How to determine the intersection between the came

Question

I have a 3D scene with an infinite horizontal plane (parallel to the xz coordinates) at a height H along the Y vertical axis.

I would like to know how to determine the intersection between the axis of my camera and this plane.

The camera is defined by a view-matrix and a projection-matrix.

Answer 1

For a general line-plane intersection there are lot of answers and tutorials.

Your case is simple due to the plane is horizontal.

I suppose the camera is at C(cx, cy, cz) and it looks at T(tx, ty,tz).
Then the line camera-target can be defined by:

cx - x     cy - y     cz - z
------  =  ------  =  ------        /// These are two independant equations
tx - cx    ty - cy    tz - cz

For a horizontal plane, only a equation is needed: y = H. Substitute this value in the line equations and you get

(cx-x)/(tx-cx) = (cy-H)/(ty-cy)
(cz-z)/(tz-cz) = (cy-H)/(ty-cy)

So

x = cx - (tx-cx)*(cy-H)/(ty-cy)
y = H
z = cz - (tz-cz)*(cy-H)/(ty-cy)

Of course if your camera looks in an also horizontal line then ty=cy and there is not solution.

Answer 2

There are two sub-problems here: 1) Extracting the position and view-direction from the camera matrix. 2) Calculating the intersection between the view-ray and the plane.

Extracting position and view-direction

The view matrix describes how points are transformed from world-space to view space. The view-space in OpenGL is usually defined such that the camera is in the origin and looks into the -z direction.

To get the position of the camera, we have to transform the origin [0,0,0] of the view-space back into world-space. Mathematically speaking, we have to calculate:

camera_pos_ws = inverse(view_matrix) * [0,0,0,1]

but when looking at the equation we'll see that we are only interrested in the 4th column of the inverse matrix which will contain ¹

camera_pos_ws = [-view_matrix[12], -view_matrix[13], -view_matrix[14]]

The orientation of the camera can be found by a similar calculation. We know that the camera looks in -z direction in view-space thus the world space direction is given by

camera_dir_ws = inverse(view_matrix) * [0,0,-1,0];

Again, when looking at the equation, we'll see that this only takes the third row of the inverse matrix into account which is given by²

camera_dir_ws = [-view_matrix[2], -view_matrix[6], -view_matrix[10]]

Calculating the intersection

We now know the camera position P and the view direction D, thus we have to find the x,z value along the ray R(x,y,z) = P + l * D where y equals H. Since there is only one unknown, l, we can calculate that from

y = Py + l * Dy
H = Py + l * Dy
l = (H - Py) / Dy

The intersection point is then given by pasting l back into the ray equation.

Notes

¹ The indices assume that the matrix is stored in a column-major linear array.

² Note, that the inverse of a matrix of the form

M = [  R T ]
       0 1

, where R is a orthogonal 3x3 matrix, is given by

inv(M) = [ transpose(R)  -T ]
                0         1