The shortest distance between the lines `vecr-veca+kvecb and vecr=veca+lvecc ` is (`vecb and vecc` are non collinear) (A) 0 (B) `|vecb.vecc|` (C) `(|vecbxxvecc|)/(|veca|)` (D) `(|vecb.vecc|)/(|veca|)`

If vecr.veca=vecr.vecb=vecr.vecc=0 " where "veca,vecb and vecc are non-coplanar, then

The distance of the point A(veca) from the line vecr=vecb+tvecc is (A) |(veca-vecb)xxvecc| (B) (|(veca-vecb)xxvecc|)/(|(veca-vecb)|) (C) (|(veca-vecb)xxvecc|)/(|vecc|) (D) (|vecaxx(vecb-vecc)|)/(|veca|)

If veca, vecb, vecc are any three non coplanar vectors, then (veca+vecb+vecc).(vecb+vecc)xx(vecc+veca)

If vecc=vecaxxvecb and vecb=veccxxveca then (A) veca.vecb=vecc^2 (B) vecc.veca.=vecb^2 (C) veca_|_vecb (D) veca||vecbxxvecc

Solve veca.vecr=x, vecb.vecr=y, vecc.vecr=z , where veca,vecb,vecc are given non coplanar vectors.

The two lines vecr=veca+veclamda(vecbxxvecc) and vecr=vecb+mu(veccxxveca) intersect at a point where veclamda and mu are scalars then (A) veca,vecb,vecc are non coplanar (B) |veca|=|vecb|=|vecc| (C) veca.vecc=vecb.vecc (D) lamda(vecb xxvecc)+mu(vecc xxveca)=vecc

If veca, vecb, vecc are vectors such that |vecb|=|vecc| then {(veca+vecb)xx(veca+vecc)}xx(vecbxxvecc).(vecb+vecc)=

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 vecr, veca, vecb and vecc be four non-zero vectors such that vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc| then [(veca, vecb, vecc)]=

If vecb and vecc are two non-collinear such that veca ||(vecbxxvecc) . Then prove that (vecaxxvecb).(vecaxxvecc) is equal to |veca|^(2)(vecb.vecc)