 How to determine where a line intersects a planeSuppose you have a line defined by two 3dimensional points and a plane defined by three 3dimensional points. How do you tell where the line intersects the plane?Step 1: Convert the plane into an equationThe equation of a plane is of the form Ax + By + Cz = D.To get the coefficients A, B, C, simply find the cross product of the two vectors formed by the 3 points. This will give you a vector that is normal to the triangle. The components of this vector are, coincidentally, the coefficients A, B, and C. Let's call the 3 points in the plane Q, R, and S. QR = <R_{x}  Q_{x}, R_{y}  Q_{y}, R_{z}  Q_{z}> QS = <S_{x}  Q_{x}, S_{y}  Q_{y}, S_{z}  Q_{z}> QR × QS = Normal Vector N. A = N_{x} B = N_{y} C = N_{z} Ax + Bx + Cx = 0 is the equation of the line with a similar tilt but shifted such that it goes through the origin (0, 0, 0). To get the value of D, shift the equation with one of the Q, R, S points. A(x  Q_{x}) + B(y  Q_{y}) + C(z  Q_{z}) = 0 You now have a Ax + By + Cz = D equation. Step 2: Find the equation for the lineSuppose your two points are A and B (different A and B from the plane)P(t) = A + (B  A) * t This equation can be broken up into 3 equations for each component of the vector: x = A_{x} + (B_{x}  A_{x}) * t y = A_{y} + (B_{y}  A_{y}) * t z = A_{z} + (B_{z}  A_{z}) * t Step 3: Combine them and solve for tNow you have definitions of x, y, and z in terms of a variable t. Plug these equations in for the values of x, y, and z in the equation for the plane, and solve for t. Once you get a value for t, plug it back into the equation for the line, and you'll have your point of intersection.Potential Error/Edge Cases
