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How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments.

I have two points, $p_1$ and $p_2$ that are known $x,y$ locations. I also have line segments with known lengths $a$ and $b$. I made a diagram below to illustrate the problem.

If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.

In two dimensions...

Let $vec i$ be the unit vector in the direction of the
vector $overrightarrowp_1p_2=dfracp_2-p_1$ Let $vec j$ be a unit vector perpendicular to $vec i$. (If $vec i = langle u,vrangle$, then, for example, $vec j = langle-v, urangle$.) Then

$$overrightarrowp_1p_3 =
dfraca^2sqrta^2+b^2 vec i pm
dfracabsqrta^2+b^2 vec j$$

See here for details.

In three dimesions, there will be a circle of possible points.

$$overrightarrowp_1p_3 =
dfraca^2sqrta^2+b^2 vec i +
dfracabsqrta^2+b^2cos(theta) vec j +
dfracabsqrta^2+b^2sin(theta) vec k$$

Sometimes a figure is worth a 1000 words:

In three dimensions:

• Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.




• The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.




• We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



• 
  

  If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
  $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$

  
  
  

  Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.

  
  
  

  In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.

  
  
  

  Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.

  
  



• But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.

[*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.

As I mentioned in my comment, there is no solution unless $a^2+b^2=lvert P_1P_2rvert^2$. Let $Q$ be the point $(x_3,y_3)$, $R$ the foot of the perpendicular from $Q$ to $overlineP_1P_2$, and $c = lvert P_1P_2rvert$. $triangleP_1QP_2$, $triangleP_1RQ$ and $triangleQRP_2$ are all similar, therefore $lvert P_1Rrvert = a^2/c$, which means that $$R = left(1-a^2over c^2right)P_1+a^2over c^2 P_2.$$ Similarly, $lvert QRrvert = (ab)/c$ and a unit vector perpendicular to $overlineP_1P_2$ is $frac1c(y_1-y_2,x_2-x_1)$, therefore the two solutions are $$Q=Rpmabover c^2(y_1-y_2,x_2-x_1).$$

In three dimensions, there is an infinite number of solutions that lie on a circle of radius $(ab)/c$ centered at $R$ and in a plane perpendicular to $overlineP_1P_2$.