- #1
joinforfun89
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Hi ,Given 3 points A[x1,y1,z1], B[x2,y2,z2] and C, and given the distance between B and C is known and the angle ABC is supplied. How do we ascertain the coordinates of the point C ?
I can calculate the dot product using the angle, by defining vectors BC and AC and then use the distance formula to get another equation for the coordinates of C. But I have a hunch that there is a simpler way to do this. I need to program this, and therefore the requirement for a simpler solution.
If there is a derivation that you know please point me towards it.
Thanks a lot in advance.MORE Information Added:
I didn't include all the information earlier and yes Mathman is right that as stated this problem doesn't have a unique solution. This is actually a smaller part of the bigger problem I have. Here it goes.
A-B-C-D are 4 points. We define rx = length(B-C), angle, eta = (A-B-C) and angle theta = (B-C-D) and the torsion angle omega= (A-B-C-D). What I really need to do is to find the coordinates of C and D provided that I have the new values of rx, eta, theta and omega. Any pointers would be welcome.
Thanks again.
I can calculate the dot product using the angle, by defining vectors BC and AC and then use the distance formula to get another equation for the coordinates of C. But I have a hunch that there is a simpler way to do this. I need to program this, and therefore the requirement for a simpler solution.
If there is a derivation that you know please point me towards it.
Thanks a lot in advance.MORE Information Added:
I didn't include all the information earlier and yes Mathman is right that as stated this problem doesn't have a unique solution. This is actually a smaller part of the bigger problem I have. Here it goes.
A-B-C-D are 4 points. We define rx = length(B-C), angle, eta = (A-B-C) and angle theta = (B-C-D) and the torsion angle omega= (A-B-C-D). What I really need to do is to find the coordinates of C and D provided that I have the new values of rx, eta, theta and omega. Any pointers would be welcome.
Thanks again.
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