Building view transform matrices

Question

For a 3D scene in the world coordinates, its View Reference Point $\mathrm{VRP}$ is at $(5,-2,1)$, and a viewer is looking towards point $A=(1,1,1)$. Construct a transform matrix which will map world coordinate points to a right-handed $(UVN)$ viewing space, so that $\mathrm{VRP}$ is the origin, the line joining $\mathrm{VRP}$ to $A$ is the positive $N$ axis, and the view-up Vector is $(0,0,1)$.

To my knowledge, the final answer should be a single transform matrix. To calculate the final matrix, I first find the matrix which translates the coordinate system to the to the origin in homogeneous matrix form:

Translation matrix:

$$ T= \begin{bmatrix} 1 &0 &0 &-5 \\ 0 &1 &0& 2 \\ 0 &0 &1 &-1 \\ 0 &0 &0 &1 \\ \end{bmatrix} $$

Using the unit vectors of the coordinate axes, the resulting rotation matrix is:

$$ R= \begin{bmatrix} -0.6 &-0.8 &0 &0 \\ 0 &0 &0 &1 \\ -0.8 &0.6 &0 &0 \\ 0 &0 &0 &1 \\ \end{bmatrix} $$ Then the final matrix (the final answer for full marks) would be the rotation matrix multiplied by the translation matrix:

$$ M= \begin{bmatrix} -0.6 &-0.8 &0 &1.4 \\ 0 &0 &0 &1 \\ -0.8 &0.6 &0 &5.2 \\ 0 &0 &0 &1 \\ \end{bmatrix} $$ Is my method and final answer correct?

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Following the answer to this question, I have calculated the translation matrix:

$$ T= \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

I have calculated the rotation matrix:

$$ R= \begin{pmatrix} -0.6 & -0.8 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -0.8 & 0.6 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

But what would the final transformation matrix $M$ be? My lecture slides tell me it would be rotation matrix multiplied by translation matrix, with the last column multiplied by $-1$ except for the $1$. This would equal:

$$ M = R \times T = \begin{pmatrix} -0.6 & -0.8 & 0 & -1.4 \\ 0 & 0 & 1 & 1 \\ -0.8 & 0.6 & 0 & -5.2 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

But the answer here just adds them together somehow. Which one is correct?

Answer 1

To solve the problem, we will flow three steps, first, calculate translation matrix, second, calculate rotation matrix, third, get transformation matrix.

Because translation is a affine transformation, we need to use a 4x4 matrix to represent translation. I find you use column vector post-multiply matrix. the result of the translation matrix is same as yours.

Translation matrix =

Then, same as your idea, we use basis vectors(same as unit vectors of the coordinate axes) to calculate rotation matrix.

1. basis z vector: we can get rotated basis z vector by using a vector from VRP to point A, and normalize it.

   z = b(VRP) - a = [5,-2,1] - [1,1,1] = [4,-3,0] ,then, normalize to [0.8,-0.6,0]

2. basis x vector: we know basis vector is a vector perpendicular to z-up plane, we get the x vector by cross product up and z

   x = up x z = [1,1,1] x [0.8 ,-0.6 ,0] = [0.6, 0.8, 0]

3. basis y vector, y is prependicular to z-x plane.

   y = z x x = [0.8 ,-0.6, 0] x [0.6, 0.8, 0] = [0, 0, 1]

   finally, the rotation matrix is:

   Rotation Matrix =

   The 4x4 matrix is:

   Transformation matrix =

   The transformation matrix is a matrix from Object Space to World Space, we should invert this matrix to get World Space to Camera Space matrix.(Thanks for @JarkkoL correcting me).

   Refer to 3D Math Primer for Graphics and Game Developmen.

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Following your request for clarification: I just combine the translation matrix and the rotation matrix, actually, It's the result translation matrix(T) multiply rotation matrix(R).

The rotation is a 3x3 matrix, we can extend the matrix to a 4x4 matrix with adding a dimension, the 4x4 rotation matrix is:

And the translation matrix is:

So, now we can calculate the transform matrix by translation matrix multiply(T) rotation matrix(R):

Cause the matrix multiply formula is too big for the 4x4 matrix, we just look at the 1st row, In the transform matrix(M), the M11 element equals to T11*R11+T12*R21+T13*R31+T14*R41, cause T12, T13, R41 is 0, so, the M11 equals to R11 right? And in the same way, M14 equals to T11*R14+T12*R24+T13*R34+T14*R44, only R44 equals to 1, so, the M14 equals to T14. so, I just simply put them together in my last answer. For the reason why we put the translation matrix on the left, you can check on the book that I gave you in my previous answer on page 91(translating first and then rotating).