Getting perpendicular distance by Gram Schmidt Process Use the Gram-Schmidt process to find the perpendicular distance from the points to the corresponding lines in the problems.
a. point $(0,0)$ to the line through $(1,1)$ and $(3,0)$
b. point $(-1,0)$ to the line $y=x$
c. point $(1,1)$ to the line through $ (-1, -1)$ and $(0,2)$
MY ATTEMPT: 
I know how to use Gram-Schmidt process to find the orthonormal bases by given two vectors. 
Should I convert given two points to get a linear equation and perform Gram Schmidt process, regarding the linear equation as the inner product? 
If I get the orthonormal basis by GS process, is the length of that orthonormal basis is the perpendicular distance from the corresponding point?
 A: To use something like Gram-Schmidt to find the distance from a point $x$ to the line $\ell$ through points $p$ and $q$, start with the vectors $v_{1} = (q - p)/\|q - p\|$ (a unit direction vector for the line) and $v_{2} = x - p$ (a displacement vector from the line to $x$). "Orthogonalize" $v_{2}$, i.e., replace $v_{2}$ by
$$
w_{2} = v_{2} - \langle v_{2}, v_{1}\rangle v_{1}.
$$
The magnitude of $w_{2}$ is the distance from $x$ to $\ell$.
A: Suppose we have a line given by a point $A$ and vector $\vec u$, i.e. the points on the line are points of the form $A+t\vec u$, where $t\in\mathbb R$.
We want to find the distance of $P$ from this line. This is the same thing as finding the length of the orthogonal projection of the vector $\overrightarrow{AP}$ into the subspace $[\vec u]^\bot$.
If you apply Gram-Schmidt to the vectors $\vec u$ and $\overrightarrow{AP}$, you will in fact find this orthogonal projection. (Using orthogonalization you find vector $\vec v=\overrightarrow{AP}-t\vec u$ which is orthogonal to $\vec u$. This means that $\overrightarrow{AP}=t\vec u+\vec v$, i.e., $t\vec u$ is the orthogonal projection of $\overrightarrow{AP}$ to the subspace $[\vec u]$ and $\vec v$ is the orthogonal projection of $\overrightarrow{AP}$ on the subspace $[\vec u]^\bot$.) 
