Find distance between line x-y-z =2, 2x+y+z=4 and the plane x-y=5
My problem is that the book is saying the minimum distance between the plane and line is zero but l am getting . Can someone tell me where did l get this one wrong
Attempt to solution :
If l intersect the first two planes given l will get the equation of the line :
x-y-z=2................(1)
2x+y+z=4...............(2)
Add (1) and (2) ===> 3x+2z=6
z=3-3x/2.............(3)
Substitute (3) into (1)
y=1-x/2
and x=x
Equations of the line :
x=t
y=1-t/2
z=3-3t/2
Get point on the line and the normal vector for plane :
(0,1,3)
Formula for distance between point and plane x-y=5 :
normal vector for the plane : (1,-1,0)
when l solve for distance l get -6/sqrt(2).