Show that the points whose position vectors are `veca,vecb,vecc,vecd` will be coplanar if `[veca vecb vecc]-[veca vecb vecd]+[veca vecc vecd]-[vecb vecc vecd]=0`

Prove that the perpendiculasr distanceof as point with position vector veca from the plane thorugh three points with position vectors vecb,vecc, vecd is ([veca vecc vecd]+[veca vecd vecb]+[veca vecb vecc]-[vecb vecc vecd])/(|vecbxxvecc+veccxxvecd+vecdxvecb|)

If the vectors veca, vecb, vecc, vecd are coplanar show that (vecaxxvecb)xx(veccxxvecd)=vec0

[ veca + vecb vecb + vecc vecc + veca ]=[ veca vecb vecc ] , then

Statement 1: veca, vecb and vecc arwe three mutually perpendicular unit vectors and vecd is a vector such that veca, vecb, vecc and vecd are non- coplanar. If [vecd vecb vecc] = [vecdvecavecb] = [vecdvecc veca] = 1, " then " vecd= veca+vecb+vecc Statement 2: [vecd vecb vecc] = [vecd veca vecb] = [vecdveccveca] Rightarrow vecd is equally inclined to veca, vecb and vecc .

for any four vectors veca,vecb, vecc and vecd prove that vecd. (vecaxx(vecbxx(veccxxvecd)))=(vecb.vecd)[veca vecc vecd]

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 and veca.vecb\'=veca.vecc\'=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

For any four vectors, prove that ( veca × vecb )×( vecc × vecd )=[ veca vecc vecd ] vecb −[ vecb vecc vecd ] veca

Prove that [lambda veca +mu vecb" "vecc" "vecd]=lambda [veca" "vecc" "vecd]+mu[vecb" "vecc" "vecd] .

If the vectors veca, vecb, and vecc are coplanar show that |(veca,vecb,vecc),(veca.veca, veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|=0

If veca, vecb, vecc are non-null non coplanar vectors, then [(veca-2vecb+vecc, vecb-2vecc+veca, vecc-2veca+vecb)]=