This is a projection excercise I'm stuck on. Lets say we have two planes in R3, $A$ and $B$, which both go through the origin. We also have two orthogonal projections, $ProjA$ and $ProjB$, projecting on plane $A$ and $B$ respectively. $ProjA$ is given by the projection matrix $P1$, so that $$ProjA(v) = P1v$$ $ProjB$ is specified by the following vector $d2$ = [knownx; d; known z] that is parallel to the projection direction. Note that $d2$ is parametric in $d$. The matrix $P1$ and the known $x$ and $z$ values in $d2$ are given in the task.
The task is is to find the only possible value of $d$ such that the following holds for all $v ∈ R3: ProjA(ProjB(v)) = ProjB(ProjA(v))$.
My thought process was this:
Since d2 is parallell to projection direction, then d2 is orthogonal to plane B. So,
$$ProjB(v) = v - Projd2(v)$$
where $Projd2(v)$ is the projection of v onto d2: $d2.v\over||d2||^2$$d2$
Continuing,
$$ProjA(ProjB(v)) = ProjA(v-Projd2(v)) = P1(v-Projd2(v))$$
$$ProjB(ProjA(v)) = ProjB(P1v) = P1v - Projd2(P1v)$$
Thus,
$$P1(v-Projd2(v)) = P1v - Projd2(P1v)$$
And finally,
$$P1(Projd2(v)) = Projd2(P1v) $$
Am I on the right path here?
Any tips would be appreciated.
If $a$ is the unit normal vector to plane $A$ and $b$ is the unit normal vector to plane $B$. Then
$\text{ Proj_A } (v) = ( I_3 - {aa}^T ) v $
And similarly,
$ \text{ Proj_B }(v) = (I_3 - {bb}^T ) v $
Now using the basis $ a, b, a \times b$, we can write
$ v = c_1 a + c_2 b + c_3 (a \times b) $
So that
$ \text{ Proj_A } (v) = ( I_3 - {aa}^T ) v = v - c_1 a - c_2 a (a^T b) $
$ \text{ Proj_B } (v) = ( I_3 - {bb}^T ) v = v - c_1 b (a^T b) - c_2 b$
And
$ \text{ Proj_B }( \text{ Proj_A }) (v) = (I - {bb}^T) ( v - c_1 a - c_2 a (a^T b) ) \\ = v - c_1 b (a^T b ) - c_2 b - (I - {bb}^T) ( c_1 a + c_2 a (a^T b) ) \\ = v - c_1 b (a^T b) - c_2 b - c_1 a - c_2 a (a^T b) + c_1 b (a^T b) + c_2 b (a^T b)^2 $
Exchanging $a$ and $b$, and $c_1$ and $c_2$, we get
$ \text{ Proj_A }( \text{ Proj_B }) (v) = v - c_2 a (a^T b) - c_1 a - c_2 b - c_1 b (a^T b) + c_2 a (a^T b) + c_1 a (a^T b)^2 $
Since $a,b$ are linearly independent, (otherwise we have the trivial case $a= \pm b$), then the equality of the two expressions implies by equating coefficients of $a$ and $b$:
$ - c_1 - c_2 (a^T b) = - c_2 (a^T b) - c1 + c_2 (a^T b) + c_1 (a^T b)^2 $
which reduces to
$ c_2 (a^T b) + c_1 (a^T b)^2 = 0 \hspace{20pt} \cdots (1)$
And similarly from the coefficients of $b$, we get
$ - c_1 (a^T b) - c_2 + c_1 (a^T b) + c_2 (a^T b)^2 = - c_2 - c_1 (a^T b) $
which reduces to
$ c_1 (a^T b) + c_2 (a^T b)^2 = 0 \hspace{20pt} \cdots (2)$
Since $c_1$ and $c_2$ are arbitrary, then equations (1) and (2) can only be satisfied if $a^T b = 0 $ , i.e. the two planes $A$ and $B$ are perpendicular to each other.
Therefore, from the given matrix $P_1 = I - {aa}^T$ , we have $Q = {a a}^T = I - P_1 $
From $Q = I - P_1$ we can compute the ratios $\alpha = a_2/a_1 = Q_{21}/Q_{11} $ and $\beta=a_3/a_1 = Q_{31}/Q_{11}$ , then we will $ a = t [ 1, \alpha, \beta]$, and since $a^T a = 1$ we find $t$, and hence we now have $a$.
Since $ b = [x_0, d , z_0 ] $ , then we want
$ a_1 x_0 + a_2 d + a_3 z_0 = 0 $
From which $d$ follows.