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 projection of point P(vecp) on the plane vecr.vecn=q is (vecs) , then

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

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

Find the equation of the plane through the line of intersection of the planes vecr.(2hati-3hatj+4hatk)=1 and vecr.(hati-hatj)+4=0 and perpendicular to the plane vecr.(2hati-hatj+hatk)+8=0 .

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

The equation of the plane contaiing the lines vecr=veca_(1)+lamda vecb and vecr=veca_(2)+muvecb is

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