Consider the set of all unit vectors of the form $ S = (0, x_1, ..., x_n)$. If $\mathbf{e}_1$ and $\mathbf{e}_2$ are members of an orthonormal set, does $\alpha_1 \mathbf{e}_1+ \alpha_2 \mathbf{e}_2$ always belong to $S$ for some $\alpha_1, \alpha_2$ in $\mathbb{R}$?
Since $S$ is comprised of unit vectors, we will need $\alpha_1^2 + \alpha_2^2 = 1$ by considering the dot product $(\alpha_1 \mathbf{e}_1+ \alpha_2 \mathbf{e}_2) \cdot (\alpha_1 \mathbf{e}_1+ \alpha_2 \mathbf{e}_2) $. If both first components of $\mathbf{e}_1$ and $\mathbf{e}_2$ are 0, then $\alpha_1 \mathbf{e}_1+ \alpha_2 \mathbf{e}_2$ clearly belongs to $S$. If one first component of $\mathbf{e}_1$ or $\mathbf{e}_2$ is 0, then the other $\alpha$ can be set to $0$ to show that the $\alpha_1 \mathbf{e}_1+ \alpha_2 \mathbf{e}_2$ belongs to $S$. What about the case where the first components of both $\mathbf{e}_1$ and $\mathbf{e}_2$ are nonzero?
Let $x_1$ be the first component of $\mathbf{e}_1$ and let $x_2$ be the second component. Take$$\lambda=\frac1{\sqrt{x_1^{\,2}+x_2^{\,2}}}.$$Then $(-\lambda x_1)^2+(\lambda x_2)^2=1$, and the first component of $\lambda x_2\mathbf{e}_1-\lambda x_1\mathbf{e}_2$ is $0$.