In 3D, I'm finding the perpendicular distance of a point to a line (consisting of a point and a direction vector).
I've found two different ways,
one using the dot product, distance between point on the line and using sin to find the length of the side aka the distance.
the second was with the distance forumulae combined with the vector multiplied by an unknown, and i derive the the equation to find the min of the unknown which subbed into the dist forumale to get the distance.
My problem is I feel there should be much a simpler solution to this. Is there another simpler way of doing this?
thx

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# In 3D, finding the perpendicular distance of a point to a line?

Started by johnnyBravo, Jul 12 2007 09:33 PM

7 replies to this topic

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#2
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Posted 12 July 2007 - 09:55 PM

1). Find slope of line perpendicular to the line, call it pm

2). Create a line that goes through the point and has a slope of pm

3). Intersect this new line and the original line

4). Find distance between the intersection point and the original point

Of course, it's a bit more than four steps when you do it out in code, but it shouldn't be bad.

Edit: Oh gods, I forgot step 4! How could I do that?

2). Create a line that goes through the point and has a slope of pm

3). Intersect this new line and the original line

4). Find distance between the intersection point and the original point

Of course, it's a bit more than four steps when you do it out in code, but it shouldn't be bad.

Edit: Oh gods, I forgot step 4! How could I do that?

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#4
Members - Reputation: **2051**

Posted 13 July 2007 - 02:02 AM

Quote:Yeah (also, even in 2D I wouldn't recommend approaching this sort of problem in terms of slopes).

Original post by johnnyBravo

Isn't that just for 2D?

The equation to find the closest point on a line to a point (in any dimension) is:

closest = O + ((P-O).D)/(D.D) * DThere are several ways to derive this, one of which is as a minimization problem (which I assume is what you were referring to earlier). However you derive it though, I don't think there's any simpler solution than the above.

Where:

P is the query point

O is the line origin

D is the line direction

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#6
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Posted 13 July 2007 - 02:28 AM

Quote:Yeah, it looks like you've got an extra '2' in there.

Original post by johnnyBravo

Ah ok.

Quote:

Original post by jyk

[code=auto:0]((P-O).D)/(D.D)

I had pretty much the same thing except I had ((P-O).D)/(2*(D.D))

The 2 was from in the distance formulae part eg (2*b)*t + (c^2)*t^2

Or am I wrong?

thx

I'll sketch out the derivation here (very informally) so you can compare it to yours:

f(P) = (P-(O+tD))^2

f(P) = (P-O-tD)^2

d = P-O

f(P) = (d-tD)^2

f(P) = D^2t^2-2dDt+d^2

f'(P) = 2D^2t - 2dD

2D^2t - 2dD = 0

2D^2t = 2dD

D^2t = dD

t = dD/D^2

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#7
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Posted 13 July 2007 - 02:33 AM

The above equation Jyk made is the simplest one I could imagine.

What the equation does is this:

1. Get vector from O to P

2. Project the OP vector to line(vector) D , which would drop a perpendicular from point P to line D.

3. Now the projection would give you the component(a distance) of the OP vector from O to P parallel to D. I hope you could understand me.

4. The return depends on where P lies. Left or right of the line.

What the equation does is this:

1. Get vector from O to P

2. Project the OP vector to line(vector) D , which would drop a perpendicular from point P to line D.

3. Now the projection would give you the component(a distance) of the OP vector from O to P parallel to D. I hope you could understand me.

4. The return depends on where P lies. Left or right of the line.