missing basis and finding coordinates of vector with respect to basis Vectors $w_1, w_2, w_3$ form an orthogonal basis for $R^3$. Given that $w_1 = \begin{pmatrix} 2\\3\\5 \end{pmatrix}$, what are the coordinates of the vector $v=\begin{pmatrix}0\\1\\2\end{pmatrix}$ with respect to the basis?
I am not sure how to begin this problem. I believe I should find $w_2$ and $w_3$ first, and I know that $w_1 \cdot w_2 = 0$, $w_2 \cdot w_3 = 0$ and $w_3 \cdot w_1 = 0$, but I'm not quite sure how that helps me exactly, given that there are 6 unknown variables (entries) from the basis. Any help would be really appreciated.
 A: Hint: Can you think of any nonzero vector that is orthogonal to $(2,3,5)$? Let this be $w_2$. 
Then take $w_3=w_1\times w_2$.
A: As stated, the problem cannot be solved. Suppose that $v=\alpha_1w_1+\alpha_2w_2+\alpha_3w_3$. Then the coefficients $\alpha_2$ and $\alpha_3$ cannot be both different from $0$ (since $v$ is not a multiple of $w_1$). If $\alpha_2\ne0$, then$$v=\alpha_1w_1+\alpha_2w_2+\alpha_3w_3=\alpha_1w_1-\alpha_2(-w_2)+\alpha_3w_3$$and $\{w_1,w_2,w_3\}$ is still an orthogonal basis, but now you have different coefficients. And the same thing happens if $\alpha_3\ne0$.
A: For $w_2$ you have that
$$
\eqalign{
  & w_{\,2}  = \left( {\matrix{   a  \cr   b  \cr    c  \cr } } \right)\;\;
 :\quad 2a + 3b + 5c = 0\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad w_{\,2}
   = \left( {\matrix{   a  \cr   b  \cr    { - \left( {2a + 3b} \right)/5}  \cr 
 } } \right) \cr} 
$$
meaning that you are left two degree of freedom as it should be.
Then for $w_3$ you can take
$$
w_{\,3}  = \lambda \,w_{\,1}  \times w_{\,2} 
$$
where you can limit $c$ to be positive if you need to keep the chirality according to the "right-hand" rule.
You have in total two degree of freedom and a ascale parameter.
After that you can express $v$ in such a basis, leaving the parametrs free.
