRvecalphavecbeta])/(|vecalphaxxvecbeta|^(2))`
Take dot product of (i) with `vecalphaxx(vecalphaxxvecbeta)` we have
`vecR.(vecalphaxx(vecalphaxxvecbeta))=k_(2)(vecalphaxx(vecalphaxxvecbeta)).vecbeta`
`= k_(2)[(vecalpha.vecbeta)vecalpha-(vecalpha.vecalpha)vecbeta].vecbeta=k_(2)[(vecalpha.vecbeta)^(2)-|vecalpha|^(2)|vecbeta|^(2)]`
`=-k_(2)|vecalphaxxvecbeta|^(2)`
`k_(2)=(-[vecR.(vecalphaxx(vecalphaxxvecbeta))])/(|vecalphaxxvecbeta|^(2)) " simiarly "k_(1)=-([vecR.(vecbetaxx(vecbetaxxvecalpha))])/(|vecalphaxx vecbeta|^(2))`
`Rightarrow vecR=(-[vecR.[vecbetaxx(vecbetaxxvecalpha))]vecalpha)/(|vecalphaxxvecbeta|^(2))-([vecR.(vecalphaxx(vecalphaxxvecbeta))]vecbeta)/(|vecalphaxxvecbeta|^(2))+(([vecR.(vecalphaxxvecbeta))](vecalphaxxvecbeta))/(|vecalpha xx vecbeta|^(2))`
`Rightarrow vecR=(-[vecR.[vecbetaxx(vecbetaxxvecalpha))]vecalpha)/(|vecalphaxxvecbeta|^(2))-([vecR.(vecalphaxx(vecalphaxxvecbeta))]vecbeta)/(|vecalphaxxvecbeta|^(2))+(([vecR.(vecalphaxxvecbeta))](vecalphaxxvecbeta))/(|vecalpha xx vecbeta|^(2))`