jxu

03-22-2001, 07:24 AM

Does anybody know how to calculate the distance between a line and a point in 3d (not 2d) space? Please give a closed formula or code if possible. Thanks

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jxu

03-22-2001, 07:24 AM

Does anybody know how to calculate the distance between a line and a point in 3d (not 2d) space? Please give a closed formula or code if possible. Thanks

Gavin

03-22-2001, 07:32 AM

Do you mean the shortest distance, i.e. the perpendular distance from the line?

gav

gav

jxu

03-22-2001, 07:34 AM

Yes. I meant the shortest distance.

Randolph

03-22-2001, 08:11 AM

If we have line in 3D space, which is defined by

x=x1+l*t

y=y1+m*t

z=z1+n*t

and point M0=(x0,y0,z0) then distance between them is given by a formulae:

d = sqrt(det(A)^2 + det(B)^2 + det(C)^2)/sqrt(l^2+m^2+n^2)

A,B and C are 2x2 matrices of following form:

A=

[ vy vz]

[ m n]

B=

[ vz vx]

[ n l]

C=

[ vx vy]

[ l m]

vx = x1-x0 vy = y1-y0 vz = z1-z0

det(M) means determinant M where M is a square matrix. In our occasion if M is 2x2 matrix

[a b]

[c d]

then det(M)=a*d-b*c. Hope, this will help

[This message has been edited by Randolph (edited 03-22-2001).]

x=x1+l*t

y=y1+m*t

z=z1+n*t

and point M0=(x0,y0,z0) then distance between them is given by a formulae:

d = sqrt(det(A)^2 + det(B)^2 + det(C)^2)/sqrt(l^2+m^2+n^2)

A,B and C are 2x2 matrices of following form:

A=

[ vy vz]

[ m n]

B=

[ vz vx]

[ n l]

C=

[ vx vy]

[ l m]

vx = x1-x0 vy = y1-y0 vz = z1-z0

det(M) means determinant M where M is a square matrix. In our occasion if M is 2x2 matrix

[a b]

[c d]

then det(M)=a*d-b*c. Hope, this will help

[This message has been edited by Randolph (edited 03-22-2001).]

Korval

03-22-2001, 09:44 AM

Is there, perhaps, a shorter formula? One requiring less computation?

DFrey

03-22-2001, 10:03 AM

Given any point q on the line, and the test point p, let P=p-q. Let l be a unit vector parallel to the line. Then d, the shortest distance from p to the line is:

d=sqrt(P•P-(l•P)^2)

[This message has been edited by DFrey (edited 03-22-2001).]

d=sqrt(P•P-(l•P)^2)

[This message has been edited by DFrey (edited 03-22-2001).]

EricK

03-22-2001, 11:00 AM

Originally posted by DFrey:

Given any point q on the line, and the test point p, let P=p-q. Let l be a unit vector parallel to the line. Then d, the shortest distance from p to the line is:

d=sqrt(P•P-(l•P)^2)

[This message has been edited by DFrey (edited 03-22-2001).]

Hmmm. Doesn't that assume the line passes through the origin?

Given any point q on the line, and the test point p, let P=p-q. Let l be a unit vector parallel to the line. Then d, the shortest distance from p to the line is:

d=sqrt(P•P-(l•P)^2)

[This message has been edited by DFrey (edited 03-22-2001).]

Hmmm. Doesn't that assume the line passes through the origin?

EricK

03-22-2001, 11:25 AM

Originally posted by EricK:

Hmmm. Doesn't that assume the line passes through the origin?

Oops. Let me answer my own question, NO.

Hmmm. Doesn't that assume the line passes through the origin?

Oops. Let me answer my own question, NO.

Randolph

03-23-2001, 09:27 AM

Of course, if you have UNIT vector, parallel to a line, you may simplify that formula! I gave formula with most common requirements. Normalizing vector also requires computation, although if you have to find distances from multiple points to one line this will give you huge boost of effectiveness.

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