<s>[INST] How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
 [/INST] I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$. 
We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = \cos(\alpha - \beta) = 0$.
You have 
$ab+cd = \cos(\alpha) \sin(\alpha) + \cos(\beta) \sin(\beta)\\
= \frac{1}{2} ( \sin(2 \alpha) + \sin(2 \beta) ) \\
= \sin(\alpha+\beta) \cos(\alpha - \beta) \\
= 0 $
The answer should be 0.
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