Given: two planes \varepsilon_i: \quad \vec n_i\cdot\vec x=d_i, \quad i=1,2, \quad \vec n_1,\vec n_2
linearly independent, i.e. the planes are not parallel. Wanted: A parametric representation \vec x= \vec p + t\vec r of the intersection line. The direction of the line one gets from the
crossproduct of the normal vectors: \vec r=\vec n_1\times\vec n_2. A point P:\vec p of the intersection line can be determined by intersecting the given planes \varepsilon_1, \varepsilon_2 with the plane \varepsilon_3: \vec x = s_1\vec n_1 + s_2\vec n_2, which is perpendicular to \varepsilon_1 and \varepsilon_2. Inserting the parametric representation of \varepsilon_3 into the equations of \varepsilon_1 und \varepsilon_2 yields the parameters s_1 and s_2. P: \vec p= \frac{ d_1(\vec n_2\cdot\vec n_2)-d_2(\vec n_1\cdot \vec n_2)} {(\vec n_1\cdot\vec n_1)(\vec n_2\cdot\vec n_2)-(\vec n_1\cdot\vec n_2)^2} \vec n_1 + \frac{ d_2(\vec n_1\cdot\vec n_1)-d_1(\vec n_1\cdot\vec n_2)}{(\vec n_1\cdot\vec n_1)(\vec n_2\cdot\vec n_2)-(\vec n_1\cdot\vec n_2)^2} \vec n_2\ .
Example: \varepsilon_1:\ x+2y+z=1, \quad \varepsilon_2:\ 2x-3y+2z=2 \ . The normal vectors are \vec n_1=(1,2,1)^\top, \ \vec n_2=(2,-3,2)^\top and the direction of the intersection line is \vec r=\vec n_1\times\vec n_2=(7,0,-7)^\top. For point P:\vec p, one gets from the formula above \vec p=\tfrac{1}{2}(1,0,1)^\top \ . Hence :\vec x=\tfrac{1}{2}(1,0,1)^\top + t (7,0,-7)^\top is a parametric representation of the line of intersection.
Remarks: • In special cases, the determination of the intersection line by the
Gaussian elimination may be faster. • If one (or both) of the planes is given parametrically by \vec x= \vec p + s\vec v + t \vec w , one gets \vec n = \vec v \times \vec w as normal vector and the equation is: \vec n\cdot \vec x = \vec n\cdot \vec p. == Intersection curve of a plane and a quadric ==