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Prove that [vecaxxvecbvecbxxveccveccxxveca] = [vecavecbvecc]^2 |
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Answer» Solution :`[vecaxxvecbvecbxxveccveccxxveca] = (VECAXXVECB).[vecbxxvec)xx(veccxxveca)]` [using VECTOR triple product. =`(vecaxxvecb).{(vecbxxvecc)xxveca)vecc-(vecbxxvecc).vecc)veca}` = `(vecaxxvecb).{((vecbxxvecc).veca)vecc}[because (vecbxxvecc).vecc = 0]` =`{vecaxxvecb).vecc} {vecbxxvecc).veca}` =`{veca.(vecbxxvecc} {veca.(vecbxxvecc)}` [therefore In scalar triple product dot and cross can be interchanged and dot product is COMMUTATIVE.] =`{vecavecbvec][vecavecbvecc] = [vecavecbvecc]^2`(Proved). |
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