In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

   (a)     z = 2 

Answers (1)

Equation of plane Z=2, i.e.  0x+0y+z=2

The direction ratio of normal is 0,0,1

\therefore \, \, \, \sqrt{0^2+0^2+1^2}=1

Divide equation 0x+0y+z=2 by 1 from both side 

We get,      0x+0y+z=2

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

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