Given a probability distribution: $$P(\pmb r(t); \pmb\mu, \Sigma) = \frac{1}{\sqrt{(2\pi)^n |\Sigma|}} \exp(-\frac{1}{2}(\pmb r(t) - \pmb\mu)^T\Sigma^{-1}(\pmb r(t) - \pmb\mu))$$ $$\pmb r(t) = \pmb \mu_0 + t\pmb{\hat{u}}$$ where $\pmb{\hat{u}}$ is a unit vector, I wonder what kind of distribution is this. Is this still a normal distribution? If it is, what is its mean and standard deviation?
I have been trying to work out the case where $\pmb\mu_0 = \pmb\mu$, but to no avail:
\begin{align*} P(\pmb y = t\pmb{\hat{u}} + \pmb\mu) &= \frac{1}{(2\pi)^{n/2} |\Sigma|^{1/2}} \exp \left( -\frac{t^2}{2} \pmb{\hat{u}}^T \Sigma^{-1} \pmb{\hat{u}} \right) \end{align*}
I tried to eigen-decompose $\Sigma^{-1}$ but cannot produce a $|\Sigma|^{-1}$ on the exponent.
With $\mu'=\mu-\mu_{0}$ we have \begin{eqnarray*} p(t)&=&\frac{1}{Z}\exp\left\{-\frac{1}{2}\left(tu-\mu'\right)^{T}\Sigma^{-1}\left(tu-\mu'\right)\right\}\\ &=& \frac{1}{Z'}\exp\left\{-\frac{1}{2}\left( \left(u^{T}\Sigma^{-1}u\right)t^{2}-2\left(u^{T}\Sigma^{-1}\mu'\right)t\right)\right\}\\ &=&\frac{1}{Z''}\exp\left\{-\frac{1}{2}\left(u^{T}\Sigma^{-1}u\right) \left(t-\frac{u^{T}\Sigma^{-1}\mu'}{u^{T}\Sigma^{-1}u}\right)^{2}\right\}. \end{eqnarray*} This is the form of a Normal distibution with mean \begin{equation*} \frac{u^{T}\Sigma^{-1}\mu'}{u^{T}\Sigma^{-1}u} \end{equation*} and variance \begin{equation*} \frac{1}{u^{T}\Sigma^{-1}u}. \end{equation*} The normalising factor $Z''$ is of course \begin{equation*} \sqrt{\frac{u^{T}\Sigma^{-1}u}{2\pi}}. \end{equation*}