Find the projection of the line `vecr=veca+tvecb` on the plane given by `vecr.vecn=q`.

The equation of the plane containing the line vecr= veca + k vecb and perpendicular to the plane vecr . vecn =q is :

The angle theta the line vecr=vecr+lamdavecb and the plane vecr.hatn=d is given by (A) sin^-1((vecb.hatn)/(|vecb|)) (B) cos^-1((vecb.hatn)/(|vecb|)) (C) sin^-1((veca.hatn)/(|veca|)) (D) cos^-1((veca.hatn)/(|veca|))

Find the radius of the circular section of the sphere |vecr|=5 by the plane vecr*(veci+vecj-veck)=4sqrt(3) .

Let |A|=1, |vecb|=4 and veca xxvecr +vecb=vecr . If the projection of vecr along veca is 2, then the projection of vecr along vecb is

The equation of the plane which contains the origin and the line of intersectio of the plane vecr.veca=d_(1) and vecr.vecb=d_(2) is

Distance of the point P(vecc) from the line vecr=veca+lamdavecb is

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

The line vecr= veca + lambda vecb will not meet the plane vecr cdot n =q, if