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If have $n$-dimensional vectors in Euclidean space $\mathbf{a},\mathbf{b},\mathbf{c}$ such that $\mathbf{a}+\mathbf{b}=\mathbf{c}$.

If I were to rotate $\mathbf{c}$ to $\tilde{\mathbf{c}}$, an arbitrary point on the $n-1$ sphere of radius $\vert \mathbf{c} \vert$ can I form $\tilde{\mathbf{c}}=\tilde{\mathbf{a}}+\tilde{\mathbf{b}}$ such that $$ \vert\tilde{\mathbf{a}}\vert=\vert\mathbf{a}\vert \\\vert\tilde{\mathbf{b}}\vert=\vert\mathbf{b}\vert \\\langle \tilde{\mathbf{a}}, \tilde{\mathbf{b}}\rangle=\langle \mathbf{a}, \mathbf{b}\rangle $$

It seems self-evident that I should be able to construct a congruent triangle after the rotation of a side, but not sure how to prove that.

Pablitorun
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    Just apply the same rotation to a,b that you applied to c. Does this answer your question? – Martin Brandenburg Jan 30 '25 at 11:20
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    What is your definition of rotation in $n$-dimensional space where $n>3$? – CroCo Jan 30 '25 at 11:38
  • Using the definition from https://math.stackexchange.com/a/3699057/7605 for a Euclidean space.

    I guess the question I really have then is does the transformation exist for $\mathbf{c}$ to every point on the $(n-1)$-sphere of radius $\vert c \vert$, again this seems obvious, but trying to formalize it.

     – Pablitorun Jan 30 '25 at 11:45

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