The distance between the line `x=2+t,y=1+t,z=-1/2-t/2` and the plane `vecr.(hati+2hatj+6hatk)=10`, is

Find the angle between the line vecr.(3hati+hatk)+lambda(hatj+hatk) and the plane vecr.(2hati-hatj+2hatk)=1 .

Find the distance between the planes 2x - 3y + 6z+8= 0 and vecr.(2hati-3hatj+6hatk)=-4

The distance from the line x=2+t,y=1+t,z= -(1)/(2)-(1)/(2)t to the plane x+2y+6z=10 is (lambda)/(sqrt(mu)) . Then 5lambda-mu=

The square of the shortest distance between the lines : vecr=s(hati-hatj-hatk) and vecr=3hatj+t(hati-hatk) is :

Find the distance between the planes vecr*(hati-2hatj-2hatk)=7 and vecr*(5hati-10hatj-10hatk)=15.

The angle between the line vecr=(hati+2hatj-3hatk)+t(2hati+hatj-2hatk) and the plane vecr*(hati+hatj)+4=0 is :

Find the distance between the parallel planes vecr.(2hati-hatj-2hatk)=6 and vecr.(6hati-3hatj-6hatk)=27