Distance between line and point in vector form 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.
 A: For the expression on the right hand side to be correct (as the square of the distance), you have to assume that $\vec{a}$ is a unit vector. Then as Rahul wrote, your left-hand-side becomes simply
$$ [(\vec{p}-\vec{b})\cdot\vec{a}]^2 + \|\vec{p}-\vec{b}\|^2 $$
And this expression is still incorrect, but could be easily fixed, if you insert a minus sign
$$ -[(\vec{p}-\vec{b})\cdot\vec{a}]^2 + \|\vec{p}-\vec{b}\|^2 $$
That the left and right hand sides equal is just the following statement: given an orthonormal basis $\vec{i}, \vec{j}, \vec{k}$, any vector can be written as
$$ \vec{w} = w_i\vec{i} + w_j\vec{j} + w_k\vec{k} $$
The norm satisfies
$$ \|\vec{w} \|^2 = w_i^2 + w_j^2 + w_k^2 $$
and note that $w_i = \vec{w}\cdot \vec{i}$. Now, consider $\vec{w}\times \vec{i} = - w_j \vec{k} + w_k \vec{j}$. So you have that 
$$\|\vec{w}\times \vec{i}\|^2 = w_j^2 + w_k^2 $$
So 
$$ \|\vec{w}\|^2 = \|\vec{w}\cdot \vec{i}\|^2 + \|\vec{w}\times \vec{i}\|^2 $$
holds for any unit vector $\vec{i}$. 
Geometrically this idea is simple: give the vector $\vec{w}$ and a unit vector $\vec{a}$, you can always decompose $\vec{w}$ into a vector that is parallel to $\vec{a}$ and one is orthogonal. And the squared norm of $\vec{w}$ will be equal to the sum of the square norms of the two components by Pythagorean theorem. 

For actually deriving this statement using multivariable calculus, I would recommend following Didier's hint and minimize the square of the distance instead. 
