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Consider vectors `veca,vecb,vecc,vecp=(vecb.vecc)veca-(vecc.veca)vecb,vecq=(veca.vecc)vecb-(veca.vecb)vecc,vecr=(vecb.veca)vecc-(vecb.vecc)veca`, thenA. `vecp.vecc=0`B. `vecp,vecq,vecr` can form a triangleC. `/_A(veca)B(vecb)C(vecc)` and `/_P(vecp)Q(vecq)R(vecr)` are similarD. `vecp,vecq,vecr` are collinear |
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Answer» Correct Answer - A::B::C `vecp+vecq+vecr=0` So, `vecp, vecq, vecr` can form a triangle `vecp.ecc=(vecb.vecc)(veca.vecc)-(vecc.veca)(vecb.vecc)=0` `:.vecp_|_vecc` `impliesvecq--veca` and `vecr_|_vecb` |
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