I know we need to get the inverse of transformation matrix first to get the coordinate relative to other frames. Explanation is in this video.

I am reading directx11 book. And the following ones are the problem and solution in this book.

  1. Suppose that we have frames \$A\$ and \$B\$. Let \$p_A=(1,-2,0)\$ and \$q_A=(1,2,0)\$ represent a point and force, respectively, relative to frame \$A\$. Moreover, let \$Q_B=(-6,2,0)\$, \$u_B=\left(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right)\$, \$v_B=\left(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right)\$ and \$w_B=(0,0,1)\$ describe frame \$A\$ with coordinates relative to frame \$B\$. Build the change of coordinate matrix that maps frame \$A\$ coordinates into frame \$B\$ coordinates, and find \$p_B=(x,y,z)\$ and \$q_B=(x,y,z)\$. Draw a picture on graph paper to verify that your answer is reasonable.

$$\begin{align} p_B &= \begin{bmatrix}1&-2&0&1\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&0&0 \\ -\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&0&0 \\ 0&0&1&0 \\ -6&2&0&1\end{bmatrix} \\ &= \begin{bmatrix}\frac{3}{\sqrt 2}-6&-\frac{1}{\sqrt 2}+2&0&1\end{bmatrix} \\ &\approx\begin{bmatrix}-3.88&1.29&0&1\end{bmatrix} \end{align}$$

I cannot understand how the coordinate relative to frame \$B\$ can be figured out by multiplying the coordinate relative to frame \$A\$ to the transformation matrix, not a inverse one.

Is there any material to study more or did I understand the problem in wrong way?


1 Answer 1


Note that the given \$Q_B, u_B, v_b, w_b\$ vectors are defined "with coordinates relative to B". That means they're already baking-in the inverse of frame B.

Whoever wrote down that \$Q_B\$ position vector in "coordinates relative to B" had to find the origin of A in the world, and inverse-transform that point by frame B to find the corresponding point in frame B's coordinates, and similarly for the other offset vectors.

All this problem has really done is just give you the rows of the matrix \$M = A(B^{-1})\$ and asked you to assemble them and use the resulting matrix to transform two vectors.

They don't give you any description of A and B in terms of a third coordinate system, so you don't have the inputs you'd need to compute the inverse of B on your own anyway. All you have is the relative transformation from points in A to B.

If you took the inverse of matrix \$M\$, you'd get \$M^{-1} = B(A^{-1})\$, the relative transformation from points in B back to A.


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