Determine Point Coordinates In 3D
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08-07-2019 - |
Question
I have a line that exists in 3d that is between two known points: {X1, Y1, Z1} and {X2, Y2, Z2}.
I also know that I am a certain distance for one of the points: D
How can I determine what the coordinates of the point where I am after moving D from {X1, Y1, Z1}?
Thanks
Solution
Assuming you want to move the distance D from point 1 to point 2 :
P1 = [ X1, Y1, Z1 ]
P2 = [ X2, Y2, Z2 ]
The line vector can be described as :
V = P2 - P1 = [ Xv = X2 - X1, Yv = Y2 - Y1, Zv = Z2 - Z1 ]
The line's length can be determined as :
VL = SQRT(Xv^2 + Yv^2 + Zv^2) // ^2 = squared
The line's versor aka the unit vector can be determined as :
v = V / VL = [Xv / VL, Yv / VL, Zv / VL]
The target point PD can be determined as :
Pd = P1 + D * v // Starting from P1 advance D times v
Please note that P1 and v are vectors and D is a scalar
OTHER TIPS
First, determine the length of the line segment:
d=sqrt((X1-X2)^2+(Y1-Y2)^2+(Z1-Z2)^2))
You are moving D from P1=(X1,Y1,Z1) toward P2=(X2,Y2,Z2). This puts you at the point (X3,Y3,Z3):
{XYZ}3={XYZ}1+(D/d)*({XYZ}2-{XYZ}1})
Where you expand that into 3 equations, one for each of X, Y, and Z.
This works because you are D/d of the way between P1 and P2. Check: Say D=d. Then you should be at exactly P2.
Take the vector between the two points
<X2-X1, Y2-Y1, Z2-Z1>
Turn that into a unit vector pointing in the same direction but with length 1. You do that by dividing by the distance between the two points:
<X2-X1, Y2-Y1, Z2-Z1>
---------------------------------------
sqrt((X2-X1)^2 + (Y2-Y1)^2 + (Z2-Z1)^2)
Then multiply that by D and add to your original point to get the new point.
<X2-X1, Y2-Y1, Z2-Z1>
(X1, Y1, Z1) + D * ---------------------------------------
sqrt((X2-X1)^2 + (Y2-Y1)^2 + (Z2-Z1)^2)
This is a linear combination problem:
dist = distance(p1, p2)
distance D is given
f = D / dist (fractional coordinate of point D within LineSeg (p1, p2)
pD = LinearCombo (1-f, p1, f, p2) (coordinates of point distance D from p1)