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Please only use the following definition of the dot product:

$u \dot{} v = u^{T}v$

Using the above definiton only (not the cosine definition) why would the dot product being zero imply the angle between the vectors is 90/-270?

This isn't homework. Please avoid use of the Pythagorean theorem, since my book uses the above result to prove it.

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  • $\begingroup$ Do you mean $\pm\pi/2$? $\endgroup$ – Olivier Bégassat Dec 3 '13 at 2:22
  • $\begingroup$ @OlivierBégassat Yes I do. Sorry about that. $\endgroup$ – dfg Dec 3 '13 at 2:34
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By assumption, we have $$||u+v||^2=(u+v)^T(u+v)=||u||^2+2u\cdot v+||v||^2=||u||^2+||v||^2$$ And then the orthogonality results from Pythagorean theorem.

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  • $\begingroup$ Could this be proved without the Phythagorean theorem? Because my book uses this result to prove the theorem. Sorry, I should have included that in my question. $\endgroup$ – dfg Dec 3 '13 at 2:35
  • $\begingroup$ @dfg Well, how do you definiton of orthogonality $\endgroup$ – Shuchang Dec 3 '13 at 2:37
  • $\begingroup$ Two vectors are orthogonal if the angle between them is 90. $\endgroup$ – dfg Dec 3 '13 at 2:38
  • $\begingroup$ @dfg But since cosine is not allowed here, how do you measure the angle? $\endgroup$ – Shuchang Dec 3 '13 at 2:40
  • $\begingroup$ Good point, forget what I said. Thanks for the help. $\endgroup$ – dfg Dec 3 '13 at 2:49

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