Suppose $v$ is a fixed vector in $\mathbb R^n$, and let $u\in S^{n-1}$ (unit sphere in $n$ dimensions; $S^{n-1}=\{x\in\mathbb R^n:\|x\|=1\}$) be uniformly generated. What is the distribution of $\langle v, u\rangle u$?
My hunch is that this is uniformly distributed on $D^{n}(\|v\|)$, the disk on $n$ dimensions with radius $\|v\|$. Here, $D^n(r)=\{x\in\mathbb R^n: \|x\|\leq r\}$. I am not entirely sure how to address this though.
How does this generalize for $\langle u_1, v\rangle u_1 + \langle u_2, v\rangle u_2$? Here $u_1,u_2$ are randomly picked orthonormal vectors. So, $\langle u_1, u_2\rangle =0$.
It would be great if there is a resource or reference for these kinds of distribution theories.
EDIT: I think my conjecture was wrong. Suppose for simplicity $\|v\|=1$. For $n=2$, write $v=(\cos(\alpha),\sin(\alpha))$ and $u=(\cos(\theta),\sin(\theta))$ where $\theta\sim Unif(-\pi, \pi)$ and $\alpha$ is fixed. Then, $\langle u, v\rangle u=(\cos(\theta-\alpha)\cos(\theta), \cos(\theta-\alpha)\sin(\theta)$. This doesn't seem to be uniformly distributed on $D^n$ (take $\alpha = 0$) but I am not sure.
It’s enough to consider the case $v=(1,0,\ldots,0)$ since the general case can be obtained from that by rotating and scaling.
Then $w=\langle v,u\rangle u=u_1u$. To find the resulting shape, invert $w_1=u_1^2$ to $u_1=\sqrt{w_1}$ and consider the inverse map $u=\frac w{u_1}=\frac w{\sqrt{w_1}}$. Since $\|u\|^2=1$, this yields $\frac{\|w\|^2}{w_1}=1$ and thus $\sum_i{w_i^2}=w_1$, or
$$\left(w_1-\frac12\right)^2+\sum_{i\ne1}w_i^2=\left(\frac12\right)^2\;.$$
Thus, the resulting vectors lie on the sphere of radius $\frac12$ centred at $\left(\frac12,0,\ldots,0\right)$. To obtain the density, consider the Jacobian $\left|\frac{\partial w_i}{\partial u_j}\right|$. This is a lower triangular matrix with non-zero entries in the first column and on the diagonal, so its determinant is just the product of the diagonal entries, which are $2u_1$ in the first entry and $u_1$ in the remaining ones, yielding the determinant $\left|\frac{\partial w_i}{\partial u_j}\right|=2u_1^n$. Since the distribution on the original sphere is uniform, the density on the transformed sphere is proportional to
$$\frac1{2u_1^n}=\frac12w_1^{-\frac n2}\;.$$