Equality of vectors

Analytic Geometry
Post Reply
User avatar
Riemann
Posts: 178
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Equality of vectors

#1

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
quicklatex.com-4a9d5cf71853d480c0d967a5686094ba_l3.png (3.65 KiB) Viewed 571 times

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

Tags:
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 18 guests