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 $ab+cd$ ?

If $b=0,ac=0\implies c=0\implies ab+cd=0$
Else $ac+bd=0\iff ac=-bd\iff\dfrac ab=\dfrac{-d}c=k$(say)
$\implies a=bk, d=-ck$
If $a^2+b^2=c^2+d^2,$ not necessarily $=1$
$b^2(1+k^2)=c^2(1+k^2)\implies b^2=c^2$ if $1+k^2\ne0$
Now $ab+cd=(bk)b+c(-ck)=k^2(b^2-c^2)=?$
