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Show there is a point of the plane $\{x \in \mathbb{R^3} \mid x_1 + 2x_2 + 3x_3 = 13\}$ closest to the point $(1, 1, 1)$.

Let a function $f: A \rightarrow \mathbb{R}$ defined for all $x \in A$ by $f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$.

We can take the domain $A = x_1 + 2x_2 + 3x_3 = 13 \cap B[(1, 1, 1), r]$.

If we take $r = (1, 1, \frac{10}{3})$, the plane intersects the ball.

So, this domain is bounded and closed. We show that $f$ is continuous.

So, by the extreme value theorem, there is $a \in A$ and $b \in A$ such that for all $x \in A$ : $f(a) \le f(x) \le f(b)$.

We showed that there is a point $a \in A$ that is closest to the point $(1, 1, 1)$.

The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.

Is it correct ?

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  • $\begingroup$ Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ ) $\endgroup$ – Maksim Mar 18 at 14:59

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