<s>[INST] Consider $v=v_1+v_2$, where $v_1 \in M$ and $v_2 \in M^{\perp}$ $M = span\{\begin{pmatrix}8\\0\\-6\end{pmatrix}, \begin{pmatrix}8\\6\\-6 \end{pmatrix}\}$
I am trying to calculate $v_1$ and $v_2$ when $v=\begin{pmatrix}2 \\ 4 \\ 6 \end{pmatrix}$.
I know that since $v_1 \,\in \, M$ and $v_2 \, \in \, M^{\perp}$, $v_1\cdot v_2 = 0$.
Let $v_1 = a_1\begin{pmatrix}8\\0\\-6\end{pmatrix} + a_2\begin{pmatrix}8\\6\\-6 \end{pmatrix}$. The orthogonal basis of $M = \begin{pmatrix}8\\0\\-6\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix}$. From this, we can say that $v=a_1\begin{pmatrix}8\\0\\-6\end{pmatrix} + a_2\begin{pmatrix}8\\6\\6 \end{pmatrix}+a_3\begin{pmatrix}8\\0\\-6\end{pmatrix}+a_4\begin{pmatrix}0\\1\\0\end{pmatrix}$
But if we put this into an augmented matrix, it gives an inconsistent system (which is clearly wrong). What is the correct way to solve this?
 [/INST] It is quite easy to see that $w=\begin{bmatrix}6\\0\\8\end{bmatrix}$ is orthogonal to the basis vectors given for $M$. Thus $w \in M^{\perp}$. Moreover $\text{dim}(M)=2$ and $M \subset \Bbb{R}^3$, so $\text{dim}(M^{\perp})=1$. This means we can say that $\{w\}$ is a basis for $M^{\perp}$.
So we want to solve for $a,b,c$ such that
$$\begin{bmatrix}2\\4\\6\end{bmatrix}=\underbrace{a\begin{bmatrix}8\\0\\-6\end{bmatrix}+b\begin{bmatrix}8\\6\\-6\end{bmatrix}}_{v_1 \in M}+\underbrace{c\begin{bmatrix}6\\0\\8\end{bmatrix}}_{v_2 \in M^{\perp}}.$$
This yields the system
\begin{align*}
4a+4b+3c&=1\\
3b&=2\\
-3a-3b+4c&=3
\end{align*}
Upon solving this, we get $a=-\frac{13}{15}, b=\frac{2}{3}$ and $c=\frac{3}{5}$.
Thus
$$v_1=\color{red}{\frac{-13}{15}}\begin{bmatrix}8\\0\\-6\end{bmatrix}+\color{red}{\frac{2}{3}}\begin{bmatrix}8\\6\\-6\end{bmatrix},$$
and
$$v_2=\color{red}{\frac{3}{5}}\begin{bmatrix}6\\0\\8\end{bmatrix}$$
</s>