# Thread: Mathematics for aplications development.

1. ## Mathematics for aplications development.

Hi,

im doing test revision. Could someone take me through the steps to solving these two questions, the actuall answers and working isnt important, these are just random questions on a revision sheet and having left my notes at uni i have nothing to look at to help me.

The plane 3x-y+z=0 passes through the origin. Find the reflection, in this plane, of the point P(4,-3,0)

and

Calculate the perpendicular distance between the planes r.(4i-3j+k)=10 and 4x-3y+z=8

thank you very much for your time!

2. Originally Posted by EpicOfX
Hi,

im doing test revision. Could someone take me through the steps to solving these two questions, the actuall answers and working isnt important, these are just random questions on a revision sheet and having left my notes at uni i have nothing to look at to help me.

The plane 3x-y+z=0 passes through the origin. Find the reflection, in this plane, of the point P(4,-3,0)
Call the reflection point Q= (a, b, c). Then the line PQ is perpendicular to the plane and P and Q are equi-distant from the plane.

The plane has "normal vector" <3, -1, 1> and so the perpendicular line through P has parametric equations x= 3t+ 4, y= -t- 3, z= t. Put those for into the equation of the plane to find the point where the line intersects the plane. Find the distance from P to that point and then find the point on the line the same distance on the other side of the plane.

and

Calculate the perpendicular distance between the planes r.(4i-3j+k)=10 and 4x-3y+z=8

thank you very much for your time!
I take it that r= xi+ yj+ zk so "r.(4i-3j+ k)= 10" is 4x- 3y+ z= 10. Notice that the coefficients of the two planes are the same so the two planes are parallel and there is a common perpendicular. Take any point on the first plane. Since z= 10- 4x+ 3y, taking x= 1, y= -1 gives z= 10-4-3= 3 so (1, -1, 3) is a point on the first plane. A line perpendicular to the plane, through that point, has parametric equations x= 4t+ 1, y= -3t- 1, z= t+ 3. Put those into the equation for the second plane, 4x- 3y+ z= 8, to find