Does projection of 3D points reduce distances by exactly 1/3?

Question

Let $P$ be a set of $n$ random points uniformly distributed inside a unit-radius sphere centered on the origin. Orthogonally project $P$ to a random plane through the origin; call the projected points $P_{\bot}$. Let $A$ be the distance matrix for $P$, and $B$ the distance matrix for $P_{\bot}$, where an $n \times n$ distance matrix records the Euclidean distance between pairs of points. So these matrices are symmetric across a diagonal of zeros.

Finally, define the distance between matrices $A$ and $B$ as $$ d(A, B) = \frac{\sqrt{\sum_{i=1}^n \sum_{j=1}^n (a_{ij} - b_{ij})^2}}{n} \;. $$ (This matrix distance likely has a name, but I don't know it.)

Simulations suggest that $d(A,B)$ is close to $\frac{1}{3}$, independent of $n$. For example, for $n{=}200$, $50$ random trials led to $0.332$.

Q1. Is it true $d(A,B) = \frac{1}{3}$ exactly?

Q2. If so, is there some intuitive way to see this without calculation?

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          ^{$n=200$ points $P$ (blue) in $\mathbb{R}^3$ projected to $P_{\bot}$ (red).}

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Update 1. Following @Henry.L's suggestion in the comments, for a sphere of radius $r{=}2$, I find $d(A,B) \approx \frac{2}{3}$. This suggests the matrix distance might be $\frac{r}{3}$.

Update 2. A histogram of $d(A,B)$ for $n{=}100$, $5000$ random trials:

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Answer 1

No. In the large $n$ limit, this is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$.

We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute $$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ Putting these together gives $E[(a_{ij}-b_{ij})^2]=2-\frac{3\pi}{5}$.