I am not asking anyone to do this for me. This question pops out of the blue, the ones before and after are trivial in comparison. I need hints:
If $\vec{p}$ is a fixed point and $\vec{x}(t) = \vec{a}t+\vec{b} $ is a line then show that the distance between the $\vec{p}$ and the line is
$\left( (\vec{p}-\vec{b})\cdot \vec{a}\right)^2 \left( 1 - \frac{2}{\|a\|} \right)^2 + \left\|\vec{p}-\vec{b}\right\|^2 = \left\| \vec{a}\times (\vec{p}-\vec{b})\right\|^2$
The expression on the RHS is intuitively straightforward. But I cannot call in any intuition, or even a method, for the LHS.
If this weren't such a formidable looking purely vector expression, I would have proceeded to find the minimum of $\left\|\vec{x}(t)-\vec{p}\right\|$. But I don't think that would work here, as this is a purely vector equation.
Update, I did it and get the expression,
$\sqrt{ ||\vec{p}-\vec{b}||^2 - \left(\frac{(\vec{p}-\vec{b} )\cdot \vec{a}}{||\vec{a}||}\right)^2} = \frac{||\vec{a}\times (\vec{p}-\vec{b})||}{||\vec{a}||}$
I am pretty sure this is the correct version of the above quoted expression.