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I am reading a book on the math behind game engine. At the end of the chapter that talks about quaternions they are some exercises. I need to prove that $||q_1q_2||^2 = ||q_1||^2||q_2||^2$

It should be farely simple but I'm stuck on probably the dumbest thing. How I proceed :

Let $q_1 = v_1 + s_1$ and $q_2 = v_2 + s_2$. V is a 3d vector and s a scalar.

Then $||q_1q_2||^2 = (v_1\times v_2 +s_1v_2+s_2v_1)^2+(s_1s_2-v_1\cdot v_2)^2$

I can already see the LaGrange identity simplifying the annoying part but I apparently clearly don't understand how the distribution works in this case.

For example if I take the last part I would like to expand it as $(s_1s_2)^2+(v_1\cdot v_2)^2-2 (s_1s_2(v_1\cdot v_2))$ but the book expands at as $(s_1s_2)^2+(v_1\cdot v_2)^2$.

Of course how I distribute the terms makes me unable to give the proper answer but I don't understand why it doesn't work how I think. I am clearly missing some knowledge or I am very tired but I would need help to understand why I can't do what I want here.

Screenshot of the official answer : Answer from the book

rXp
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  • This looks strange. What is your definition of the norm? (In the standard one, this is a straightforward calculation.) – Randall Feb 17 '23 at 13:42
  • Can you do it for complex numbers? – Kurt G. Feb 17 '23 at 13:42
  • @Randall let q* be the conjugate of q : $q^* = -v + s$ then

    $qq^=q^q=v^2 +s^2$ so

    $||q||=\sqrt{qq^*}=\sqrt{v^2+s^2}$

     – rXp Feb 17 '23 at 14:44
  • The simplest proof of $|q_1q_2|^2=|q_1|^2|q_2|^2$ is obtained by representing the quaternions as complex $2\times 2$-matrices and using the fact that quaternion multiplication is ordinary matrix multiplication and $\det q$ is nothing else than the squared length of the quaternion. – Kurt G. Feb 17 '23 at 15:14
  • @KurtG. Yes I understand that, I searched on the internet the other ways of proving that but my issue is understanding how the book did it. Specifically how they expanded the second part of the equation. I added to the question a screenshot of the official answer. – rXp Feb 17 '23 at 15:22
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    What an insanely complicated proof for such a simple thing... Why not just use $| q_1 q_2 |^2 = (q_1 q_2)(q_1 q_2)^* = q_1 q_2 q_2^* q_1^* = \cdots$ instead? – Hans Lundmark Feb 17 '23 at 15:26
  • The book does not expand $(s_1s_2-v_1\cdot v_2)^2$ as $s_1^2s_2^2+(v_1\cdot v_2)^2$. Rather, the book cancels the $-2s_1s_2v_1\cdot v_2$ we get from expanding the square $(s_1s_2-v_1\cdot v_2)^2$ with the $+2s_1s_2v_1\cdot v_2$ we get from expanding the square $(v_1\times v_2+s_1v_2+s_2v_1)^2$. Do you see what happens to the other things we get from expanding $(v_1\times v_2+s_1v_2+s_2v_1)^2$? – anon Feb 17 '23 at 15:35
  • @HansLundmark I started from the multiplication $q_1q_2 =v_1 \times v_2 + s_1v_2 + s_2v_1+ s_1s_2 - v_1 \cdot v_2$. That is how the mult is represented in the book so that q is still a vector and a scalar since quaternion is used for rotation and other transforms. – rXp Feb 17 '23 at 15:36
  • @runway44 You mean that $(s_1v_2)(s_2v_1) = s_1s_2(v_1\cdot v_2)$ ? Implied multiplication of vectors is a dot product not a outer product ? If yes, it explains why I am lost. – rXp Feb 17 '23 at 15:39

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