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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. – Rahul Sep 2 '11 at 6:34
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. – Rahul Sep 2 '11 at 6:36
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. – user1551 Sep 2 '11 at 8:14
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. – Did Sep 2 '11 at 10:23

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

$$\|\vec{w}\times \vec{i}\|^2 = w_j^2 + w_k^2 $$


$$ \|\vec{w}\|^2 = \|\vec{w}\cdot \vec{i}\|^2 + \|\vec{w}\times \vec{i}\|^2 $$

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.

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