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 $ab+cd$ ?

A solution by Sumit Ray
$ac=-bd$
$\frac{a}{b} = -\frac{d}{c} = k$
$a=bk \text{ and }d=-ck$
$a^2+b^2=1\implies b^2 = \frac{1}{k^2+1}\implies c^2 = \frac{1}{k^2+1}$
Thus $b^2 - c^2 =0$
Now \begin{align*}ab+cd &= b^2\cdot k-c^2\cdot k\\&= k(b^2-c^2)
          = 0\end{align*}
