It is proved that L 1 and L2 are placed in a solid rectangular coordinate system, with L 1 as the X axis, L2 intersects with the Y axis and is parallel to the X and Y axes.

Let L 1 be Y=0 and z = 0；; L2 is ax+by = 0 and z = r.

Let the plane passing through L 1 be aY+bZ=0, and the plane passing through L2 be AX+BY+C(Z-R)=0.

∵ The two planes are perpendicular, and the normal vectors of the two planes are (0, a, b), (a, b, c).

∴ aB+bC=0 ∴ The intersection equation is AX+BY+C(Z-R)=0, CY-BZ=0.

Substitute C=BZ/Y into AX+BY+C(Z-R)=0.

AX+BY+(Z-R)*BZ/Y=0，AXY+BY？ +(Z-R)*BZ=0

When B≠0, AXY+BY? +(Z-R)*BZ=0 is hyperboloid; When B=0, the surface becomes a plane X=0 or Y=0.