In Prasolov's "Geometry", problemset 1, problem 1.10 in the solution the author states:
Suppose that set of vectors of unit length $a_1 ... a_n$ is linearly dependent. Then, there exist coefficients $y_1 ... y_n$ such that $\sum_{k=1}^{n}y_ka_k=0_n$ and sum of squares of the coefficients sums up to 1 (i.e. $(y_1...y_n)$ is also a vector of unit length).
The space we are talking about is $\mathbb{R}^n$. I'm having trouble proving for myself why such coefficients should exist.
Since the vectors are linearly dependent, you can always find a non-trivial solution to the following system
$$\sum_{i=1}^ny_i\bar{a_i} = \bar{0}$$
Now, let a non-trivial solution be expressed as follows
$$\bar{y_{0}} = (y_1, y_2, y_3..., y_n)$$
Then the following will also be a non-trivial solution
$$\bar{y} = \left(\frac{y_1}{\sqrt{\sum y_i^2}}, \frac{y_2}{\sqrt{\sum y_i^2}}, ...\frac{y_n}{\sqrt{\sum y_i^2}}\right)$$