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|)`

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|

If veca xx (vecbxx vecc)= (veca xx vecb)xxvecc then

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

If veca,vecb,vecc are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is (A) veca+vecb+vecc (B) veca/|veca|+vecb/|vecb|+vec/|vecc| (C) veca/|veca|^2+vecb/|vecb|^2+vecc/|vecc|^2 (D) |veca|veca-|vecb|vecb+|vecc|vecc

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 |veca|=|vecb|=|vecc|=1 and |veca-vecb|^(2)+|vecb-vecc|^(2)+|vecc-veca|^(2)=9 then |2veca+5vecb+5vecc|=?

If [ veca vecbvecc]=2 , then find the value of [(veca+2vecb-vecc) (veca - vecb) (veca - vecb-vecc)]

If [ veca vecbvecc]=2 , then find the value of [(veca+2vecb-vecc) (veca - vecb) (veca - vecb-vecc)]

If veca, vecb, vecc are any three vectors such that (veca+vecb).vecc=(veca-vecb)=vecc=0 then (vecaxxvecb)xxvecc is