# Distance from line to a plane

1. Nov 30, 2008

### Melawrghk

1. The problem statement, all variables and given/known data
Find the shortest distance between the plane 2x−6y+4z=10 and the line passing through the points (0,-15,8) and (-6,-13,14).

3. The attempt at a solution
So orthogonal vector to the plane is (2, -6, 4)T
The parametric equation of the line is:
L=(0,-15,8)+t(-3,1,3)

And I don't know after that. I can find distance from a point to a plane, but not from a line. Any help will be appreciated.

2. Nov 30, 2008

### rock.freak667

If N is the normal vector to the plane which you've found to be (2,-6,4) and you found the direction of L to be u=(-3,1,3).

Consider what N.u works out to be and what it means.

EDIT: Forgot to put in that when it comes to a line and a plane, the line is will either intersect it at one point, is parallel to the plane or lies within the plane. Can easily test for all. Best to test for the second and third one when starting the question.

Last edited: Nov 30, 2008
3. Nov 30, 2008

### Melawrghk

Well, the dot product is 0 meaning L is perpendicular to N, so it's parallel to the plane. So any point on L is equidistant from the plane. So I can just use (0, -15,8) to find the distance?

Neat, thanks!

4. Nov 30, 2008

### rock.freak667

yes, you should be able to.