The distance to a linearly independent vectors on the sphere determines a point uniquely

Question

Let $v_i \in \mathbb{S}^n \subseteq \mathbb{R}^{n+1}$ be independent vectors on the sphere.

Let $v,w \in \mathbb{S}^n \cap \text{span}\{v_i\}$ and suppose that the Euclidean distances $\|v-v_i\|=\|w-v_i\|$ for every $i$.

Then $w=v$.

Answer 1

Write $v=\sum_j x_jv_j,w=\sum_j y_jv_j$. Since $v,w,v_i \in \mathbb{S}^n$, we know that $$\langle v,v_i \rangle=\langle w,v_i \rangle. \tag{1}$$

Then $$ \sum_j x_j\langle v_j,v_i \rangle = \sum_j y_j\langle v_j,v_i \rangle, $$ or $$ \sum_j \langle v_j,v_i \rangle(x_j-y_j) =0. $$ Writing $A_{ij}=\langle v_j,v_i \rangle$, we get $A(x-y)=0$.

Since the $v_i$ are linearly independent, the Gram matrix $A$ is invertible, so $x=y$ as required.