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Let $(\cdot,\cdot)$ be a scalar product on a finite dimensional vector space and let $0\ne v,w$ be two vectors of the same length, i.e. $(v,v)=(w,w)$.

Visually, it seams to me clear that the angle formed between $v$ and $w-v$ must be obtuse, i.e. that $(v,w-v)\le 0$.

How to prove this formally? (Or is this statement false?)

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Cauchy-Schwarz. –  Gunnar Þór Magnússon Jan 17 '13 at 11:48

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up vote 3 down vote accepted

Let $\theta$ denote the angle between $v$ and $w$, then $$ (v,w-v)=(v,w)-(v,v)=|v||w|\cos(\theta)-|v|^2=|v|^2(\cos(\theta)-1)\leq 0, $$ because $|v|=\sqrt{(v,v)}=|w|$.

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+1 Slick and cunning –  DonAntonio Jan 17 '13 at 12:03

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