Prove projection is equal to combination of vectors

Question

Let $v_1, v_2 \in \mathbb{R}^3$ such that $v_1, v_2$ unit vectors and orthogonal. The plane P is defined by $P = \text{span } \{v_1, v_2\}$. Prove that for any $\overrightarrow{x} \in \mathbb{R}^3$ we have $\text{proj}_{P}(x) = (x \cdot v_1)v_1 + (x \cdot v_2)v_2$

I got $\text{proj}_{P}(x) = \text{perp}_{P}(x) = x - \text{proj}_{n}(x) = x - \frac{\overrightarrow{x} \cdot \overrightarrow{n}}{||n||^2}\overrightarrow{n}$, where $x$ is the vector and $n$ is the normal vector for $P$.

How do I proceed now?

I know that:

$n \cdot \text{proj}_{P}(x) = 0$ , but that isnt helping.

Answer 1

We have $$x=\alpha_1v_1+\alpha_2v_2+n\quad\hbox{and}\quad proj_P(x)=\alpha_1v_1+\alpha_2v_2$$ for some constants $\alpha_1,\alpha_2$ and some vector $n$ which is perpendicular to both $v_1$ and $v_2$. Take the dot product of the first equation with $v_1$ and see what you get.

Good luck!