View Full Version : Closest point on a Vector to a point
06-11-2004, 05:03 PM
I have a vector and a point, p1. I want to find the closest point along the vector to p1. I've read about finding the closest point on a line to a point, but I think choosing arbitrary points on the vector is not the way to go.
06-12-2004, 11:02 PM
i cannot figure out what do u mean by "a point on a vector". A vector is sth. having length and direction,but no inherent location. So, to express a point, say P,you can do it as following:
P = P0 + Vector
06-13-2004, 10:32 AM
The closest point on a line to another point p1 is the intersection of the line with a perpendicular line formed from the point p1 to the line. If testing with line segment rather than a line constrain to extents of segment.
Graphics GEMS has a lot of info for stuff like this. You might also look at Dave Eberley's code page at http://www.magic-software.com.
06-13-2004, 11:40 PM
For a point P, and a normalized vector V, the point closest to P along V is
P' = V * dot( P, V )
In words, it's V scaled by the projection of P onto V.
06-14-2004, 07:51 AM
Alright, to make it more school-style i give an absolutely easy explanation, hehe.
P2 is the point from which you want to have the closest point on the line.
Pr is the closest point. (your result)
P1 is a point on the line. v is the direction vector of the line L. So the parametric equation of the line is L(t) = (P1 + t*v).
The closest point on the line L is the point where the line perpendicular to L and through your Pt intersects L. This means the points P1, Pr and P2 form a rightangled triangle.
The two know edges of it are v and (P2-P1). The cosine of the angle they are enclosing is
cos(alpha) = -------------
|Pt-P1|*|v|as anyone should have learned in school, hehe. In the right angled triangle this cosine is the quotient
cos(alpha) = -----------
hypothenuseSince we know the length of the hypothenuse (|P2-P1|) we can calculate the SIGNED length of the near leg:
len = ------------- * |P2-P1|
|v|This near leg is exactly the SIGNED distance on the line from P1 to Pr (which is the near leg in our rightangled triangle at P1 after all).
So we have to scale t*|v| in our line equation to that length:
len (P2-P1)dot(v) (P2-P1)dot(v)
--- * v = ------------- * v = ------------- * v
|v| |v|*|v| (v)dot(v)|v| = sqrt(sum(vi * vi) thus |v|*|v| = sum(vi*vi) = (v)dot(v). If |v| is 1, then (v)dot(v) is 1.
In our line equation that is finally:
Pr = P1 + ------------- * v.
(v)dot(v)When v is prenormalized (v=v/|v|, set to length 1) you save the division by (v)dot(v). When P1 is additionaly the origin, then you end up at Sean's equation.
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