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Prove that vecR+([vecR.(vecbetaxx(vecbetaxxvecalpha))]vecalpha)/(|vecalphaxxvecbeta|^(2))+([vecR.(vecalphaxx(vecalphaxxvecbeta))]vecbeta)/(|vecalphaxxvecbeta|^(2))=([vecRvecalphavecbeta](vecalphaxx vecbeta))/(|vecalphaxx vecbeta|^(2)) |
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Answer» Solution :`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))` |
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