Exercise 2.2.17

$a=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$, the projection of $b$ on $a$ is thus $p=\frac{a^{T}b}{a^{T}a}a$, we can prove that $b-p$ is perpendicular to $a$. So we have $\widehat{x}=9$, $p=\widehat{x}a=\left[\begin{array}{c}\hfill 9\hfill \\ \hfill 9\hfill \\ \hfill 9\hfill \\ \hfill 9\hfill \end{array}\right]$, and $e=b-p=\left[\begin{array}{c}\hfill -9\hfill \\ \hfill -1\hfill \\ \hfill -1\hfill \\ \hfill 11\hfill \end{array}\right]$, and it’s easyto see that $e^{T}a=-9-1-1+11=0$, the shortest distance from $b$ to the line through $a$ is $\left|e\right|=81+1+1+121=204$.