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ABCD is a regular tetrahedron P & Q are the mid -points of the edges AC and AB respectively, G is the cenroid of the face BCD and theta is the angle between the vectors vec(PG) and vec(DQ), then |
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Answer» The angle between `vec(AB)` and `vec(CD)` is `90^(@)` `vec(AD).vec(BC)=(-veca).(vecc-vecb)` `=-veca.vecc+veca.vecb` `=0` Hence `vec(AD)_|_^(r) vec(BC)impliesvec(AB)_|_vec(CD)` Now, `vec(PG)=-1/6(3veca-2vecb+vecc)` & `vec(DQ)=1/2(veca+vecb)` Let `|veca|=|vecb|=|vecc|=K` `implies|vec(PG)|=K/2` & `|vec(DQ)|=(sqrt(3))/2K` `impliesvec(PG).vec(DQ)=-1/6(3veca-2vecb+vecc).1/2(veca+vecb)` `costheta=-5/(6sqrt(3))impliestheta-pi-cos^(-1)(5/(6sqrt(3)))`
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