Calculating A LookAt Matrix


Answer :

Note the example given is a left-handed, row major matrix.

So the operation is: Translate to the origin first (move by -eye), then rotate so that the vector from eye to At lines up with +z:

Basically you get the same result if you pre-multiply the rotation matrix by a translation -eye:

[      1       0       0   0 ]   [ xaxis.x  yaxis.x  zaxis.x 0 ] [      0       1       0   0 ] * [ xaxis.y  yaxis.y  zaxis.y 0 ] [      0       0       1   0 ]   [ xaxis.z  yaxis.z  zaxis.z 0 ] [ -eye.x  -eye.y  -eye.z   1 ]   [       0        0        0 1 ]    [         xaxis.x          yaxis.x          zaxis.x  0 ] = [         xaxis.y          yaxis.y          zaxis.y  0 ]   [         xaxis.z          yaxis.z          zaxis.z  0 ]   [ dot(xaxis,-eye)  dot(yaxis,-eye)  dot(zaxis,-eye)  1 ]

Additional notes:

Note that a viewing transformation is (intentionally) inverted: you multiply every vertex by this matrix to "move the world" so that the portion you want to see ends up in the canonical view volume.

Also note that the rotation matrix (call it R) component of the LookAt matrix is an inverted change of basis matrix where the rows of R are the new basis vectors in terms of the old basis vectors (hence the variable names xaxis.x, .. xaxis is the new x axis after the change of basis occurs). Because of the inversion, however, the rows and columns are transposed.


I build a look-at matrix by creating a 3x3 rotation matrix as you have done here and then expanding it to a 4x4 with zeros and the single 1 in the bottom right corner. Then I build a 4x4 translation matrix using the negative eye point coordinates (no dot products), and multiply the two matrices together. My guess is that this multiplication yields the equivalent of the dot products in the bottom row of your example, but I would need to work it out on paper to make sure.

The 3D rotation transforms your axes. Therefore, you cannot use the eye point directly without also transforming it into this new coordinate system. That's what the matrix multiplications -- or in this case, the 3 dot-product values -- accomplish.


That translation component helps you by creating an orthonormal basis with your "eye" at the origin and everything else expressed in terms of that origin (your "eye") and the three axes.

The concept isn't so much that the matrix is adjusting the camera position. Rather, it is trying to simplify the math: when you want to render a picture of everything that you can see from your "eye" position, it's easiest to pretend that your eye is the center of the universe.

So, the short answer is that this makes the math much easier.

Answering the question in the comment: the reason you don't just subtract the "eye" position from everything has to do with the order of the operations. Think of it this way: once you are in the new frame of reference (i.e., the head position represented by xaxis, yaxis and zaxis) you now want to express distances in terms of this new (rotated) frame of reference. That is why you use the dot product of the new axes with the eye position: that represents the same distance that things need to move but it uses the new coordinate system.


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