Question about D3DXVec3TransformCoord

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4 comments, last by daedalic 13 years ago
We are trying to replace this direct x method with our own method. We tried just multiplying a matrix by a vector and got some very different results. What is this method doing that's different from simple multiplication?
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If "vec" is the vector your transforming, it transforms (vec.x, vec.y, vec.z, 1.0f) by the transform matrix. It then divides the result by w.
[font="Arial"]When you multiply a 3D vector and a matrix the result is a 4D vector with w != 1. On the other hand [/font][font="Arial"]D3DXVec3TransformCoord returns a 3D vector, which is a 4D vector projected back to w = 1.[/font]
Here's an example of a transform matrix we are using:

0 0 -1 0
1 0 0 0
0 -1 0 0
0 0 0 1

When we multiply this matrix by the unit vector: (0, 0, 1) using D3DXVecTransformCoord, we get (0, -1, 0) as a result. However, when we use our multiplication method we get (-1, 0, 0) as a result.

Here's our multiplication method: It ignores the 4th dimension since it shouldn't be needed for our purposes... as you can see, the transform matrix is equal to the identity matrix in the 4th dimensional positions.


INLINE Vector3D
Matrix::operator* (const Vector3D vector)
{
return Vector3D (m11 * vector.x + m12 * vector.y + m13 * vector.z,
m21 * vector.x + m22 * vector.y + m23 * vector.z,
m31 * vector.x + m32 * vector.y + m33 * vector.z);
}
It appears you should be multiplying the row vector by the matrix columns, rather than the other way 'round. I.e.,

m11*x + m21*y + m31*z = 0

m12*x + m22*y + m32*z = -1

m13*x + m23*y + m33*z = 0

Also, if you're going to "ignore" the 4th elements, then you should be using/comparing-to D3DXTransformNormal.

Please don't PM me with questions. Post them in the forums for everyone's benefit, and I can embarrass myself publicly.

You don't forget how to play when you grow old; you grow old when you forget how to play.

The problem was that I had assumed DirectX used column vectors instead of row vectors. So that takes care of that problem. Thanks!

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