Line of intersection between two planes

How can I find the line of intersection between two planes?

I know the mathematics idea, and I did the cross product between the the planes normal vectors

but how to get the line from the resulted vector programmatically

Finding the line between two planes can be calculated using a simplified version of the 3-plane intersection algorithm.

The 2'nd, "more robust method" from bobobobo's answer references the 3-plane intersection.

While this works well for 2 planes (where the 3rd plane can be calculated using the cross product of the first two), the problem can be further reduced for the 2-plane version.

• No need to use a 3x3 matrix determinant,
  instead we can use the squared length of the cross product between the first and second plane (which is the direction of the 3'rd plane).
• No need to include the 3rd planes distance,
  (calculating the final location).
• No need to negate the distances.
  Save some cpu-cycles by swapping the cross product order instead.

Including this code-example, since it may not be immediately obvious.

// Intersection of 2-planes: a variation based on the 3-plane version.
// see: Graphics Gems 1 pg 305
//
// Note that the 'normal' components of the planes need not be unit length
bool isect_plane_plane_to_normal_ray(
        const Plane& p1, const Plane& p2,
        // output args
        Vector3f& r_point, Vector3f& r_normal)
{
    // logically the 3rd plane, but we only use the normal component.
    const Vector3f p3_normal = p1.normal.cross(p2.normal);
    const float det = p3_normal.length_squared();
    
    // If the determinant is 0, that means parallel planes, no intersection.
    // note: you may want to check against an epsilon value here.
    if (det != 0.0) {
        // calculate the final (point, normal)
        r_point = ((p3_normal.cross(p2.normal) * p1.d) +
                   (p1.normal.cross(p3_normal) * p2.d)) / det;
        r_normal = p3_normal;
        return true;
    }
    else {
        return false;
    }
}

Adding this answer for completeness, since at time of writing, none of the answers here contain a working code-example which directly addresses the question.

Though other answers here already covered the principles.