Distance:Line and Point

• Apr 26th 2010, 12:11 PM
Schdero
Distance:Line and Point
Given are the line l:m*(6/2/3) and the point P(2/1/2). Sought is the point Q on l with the shortest distance to P.
I know the solution of inventing a plane E with normal vector N which equals (6/2/3), but I'd liek to know why the following approach doesn't work as well:

The vector QP must be normal to the direction vector of the line, so (6/2/3)*QP = 0. Point Q is given with (m6/m2/m3), thus the vector QP is (2-6m)/(1-2m)/(2-3m).
Idea
(2-6m)/(1-2m)/(2-3m)*(6/2/3)=0
m=no m valid for all three

Why doesn't that lead to a correct solution?

Can anyone help?
• Apr 26th 2010, 12:20 PM
Failure
Quote:

Originally Posted by Schdero
Given are the line l:m*(6/2/3) and the point P(2/1/2). Sought is the point Q on l with the shortest distance to P.
I know the solution of inventing a plane E with normal vector N which equals (6/2/3), but I'd liek to know why the following approach doesn't work as well:

The vector QP must be normal to the direction vector of the line, so (6/2/3)*QP = 0. Point Q is given with (m6/m2/m3), thus the vector QP is (2-6m)/(1-2m)/(2-3m).
Idea
(2-6m)/(1-2m)/(2-3m)*(6/2/3)=0
m=no m valid for all three

I cannot quite parse this: what value of m do you get? In an earlier version of this question you wrote m=1/3, which is wrong. You should get m=20/49.

Quote:

Why doesn't that lead to a correct solution?
The basic idea is correct, but the solution m=1/3 of your equation was wrong.
• Apr 26th 2010, 12:36 PM
Schdero
Yes I see, I also get your result when I do it, I think i have to have made some quite stupid error in the earlier version; again: thanks failure