As we can see in this question, if $u, v, w \in \mathbb{R}^2$ are such that $u \perp v$ and $v \perp w$ then $u$ and $w$ are parallels. This occurs in inner product spaces? More precisely:
Question. Let $H=(H,(\cdot, \cdot))$ be an inner product space and $u,v,w \in H$ such that $$ (u,v)=0 \quad \text{and} \quad (v,w)=0, $$ then $u= \alpha \cdot w$, for some $\alpha \in \mathbb{R}$? If so, how to prove?
I don't see how to prove, since in $\mathbb{R}^2$ we use the fact that $\{u,v\}$ forms a basis, the we can use the orthogonality condition in order to obtain the desired. But how to recover the basis concept in $H$, since we might have that $\text{dim}(H)=\infty$.
This clearly need not be true in higher dimensions. In $\Bbb R^3$, take
$$u = (1,0,0) \qquad v = (0,1,0) \qquad w = (0,0,1)$$
with the usual inner product.