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

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

If veca, vecb, vecc, vecd are coplanar vectors, then (vecaxxvecb)xx(veccxxvecd)=

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]

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 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 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 vector veca,vecb,vecc are coplanar show that |(veca,vecb,vecc),(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|

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