1.

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, then

Answer»

<P>`vecp.vecc=0`
`vecp,vecq,VECR` can form a triangle
`/_\A(veca)B(vecb)C(vecc)` and `/_\P(vecp)Q(vecq)R(vecr)` are similar
`vecp,vecq,vecr` are collinear

Solution :`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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