Equality of vectors

Analytic Geometry
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Equality of vectors


Post by Riemann »

In the following figure $\mathrm{AE}$ is the exterior angle bisector of the angle $\mathrm{A}$ of the triangle $\mathrm{AB} \Gamma$.

quicklatex.com-4a9d5cf71853d480c0d967a5686094ba_l3.png (3.65 KiB) Viewed 3883 times

Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\mathrm{AB}} -\gamma \overrightarrow{\mathrm{A} \Gamma} \right )$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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