Sorry if the title is vague. But I don't really know the terms I'm looking for ( Nor do I know any tags that go with it). I usually end up being confused with the little( and perhaps trivial things ) in math. The following for example:



enter image description here

What are the rules that demand it's a.b / ||a||||b|| and not ||a||||b|| / a.b ?

Perhaps this is a silly question. But, What are the rules when moving from one side of the equation to the other?

  • $\begingroup$ Because you are dividing $a\cdot b$ by $\|a\|\|b\|$, not the other way around. $\endgroup$ – Alex Becker Aug 22 '12 at 3:33
  • $\begingroup$ From never seeing the other, how would I know this? I feel like an idiot because I can't be more specific on what I'm trying to ask. $\endgroup$ – Sidar Aug 22 '12 at 3:35
  • $\begingroup$ If you want to move stuff on the left to the right, you divide by stuff on the left. If you want to move stuff on the right to the left, you divide by stuff on the right. What you are dividing by dictates what is in the denominator. $\endgroup$ – Alex Becker Aug 22 '12 at 3:36
  • $\begingroup$ The image aboth implies that moving from left to right, you divide by what is on the right. Or am I reading it wrong? $\endgroup$ – Sidar Aug 22 '12 at 3:40
  • $\begingroup$ In the image above, they divide both sides by $\|a\|\|b\|$ (which is on the right), then they apply the $\arccos$ function to both sides, then they switch the two sides. $\endgroup$ – Alex Becker Aug 22 '12 at 3:43

The general rule is, you can do anything to one side of an equation, so long as you do the same thing to the other side. So, if you have the equation, $$r=st$$ you can multiply both sides by $1/s$ (assuming that $s\ne0$) to get $$(r)(1/s)=(1/s)st$$ which becomes $${r\over s}=t$$ There's no way you can go from $r=st$ to $s/r=t$ by doing the same thing to both sides of the equation.

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$x = y \cos \theta$

$x/y = \cos \theta$

$\arccos (x/y) = \theta$


$x = y \times z$

$x \div y = z$

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  • $\begingroup$ Perhaps my example was too specific. But i meant it in a general way. $\endgroup$ – Sidar Aug 22 '12 at 3:32
  • $\begingroup$ OK, in general, if $r=st$, then $t=r/s$. $\endgroup$ – Gerry Myerson Aug 22 '12 at 3:35

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