In 3D, finding the perpendicular distance of a point to a line?
Author
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
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?
Quote:Original post by johnnyBravoYeah (also, even in 2D I wouldn't recommend approaching this sort of problem in terms of slopes).
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) * DWhere:P is the query pointO is the line originD is the line direction
Author
Ah ok.
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
Quote:Original post by jyk((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
Quote:Original post by johnnyBravoYeah, it looks like you've got an extra '2' in there.
Ah ok.Quote:Original post by jyk((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))^2f(P) = (P-O-tD)^2d = P-Of(P) = (d-tD)^2f(P) = D^2t^2-2dDt+d^2f'(P) = 2D^2t - 2dD2D^2t - 2dD = 02D^2t = 2dDD^2t = dDt = dD/D^2
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.
Author
Quote:Original post by jykf(P) = D^2t^2-2dDt+d^2f'(P) = 2D^2t - 2dD
Ah I forgot to multiply it by 2 when I derived the t^2
Thanks!
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