# Thread: Distance between line and plane

1. ## Distance between line and plane

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 $\displaystyle 6/\sqrt{2}$. 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)

$\displaystyle distance=\frac{ax+by+cz+d}{\sqrt{(a^2+b^2+c^2)}}$

when l solve for distance l get -6/sqrt(2).

2. Originally Posted by nyasha
x-y-z=2................(1)
2x+y+z=4...............(2)

Add (1) and (2) ===> 3x+2z=6
Please try that addition again. Check the 'z', in particular.

when l solve for distance l get -6/sqrt(2).
Please never write that again. Distance isn't usually negative. Sometimes, I suppose, but you should be very, very clear that is what you mean.

3. Originally Posted by TKHunny
Please try that addition again. Check the 'z', in particular.

Please never write that again. Distance isn't usually negative. Sometimes, I suppose, but you should be very, very clear that is what you mean.

Okay thanks for pointing out the mistake. For the distance l forgot to put absolute value symbol