In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin.
$Z=2$

(A) The equation of the plane is given by $z = 2$,which can be written as $0x + 0y + 1z = 2$ ..........$(1)$
The direction ratios of the normal vector to the plane are $(a, b, c) = (0, 0, 1)$.
To find the direction cosines $(l, m, n)$,we divide the direction ratios by the magnitude of the normal vector $\sqrt{a^2 + b^2 + c^2} = \sqrt{0^2 + 0^2 + 1^2} = 1$.
Dividing equation $(1)$ by $1$,we get $0x + 0y + 1z = 2$.
This is in the normal form $lx + my + nz = d$,where $(l, m, n)$ are the direction cosines and $d$ is the distance from the origin.
Comparing the equations,the direction cosines are $(0, 0, 1)$ and the distance from the origin is $2$ units.