 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 5y plus 8 is equal to 0. So let's start the solution. Now the given equation of the plane is 5y plus 8 is equal to 0 or this can be written as 5y is equal to minus 8. Since the direction ratios of the normal to the plane are 0 5 0 therefore the direction cosines are 0 over root 5 square which is 0 5 over root 5 square 0 over root 5 square which is again 0 or 0 1 0. On dividing the given equation of the plane throughout root 5 square which is 5 we get y is equal to minus 8 over 5 or minus y is equal to 8 over 5. Now this equation 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 8 over 5. So the distance of the plane from the origin is 8 over 5 the direction cosines of the normal to the given plane are 0 1 0 and its distance from the origin is 8 over 5. So this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.