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|)

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]

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|=

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

Let veca, vecb and vecc are three mutually perpendicular unit vectors and a unit vector vecr satisfying the equation (vecb -vecc)xx(vecr xx veca)+(vecc -veca)xx(vecr xx vecb)+(veca-vecb)xx(vecr xx vecc)=0, then vecr is

If vecb and vecc are any two orthogonal unit vectors and veca is any vector, then (veca.vecb)vecb + (veca.vecc)vecc + (veca.(vecb xx vecc))/|vecb xx vecc|^(2) (vecb xx vecc) =

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