Write vector as the sum of w and w orthogonal

Question

Let $\mathbb{R}^3$ be given with the standard inner product and let $W$ be the subspace spanned by $\left( \begin{array}{c} 4\\ -2\\ 4\\ \end{array} \right)$ and $\left( \begin{array}{c} -2\\ 6\\ 2\\ \end{array} \right)$. Write the vector $\left( \begin{array}{c} 20\\ 10\\ -9\\ \end{array} \right)$ as $w+w^{⊥}$ (orthogonal) with $w$ in $W$ and $w^{⊥}$ in $W^{⊥}$. I don't know how to solve this problem, I tried using the Gram-Schmidt method and orthogonal projection but I keep getting the wrong solution.

Answer 1

Firstly, we should find the bais of $W^{\bot}$.

If we assume $(x,y,z)^{t}\in W^{\bot}$, we have $\begin{cases} 4x-2y+4z=0\\ -2x+6y+2z=0\\ \end{cases}$.

Then $(-7,-4,5)^{t}$ is the basis of $W^{\bot}$.

We assume that $\left( \begin{array}{c} 20\\ 10\\ -9\\ \end{array} \right) =k_1\left( \begin{array}{c} 4\\ -2\\ 4\\ \end{array} \right) +k_2\left( \begin{array}{c} -2\\ 6\\ 2\\ \end{array} \right) +k_3\left( \begin{array}{c} -7\\ -4\\ 5\\ \end{array} \right) $.

Solve this system of linear equations. We have $k_1=\dfrac{3}{4}$, $k_2=\dfrac{1}{4}$ and $k_3=\dfrac{-5}{2}$.

Finally, we get $\left( \begin{array}{c} 20\\ 10\\ -9\\ \end{array} \right) =\dfrac{3}{4}\left( \begin{array}{c} 4\\ -2\\ 4\\ \end{array} \right) +\dfrac{1}{4}\left( \begin{array}{c} -2\\ 6\\ 2\\ \end{array} \right) +\dfrac{-5}{2}\left( \begin{array}{c} -7\\ -4\\ 5\\ \end{array} \right) $.