For some vector $v$, the matrix $P=vv^T$ is an orthogonal projector that projects onto the line on which $v$ lies. What is the intuitive meaning of the matrix $M$ obtained by dividing $P$ by $v^Tv$? $$ M=\frac P{v^Tv} $$
$M$ shows up in the classical definition for Gram Schmidt orthogonalization. For linearly independent vectors $v_1,\ldots,v_n$, define: $$ u_1=\frac{v_1}{\left\Vert v_1\right\Vert} $$ and $$ u_i=\left(I-\sum\limits_{j=0}^i\color{red}{\frac{v_jv_j^T}{v_j^Tv_j}}\right)v_i $$
The set of $u$ vectors is then orthonormal.
What is the purpose of dividing $v_jv_j^T$ by $v_j^Tv_j$? The outer product matrix is already an orthogonal projector. Does dividing by $v_j^Tv_j$ cause $u_i$ to have unit magnitude? (How?)
If $v$ is not unit then $v v^T$ is not even idempotent: $(v v^T)(v v^T)=\| v \|^2 v v^T$. So the normalization is needed to even make it a projector in the first place.
To put it another way, just $v v^T$ acts on the span of $v$ by multiplying by $\| v \|^2$, rather than doing nothing as it should if it is to be a projector.