 Good morning friends and poor one today We will discuss the following question in the following determine the direction cosines of the normal to the plane and the distance from the origin And we have z is equal to 2 Now the Cartesian equation of the plane in normal form is lx plus my Plus NZ is equal to D here. We have l, m and n are the direction cosines of the normal and D is the distance of plane from origin So this is the key idea behind our question Let us begin with the solution now Now we are given the equation of the plane as z is equal to 2 Now since the direction ratios of the normal to the plane are 001 Now as the equation of the plane is 0x plus 0y plus 1z is equal to 2 So we get the direction ratios of the normal to the plane as 001 That is coefficient of x coefficient of y and coefficient of z So since the direction ratios of the normal to the plane are 001 So we have the direction cosines as 0 upon under root of 0 square plus 0 square plus 1 square That is coefficient of x square plus coefficient of y square plus coefficient of z square 0 upon under root of 0 square plus 0 square plus 1 square And 1 upon under root of 0 square plus 0 square plus 1 square That is we have 00 and 1 upon under root of 1 square Or we can write this as 001 Now dividing the equation of the plane by under root of 1 square we get So dividing this equation of the plane z is equal to 2 by under root of 1 square that is 1 we get z is equal to 2 Now by key idea we know that the Cartesian equation of the plane in normal form is lx plus my plus nz is equal to d So comparing this above equation with the equation of the plane lx plus my plus nz is equal to d We see that here d is equal to 2 thus the distance of plane from origin is because by key idea we know that this d is the distance of the plane from the origin So we have got our answer as the direction cosines of the normal to the plane are 0 0 1 and the distance of the plane from origin is 2 Hope you have understood the solution Bye and take care