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: You can interptet $(a, b)$ and $(c, d)$ as two orthogonal vectors that lie on the unit circle. Converting this into polar coordinates, this means there are angles $\phi$, $\theta$ such that $(a, b) = (\cos(\phi), \sin(\phi))$, $(c, d) = (\cos(\theta), \sin(\theta))$ and $|\phi - \theta| = \frac{\pi}{2}$. Now observe that 
$$ab + cd = \cos(\phi)\sin(\phi) + \cos(\theta) \sin(\theta) = \sin(\phi + \theta)\cos(\phi - \theta) = 0$$
because $\cos\left(\pm \frac{\pi}{2}\right) = 0$.
A: $$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$
A: HINT:
$$ac+bd=0\iff\dfrac ad=\dfrac b{-c}=\pm\sqrt{\dfrac{a^2+b^2}{d^2+(-c)^2}}$$
But if $a^2+b^2=d^2+c^2,$ not necessarily $=1$
$$\dfrac ad=\dfrac b{-c}=\pm1$$
So, either $a=d,b=-c$ or $a=-d,b=c$
The result should follow immediately.
A: 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.
A: 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 $ab+cd=(bk)b+c(-ck)=k^2(b^2-c^2)=?$
A: 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*}
A: Now since $a^2+b^2=c^2+d^2=1$ (a,b) and (c,d) are points on unit circle. You can think of these ordered pairs as vectors, since their dot product is zero. Then this means they are orthogonal. Now A x B gives area of triangle under first vector and C x D gives area under second vector. These triangles are similar since angles are all same, so we can use similarity principle (I'm not sure this is correct translation). Since hypothenuse's are same, meaning triangles are same, so their areas. But making 90 degree rotation means that in second vector x or y coordinate will change sign. So A x B = - C x D
https://i.hizliresim.com/nglMoN.jpg
A: Let $(a,b)$ and $(c,d)$ be points on the unit circle. As shown in other comments, $ac+bd$ is generally $0$. But, when the point $(a,b)=(c,d)$ is on the axes, then the answer is $1$. 
$a=c=1, b=d=0$
$ac+bd=1$
