 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says in the following case Determine the direction cosines of the normal to the plane and the distance from the origin X plus 5 plus z is equal to what? Now we know that the Cartesian equation of the plane in normal form is Lx plus MY plus nz is equal to D where Lnn are the direction cosines of the normal to the plane and these are distance of the plane from the origin So this is the key idea behind our question We will take the help of this key idea to solve the above question So let's start the solution Now given equation of plane is x plus y plus z is equal to 1 since the direction ratios of the normal to the plane are 1 1 1 therefore the direction cosines of it are 1 over under root of 1 square plus 1 square plus 1 square 1 over under root of 1 square plus 1 square plus 1 square 1 over under root of 1 square plus 1 square plus 1 square or 1 over root 3 1 over root 3 1 over root 3 So, these are the direction cosines of the normal to the plane, hence on dividing the equation x plus y plus z is equal to 1 throughout by root 3 we get x over root 3 plus y over root 3 plus z over root 3 is equal to 1 over root 3. So, this is of the form lx plus ny plus nz is equal to d where d is the distance of the plane from the origin. So, here we see that d is equal to 1 over root 3. So, the distance of the plane from the origin is 1 over root 3. Hence, the direction cosines of the normal to the plane are 1 over root 3, 1 over root 3, 1 over root 3 and the distance of the plane from the origin is 1 over root 3. So, this is the answer for the upper question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.