When you find the least squares solution you solve $$A^TA = A\vec b$$ but to find the orthogonal projection into the "subspace" A, you multiply this result (the least squares solution) with the original matrix. Why is this?
If you use the analogy with the light shining orthogonally on to the subspace and the orthogonal projection is the shadow in the subspace, isn't this shadow also the least squares solution?
You want to minimize $||Ax-b||^2$. If $A$ is a full rank matrix, the unique solution is given by $$x=(A^TA)^{-1}A^Tb$$ So $Ax=A(A^TA)^{-1}A^Tb=Hb$, with $H=A(A^TA)^{-1}A^T$ the projection matrix on the subspace formed by the columns of $A$.