Three unit vectors $a,b,c$ are given on the plane. Prove that one can choose the signs in the expression $x=\pm a\pm b\pm c$ so as to obtain a vector $x$ with $|x|\le\sqrt2$.
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Tags: Vectors, geometry, vector
29.08.2023 07:09
is it true? if $a=b=c$,then$|x|=3>\sqrt2$
29.08.2023 15:48
@bove we can change the signs; if $a=b=c$ we can choose $x = a + b - c = a$ so $|x| = |a| = 1 \le \sqrt2.$
Quite a nice problem oops solution is wrong
30.08.2023 11:56
great thanks! got your points:$|x|\leq\sqrt3$,! and my another question is how to assure $|x|\leq\sqrt2$? why $|x|\leq\sqrt2$ is always achievable?
30.08.2023 17:01
@2above $\sqrt{3}\geq\sqrt{2}$ so your bound is not tight enough. I think one also must make use of the fact that we are confined to a plane, since the bound is not true in $\mathbb{R}^3$.
31.08.2023 01:20
natmath wrote: @2above $\sqrt{3}\geq\sqrt{2}$ so your bound is not tight enough. I think one also must make use of the fact that we are confined to a plane, since the bound is not true in $\mathbb{R}^3$. oh, oops, my bad