Calculating endpoint of 3D line with start point and angle.

Aug 2019
3
0
Denmark
So I need to calculate the start point (x1, y1, z1) of a 3D line of which I know the end point (x2, y2, z2), the angle which is 45 degrees, but where I also know y1. I furthermore know that y1<y2.

So is there a simplified formula for this, given the additional info?
 
Jun 2019
493
262
USA
Which angle is 45°? You need a minimum of two angles to define a line in 3D.

Also, to find the starting point, you would need the length of the line segment (or dependent information, such as one of x1, y1, or z1).
 
Aug 2019
3
0
Denmark
English isn't my first language so I'm not sure how to describe what you ask about the angle, and I'm having a hard time wrapping my head around it all as well.

The best I can do is to provide the information that it's two points in a triangle where the third point would be (x2,y1,z2) as far as I can figure out, but like I said, I can't quite wrap my head around this so I'm not sure if that's already obvious from the previously provided info or not.
 
Jun 2019
493
262
USA
So A = (x1,y1,z1), B = (x2,y2,z2), and C = (x2,y1,z2), where <CAB = 45° (or <ABC = 45°)? And y1 < y2.
This actually makes it easy, because everything lies in a plane perpendicular to the x-z plane.

On the other hand, if you are trying to find point A, there are actually an infinite number of solutions (all on a circle).

Let R = (y2-y1) = |BC|. We have a 45-45-90 triangle, so |AC| = |BC| = R. It is also given that the y component of A is the same as the y component of C. However, we can choose x1 and z1 any way we want, so long as |AC| = R.

So x1 = x2 + R cos(t), z1 = z2 + R sin(t),
and we can choose t to be any angle we want.
 

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