Consider the distribution $\mathcal{P} = \mathcal{N}(\mu, 1)$, where the variance is known but the mean is unknown. Let $X_1,X_2\sim P$ i.i.d. In this case $T = X_1+X_2$ is a sufficient statistic.
I am trying to understand this geometrically. We know that $\begin{bmatrix}X_1\\X_2\end{bmatrix} \sim \mathcal{N} \left( \begin{bmatrix}\mu\\\mu\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix} \right)$. This is a distribution centered somewhere on the line $x_2=x_1$ with circular contours.
Now if you were given $X_1+X_2=5$ say, you know that your observation lies somewhere on the line $x_1+x_2=5$. However I don't see how the probability of a specific observation, $P(X_1=3, X_2=2|T=5; \mu)$ does not depend on $\mu$. Clearly the probability is very different if $\begin{bmatrix}\mu\\\mu\end{bmatrix} = \begin{bmatrix}3\\3\end{bmatrix}$ vs. if $\begin{bmatrix}\mu\\\mu\end{bmatrix}=\begin{bmatrix}100\\100\end{bmatrix}$.
I understand the binomial distribution case (see example 2 in http://www.stat.cmu.edu/~larry/=stat705/Lecture5.pdf) but I don't see a similar intuition for the normal distribution. What am I missing?