Show that the perpendicular distance of a point `A(veca)` from the line `vecr = vecb + tvecc` is `(|(vecb-veca)xxvecc|)/(|vecc|)`

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|

The perpendicular distance of the point vecc from the joining veca and vecb is

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|)

A line l is passing through the point vecb and is parallel to vector vecc . Determine the distance of point A( veca) from the line l in from |vecb-veca+((veca-vecb)vecc)/(|vecc|^(2))vecc|or (|(vecb-veca)xxvecc|)/(|vecc|)

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

The equation of the line throgh the point veca parallel to the plane vecr.vecn=q and perpendicular to the line vecr=vecb+tvecc is (A) vecr=veca+lamda (vecnxxvecc) (B) (vecr-veca)xx(vecnxxvecc)=0 (C) vecr=vecb+lamda(vecnxxvecc) (D) none of these

If veca, vecb,vecc are unit vectors such that veca is perpendicular to the plane of vecb, vecc and the angle between vecb,vecc is pi/3 , then |veca+vecb+vecc|=

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]

Let vecn be a unit vector perpendicular to the plane containing the point whose position vectors are veca, vecb and vecc and, if abs([(vecr -veca)(vecb-veca)vecn])=lambda abs(vecrxx(vecb-veca)-vecaxxvecb) then lambda is equal to