The unit vector normal to the plane `x+2y+3z-6=0` is `(1)/(sqrt14)hati+(2)/(sqrt14)hatj+(3)/(sqrt14)hatk`

State true or flase for the following statements : A unit normal vector to the plane x + 2y + 3 z - 6 = 0 is (1)/(sqrt(14)) hat(i) + (2)/(sqrt(14)) hat(j) + (3)/(sqrt(14)) hat(k) .

Find the normal unit vector to the plane x+2y+3z-6=0.

Find a unit vector normal to the plane : vecr*(2hati-3hatj+6hatk)+14=0 .

Magnitude of the vectors 2/(sqrt(3))hati+2/(sqrt(3))hatj+2/(sqrt(3))hatk is equal to

Magnitude of the vectors 1/(sqrt(3))hati+1/(sqrt(3))hatj+1/(sqrt(3))hatk is equal to

The distance of the plane 2x+3y-6z+14=0 from point (3, 0, 1) is :

Let vecb=-veci+4vecj+6veck, vecc=2veci-7vecj-10veck. If veca be a unit vector and the scalar triple product [veca vecb vecc] has the greatest value then veca is A. (1)/(sqrt(3))(hati+hatj+hatk) B. (1)/(sqrt(5))(sqrt(2)hati-hatj-sqrt(2)hatk) C. (1)/(3)(2hati+2hatj-hatk) D. (1)/(sqrt(59))(3hati-7hatj-hatk)

The distance between the planes 3x+2y-6z=14=0 and 3x+2y-6z+21=0 is

The distance of the plane 2x-3y-6z+14=0 from point (3, 2, 0) is :