Do you remember the solution to the following equation?

$v^T x =k$

with $v$ as a vector and $k$ as a scalar. The solution is

$x=v(v^Tv)^{-1}k+x_0$

with $v^T x_0=0$.

Today we see this problem from another view point. Rewrite $v^T x =k$ as

$v^T(x-x_1)=0$

with $x_1=v(v^Tv)^{-1}k$. So x could be x_1 or x_1+x_0. So x is a hyper-plane! This hyper-plane passes through the point $x_1$ and with normal vector as $v$. Besides among all solutions $x_1$ is the one nearest to the origin.  The geometric meaning is obvious.

In addition, if $v$ is a matrix with full row-rank, similar results can be obtained.