I am not really well experienced in linear algebra, so please apologize when the solution to my question is either trivial or unfeasible. I already tried to find a solution using Lagrange multipliers (see below), but my mathematical impotence keeps from from solving the problem.
Let $\vec{w}$ be a known d-dimensional real column vector with norm $1$ ($\vec{w}^T\vec{w} = 1$) and $c$ be a known real-valued scalar. I want to find the solution for:
$$\vec{w}^T\vec{x} = c$$ subject to the constraint $$\vec{x}^T\vec{x}=1$$ (so $\vec{x}$ should be a unit norm vector).
I came up with the Lagrangian minimzation problem
$$\min f(x, \lambda) = \frac{1}{2}(\vec{w}^T\vec{x} - c)^2 - \lambda(\vec{x}^T\vec{x} - 1)$$.
Finding the partial derivatives w.r.t. $\vec{x}$ and $\lambda$ leads to (if I am correct):
$$\frac{\partial f}{\partial \vec{x}} = \vec{w}^T(\vec{w}^T\vec{x} - c) - 2\lambda\vec{x} = \vec{0}$$ and $$\frac{\partial f}{\partial \lambda} = -\vec{x}^T\vec{x} + 1 = 0$$
But from here, I really don't know how to proceed. Maybe someone could help out?
Thanks a lot in advance!
Consider what happens when $d = 3$. The set of $\vec x$ for which $\vec w^T \vec x = c$ is a plane, and the set of $\vec x$ for which $\vec x^T \vec x = 1$. By drawing yourself a picture, you should be able to see that there are 3 possibilities:
Whether we are in situation 1,2, or 3 depends on the vector $\vec w$ and the constant $c$. In situation 1, there is no solution. In situation 2, there are infinitely many solutions. In situation 3, there is exactly one solution.
With a bit more work, we can actually come to the following conclusion: if $c = 0$, then we are in situation 2 (and the plane cuts through the center of the sphere). For $c \neq 0$, situation 1 occurs when $\|\vec w\| < c$, situation 2 occurs when $\|\vec w\| > c$, and situation 3 occurs when $\|\vec w\| = c$. This holds in all dimensions $d \geq 3$.