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I Would Thanks if u guys help me

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    $\begingroup$ Don't post images instead of formulae or text. They're not accessible to people who use screen-readers, and it means nobody can find your question in search. $\endgroup$ – Dan Hulme Feb 12 '18 at 23:00
  • $\begingroup$ thanks @DanHulme . i Did not knew that. But tell me my answer if you can see it. $\endgroup$ – saad bin sami Feb 13 '18 at 13:27
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You assume (in your expansion of $-(q\times v)q$) that $\|q\|^2 = 1$, but actually $\|q\|^2 + q_0^2 = 1$; therefore, $-(q\times v)q = (1 - q_0^2)v + q(q\cdot v)$. The result follows from there.

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