The calculation relies on fundamental principles of vector algebra. The line of intersection belongs to both planes, which imposes specific geometric constraints that we can exploit algebraically.
1. Finding the Direction Vector (v)
Let the two planes be defined by their normal vectors, N₁ = <A₁, B₁, C₁> and N₂ = <A₂, B₂, C₂>. Since the line of intersection lies in both planes, it must be perpendicular to both of these normal vectors. The cross product of two vectors yields a third vector that is perpendicular to both, so the direction vector 'v' of the line is found by computing the cross product: v = N₁ × N₂.
2. Finding a Point on the Line (P₀)
To find a specific point P₀ = (x₀, y₀, z₀) on the line, we need to solve the system of two linear equations for the planes. Since there are three variables (x, y, z) and only two equations, there are infinitely many solutions (which form the line). We can find one by setting one of the variables to a constant (e.g., z=0) and solving the remaining 2x2 system for x and y. If this system has no unique solution (which happens if the line is parallel to the xy-plane), the calculator intelligently tries setting x=0 or y=0 instead to find a point.
3. Constructing the Line Equations
With a point P₀ and a direction vector v = <l, m, n>, the line can be expressed using a vector equation r = P₀ + tv, or as a set of parametric equations: x(t) = x₀ + lt, y(t) = y₀ + mt, z(t) = z₀ + nt, where 't' is a real parameter.