Show that the plane through the points `veca,vecb,vecc` has the equation `[vecr vecb vecc]+[vecr vecc veca]+[vecr veca vecb]=[veca vecb vecc]`

If veca, vecb and vecc be any three non coplanar vectors. Then the system of vectors veca\',vecb\' and vecc\' which satisfies veca.veca\'=vecb.vecb\'=vecc.vecc\'=1 veca.vecb\'=veca.veca\'=vecb.veca\'=vecb.vecc\'=vecc.veca\'=vecc.vecb\'=0 is called the reciprocal system to the vectors veca,vecb, and vecc . The value of [veca\' vecb\' vecc\']^-1 is (A) 2[veca vecb vecc] (B) [veca,vecb,vecc] (C) 3[veca vecb vecc] (D) 0

If veca,vecb,vecc are unity vectors such that vecd=lamdaveca+muvecb+gammavecc then gamma is equal to (A) ([veca vecb vecc])/([vecb veca vecc]) (B) ([vecb vecc vecd])/([vecb vecc veca]) (C) ([vecb vecd vecc])/([veca vecb vecc]) (D) ([vecc vecb vecd])/([veca vecb vecc])

if veca + vecb + vecc=0 , then show that veca xx vecb = vecb xx vecc = vecc xx veca .

If veca,vecb and vecc are non coplnar and non zero vectors and vecr is any vector in space then [vecc vecr vecb]veca+pveca vecr vecc] vecb+[vecb vecr veca]c= (A) [veca vecb vecc] (B) [veca vecb vecc]vecr (C) vecr/([veca vecb vecc]) (D) vecr.(veca+vecb+vecc)

Let veda,vecb,vecc be three noncolanar vectors and vecp,vecq,vecr are vectors defined by the relations vecp= (vecbxxvecc)/([veca vecb vecc]), vecq= (veccxxvecca)/([veca vecb vecc]), vecr= (vecaxxvecb)/([veca vecb vecc]) then the value of the expression (veca+vecb).vecp+(vecb+vecc).vecq+(vecc+veca).vecr . is equal to (A) 0 (B) 1 (C) 2 (D) 3

veca, vecb,vecc are non-coplanar vectors and vecp,vecq,vecr are defined as vecp = (vecb xx vecc)/([vecb vecc veca]),q=(veca xx veca)/([vecc veca vecb]), vecr =(veca xx vecb)/([veca vecb vecc]) then (veca + vecb).vecp+(vecb+vecc).vecq + (vecc + veca).vecr is equal to.

Show that the perpendicular distance of any point veca from the line vecr=vecb+t vecc is (|(vecb-veca)xxvecc)|/(|vecc|)

Show that the perpendicular distance from a point A(veca) to the line vecr=vecb+tvecc is |vecb+((veca.vecb).vecc)/c^2 vecc-veca|