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If I have two vectors A and B and A.B=0 then A and B are perpendicular.

I would like to know why.

Thank You.

Jean Marie
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user2860452
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  • This is by definition for arbitrary inner product spaces. If you are talking specifically about euclidian space, then note that $\langle a,b\rangle = |a||b|\cos(\theta)$ which for nonzero $a$ and $b$ is equal to zero iff $\cos(\theta)=0$ which occurs precisely when $\theta=\pi/2$ or $3\pi/2$ – JMoravitz Apr 30 '17 at 20:01
  • I am pretty sure the OP does not mean arbitrary inner product spaces but the Euclidean space. Also, I am pretty sure the OP has an idea of orthogonality that is not based on scalar products – Bananach Apr 30 '17 at 20:09
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    This could be answered at different levels of complexity. Very very hand wavy the dot product measures the shadow one vector casts on another. Being perpendicular there is no shadow cast. I am aware this is unsatisfactory as a full explanation but it may give a flavour. – Karl Apr 30 '17 at 20:16
  • I would like to know in analytic geometry – user2860452 Apr 30 '17 at 20:28
  • As was noted, there are several different ways to show this. Give us your definition of the scalar product so that we know which one is relevant. – amd Apr 30 '17 at 22:46
  • https://googleweblight.com/?lite_url=https://en.m.wikipedia.org/wiki/Dot_product&ei=kxLbIxdk&lc=pt-BR&s=1&m=99&host=www.google.com.br&ts=1493600253&sig=AJsQQ1C1YKXnRnnDLxaeDamXnKufFrGzSQ I don't know really if that is what you want – user2860452 May 01 '17 at 00:58
  • I wanted the definition that you are starting from, not a wikipedia entry that includes several. That page that you linked answers your question, by the way. – amd May 01 '17 at 01:04

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