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` (b) `x + y + z

(a) Equation of given plane is `z = 2`.
Therefore, the direction ratios of the normal to the plane are `0,0` and 1
And `sqrt(0^(2)+0^(2)+1^(2)) = 1`.
`:.` Divide both sides of equation by,
`0.x+0.y+1.z=2`.
Which is in the form `lx+my+nz=d` where l, m, n are the direction cosines of the normal and d is the perpendicular distance from origin to the plane.
Therefore, the direction cosines of the normal to the plane are 0,0 and 1 and the distance of plane from origin is 2 units.
(b) Given equation is ,
`x+y+z=1"......."(1)`
The direction ratios of the normal to the plane are 1,1 and 1
and `sqrt(1^(2)+1^(2)+1^(2)) = sqrt(3)`.
Divide both side of equation (1) by `sqrt(3)`.
`1/(sqrt(3))x+(1)/(sqrt(3))y+(1)/(sqrt(3))z=(1)/(sqrt(3))`
Which is the form `lx + my + nz = d`.
Therefore, the direction cosines of the normal to the plane are `1/(sqrt(3)),(1)/(sqrt(3))` and `(1)/(sqrt(3))` and the distance of plane from oring is `1/(sqrt(3))` unit.
(c ) Given equation is : `2x+3y - z = 5"....."(1)`
Direction ratios of normal are `2,3` and `-1`.
and ` sqrt(2^(2)+3^(2)+(-1)^(2)) = sqrt(14)`
Divide both sides by `sqrt(14)`,
`(2/(sqrt(14)))x+((3)/(sqrt(14)))y+(-(1)/(sqrt14))z = 5/(sqrt(14))`
Which is in the form `lx+my+nz=d`.
Therefore, the direction cosines of the normal to the plane are `(2)/(sqrt(14)),(3)/(sqrt(14))` and `-(1)/(sqrt(14))` and the distance of plane from origin is `( 5)/(sqrt(14))` units.
(d) Given equation is : `5y+8 = 0`
It can be written as :
`0x+(-1)y+0z = 8/5`
Therefore, the direction cosines of the normal to the plane are `0,-1` and 0 and the distance of plane from origin is `8/5` units.