Geometric interpretation of homogeneous coordinate transformations

I have a matrix $\boldsymbol{B} = \begin{bmatrix}\boldsymbol{x}&\boldsymbol{y}&\boldsymbol{z}&\boldsymbol{o_B}\end{bmatrix}$ $\equiv$ $\boldsymbol{I}$, where $\boldsymbol{x}$, $\boldsymbol{y}$ and $\boldsymbol{z}$ are orthogonal vectors, $\boldsymbol{o_B}=\begin{bmatrix}0&0&0&1\end{bmatrix}^T$ is the origin of the basis formed by $\boldsymbol{x}$, $\boldsymbol{y}$, $\boldsymbol{z}$ and $\boldsymbol{I}$ is the identity matrix. Another matrix $\boldsymbol{A}$ is defined as $\boldsymbol{A} = \begin{bmatrix}\boldsymbol{a}&\boldsymbol{b}&\boldsymbol{c}&\boldsymbol{o_A}\end{bmatrix}$, where $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$ are again orthogonal vectors and $\boldsymbol{o_A}=\begin{bmatrix}0&0&z_A&1\end{bmatrix}^T$ is the origin of the basis formed by $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$.

Matrix $\boldsymbol{A}$ can be seen as a translation of the matrix $\boldsymbol{B}$ and therefore,

$$\boldsymbol{A} = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&z_A\\0&0&0&1\end{bmatrix}\boldsymbol{B}$$

As shown in the image, a vector $\boldsymbol{r_A}=\begin{bmatrix}a&b&0&1\end{bmatrix}^T$ has its tail at $\boldsymbol{o_A}$ and the red colored vector has its tail at $\boldsymbol{o_B}$. This red colored vector is given as $\boldsymbol{BAr_A}$, which is a transformation of $\boldsymbol{r_A}$ first by $\boldsymbol{A}$ and then by $\boldsymbol{B}$.

My question is regarding the red vector, how is this vector being given as $\boldsymbol{BAr_A}$ (transformation of $\boldsymbol{r_A}$ first by $\boldsymbol{A}$ and then by $\boldsymbol{B}$)?. I am unable to get a geometrical understanding of this transformation.

I see $\boldsymbol{r_A}$ as a linear combination of the basis vectors given by matrix $\boldsymbol{A}$ and some matrix $\boldsymbol{\alpha}$ of the coefficients: $$\boldsymbol{r_A}= \boldsymbol{A^T\alpha}$$

so I guess $\boldsymbol{BAr_A}$ becomes, $$\boldsymbol{BAr_A} = \boldsymbol{BAA^T\alpha}$$ and if $\boldsymbol{A}$ is an orthogonal matrix (which it does not look like) then it satisfies $\boldsymbol{AA^T=I}$, making $\boldsymbol{BAr_A} = \boldsymbol{B\alpha}$, which does not make sense to me.

• A brief comment on your last paragraph: $A$ is not orthogonal, although its linear part—the upper-left $3\times3$ submatrix—is. – amd Dec 6 '17 at 23:25
• Yes, A is not orthogonal. Then how is this transformation linking the vector $\boldsymbol{r_A}$ to the B basis with the red colored vector? – dykes Dec 7 '17 at 12:31