Let $U=\op{span}\pa{\mat{1\\-1\\2}, \mat{1\\0\\3}}$ $\vec b = \mat{1\\4\\1}$ Find $\vec b_U \in U$ such that $\lvert \lvert \vec b - \vec b_{U} \rvert \rvert$ is minimized. $ \begin{align} A &= \mat {1&1\\-1&0\\2&3}\\ \mathcal B &= \left\{ \mat{1\\-1\\2}, \mat{1\\0\\3} \right\}\\ \vec b_{U} &= \mat{x_{1}\\x_{2}}_{\mathcal B} \end{align} $ $\begin{align} \vec b_{U}&= \mat{1&1\\-1&0\\2&3}\mat{x_{1}\\x_{2}} \\ \vec b &= \vec b_{U} + \vec b_{\perp} \\ A^\intercal \vec b &= A^\intercal A \vec x\\ \end{align}$