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**adamsmurmur** Show that there is a point on the line through (2,0,0) & (0,5,0) and that is nearest to the line x=u, y=u, z=u.

Let's see if I understood: the line through $\displaystyle (2,0,0)\,,\,(0,5,0)\,\mbox{ is }\,\{(2,0,0)+t(-2,5,0)\,\,\slash\,\,t\in\mathbb{R}\}=\{(2-2t,5t,0)\,\,\slash\,\,t\in\mathbb{R}\}$ , and apparently the other line is just $\displaystyle ((s,s,s)\,\,\slash\,\,s\in\mathbb{R}\}$, so take a general point in the first line and evaluate its distance from a general point of the second line (Apparently you don't know yet/you can't use , the well known formula for the distance of a given point from a given line):

$\displaystyle \sqrt{(2-2t-s)^2 +(5t-s)^2+s^2}=\sqrt{3s^2+29t^2-6st-4s-8t+4}$ , but of course the max.-min. points of the above are the same without the square root (why?), so define $\displaystyle f(s,t)=3s^2+29t^2-6st-4s-8t+4$ and show there's a unique solution to $\displaystyle f_s=f_t=0$ (I think it is $\displaystyle t = \frac{3}{13}$...)

Tonio

I started by creating a function, F(x) = cx + d, where that f(c,d) = The sum from 1 to n of (cx_i + b - y_i)^(2) so I could minimize it. However, I just keep coming up blanks.