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.
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 findand
Calculate the perpendicular distance between the planes r.(4i-3j+k)=10 and 4x-3y+z=8
thank you very much for your time!