Projection of a vector onto a row space using formula

Question

$v =\left [ \begin{matrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{matrix} \right ] $ $ A =\left [ \begin{matrix} 1 & 2 & 2 & 1\\ 2 & 1 & 1 & 2\\ \end{matrix} \right ] $ Find projection of $v$ onto $RowA$

$A$~$\left [ \begin{matrix} 1 & 2 & 2 & 1\\ 0 & 1 & 1 & 0\\ \end{matrix} \right ] $ So $basis$ of $RowA$ is {$\left [ \begin{matrix} 1 \\ 2 \\ 2 \\ 1 \\ \end{matrix} \right ] $ ,$\left [ \begin{matrix} 0 \\ 1 \\ 1 \\ 0 \\ \end{matrix} \right ] $}.
$U=RowA$
$Proj_uv=(v.u_1/u_1^2)u_1+(v.u_2/u_2^2)u_2=$ $\left [ \begin{matrix} 4 \\ 8 \\ 11.5 \\ 11.5 \\ \end{matrix} \right ] $ But this is wrong, why? Also, I used projection matrix $A(A^TA)^{-1}A^T$ and got the correct answer $\left [ \begin{matrix} 5/2 \\ 5/2 \\ 5/2 \\ 5/2 \\ \end{matrix} \right ]$, but this is too complicated to use in exam.

Answer 1

Because the basis is not orthogonal. To use the formula, apply Gram-Schmidt process to find a orthogonal basis.