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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.

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    Wait a minute. Either $\vec a$ is always a unit vector, in which case the $2/\lVert \vec a \rVert$ on the left is a constant, or $\vec a$ is not a always unit vector, in which case I can make the "distance" on the right as large or small as I want by rescaling $\vec a$. Something is wrong with the question.2011-09-02
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    Anyway, you could definitely try the strategy you mention in your last paragraph. You'll get a value for $t$, which you then plug back into $\lVert \vec x(t) - \vec p \rVert$, and you should end up with a reasonable-looking vector expression.2011-09-02
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    Since $\vec{b}=\vec{x}(0)$ lies on the line, unless $\vec{b}$ is the perpendicular foot from $\vec{p}$, the quantity $\|\vec{p}-\vec{b}\|^2$ and hence LHS must be strictly greater than the (squared) distance from $\vec{p}$ to the line. So something is really wrong with the question.2011-09-02
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    One can recommand to minimize $\|x(t)-p\|^2=\|a\|^2t^2-2t\langle a,p-b\rangle+\|p-b\|^2$ with respect to $t$, instead of $\|x(t)-p\|$. These are logically equivalent but the first one is computationally trivial. // Another reason, not yet mentioned, that the expression for the minimal distance written in the post cannot be right is that it scales as the square of a distance.2011-09-02

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