Triangulate 3D Point from Two Views

Given two camera projection matrices $P_1$ (3x4) and $P_2$ (3x4), and corresponding 2D points $\mathbf{x}_1 = [u_1, v_1]$ and $\mathbf{x}_2 = [u_2, v_2]$ in each image, triangulate the 3D point $\mathbf{X} = [X, Y, Z]$ using the Direct Linear Transform (DLT) method.

Algorithm (DLT Triangulation):

For each 2D point $\mathbf{x} = [u, v]$ and its projection matrix $P$, we have: $\mathbf{x} \times (P \cdot \mathbf{X}) = 0$

This gives us a system of linear equations. From each view, we extract two independent equations:

$u_i \cdot P_i^{3T} - P_i^{1T} = 0$ $v_i \cdot P_i^{3T} - P_i^{2T} = 0$

where $P_i^{jT}$ is the $j$-th row of $P_i$.

Build the $4 \times 4$ matrix $A$:

$A = \begin{pmatrix} u_1 P_1^{3T} - P_1^{1T} \\ v_1 P_1^{3T} - P_1^{2T} \\ u_2 P_2^{3T} - P_2^{1T} \\ v_2 P_2^{3T} - P_2^{2T} \end{pmatrix}$

Solve via SVD of $A$: the solution is the last column of $V$ (last row of $V^T$), converted from homogeneous coordinates.

Round each coordinate to 4 decimal places.