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`A , B , Ca n dD`are any four points in thespace, then prove that `| vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4`(area of ` A B C`.) |
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Answer» Let P.V of A,B, C d and D be `veca, vecb,vecc and vec0` , respectively. Then `vec(AB)xxvec(CD)=(vecb-veca)xx(-vecc), vec(BC)xxvec(AD)=(vecc-vecb)xx(-veca)` `vec(CA)xxvec(BD)=(veca-vecc)xx(-vecb)` `vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)` `veccxxvecb+vecaxxvecc+vecaxxvecc+vecbxxveca-vecaxxvecb+veccxxvecb` `2(veccxxvecb+ vecbxxvecbxxveca+vecaxxvecc)` `2(veccxx(vecb-veca)-vecaxx(vecb-veca))` `= 1(vec(AC)xxvec(AB))` `|vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)|=4|1/2(vec(AC)xxvec(AB))|=4triangleABC` |
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