Given that $m^2+n^2=1$, $p^2+q^2=1$, and $mp+nq=0$, how much is $mn+pq$? without using 
$(m,n)\perp(p,q)$ 

 A: $$\frac{m}{n}=-\frac{q}{p}=k\implies m=kn,\ q=-pk\\\implies n=\frac{1}{\sqrt{k^2+1}},\ m=\frac{k}{\sqrt{k^2+1}},\ p=\frac{-1}{\sqrt{k^2+1}},q=\frac{k}{\sqrt{k^2+1}}$$Then $mn+pq=0$.
A: The answer is obtained using trigonometry. 
The condition $x^2 + y^2 = 1$ implies $x = \cos\theta$ and $y = \sin\theta$. Write
$$ m = \cos \alpha \ , \quad n = \sin\alpha $$
$$ p = \cos \beta \ , \quad q = \sin\beta $$
Then
$$ mp + nq = \cos\alpha cos\beta + \sin\alpha \sin \beta = \cos(\alpha - \beta) $$
$$ mn + pq = \cos\alpha \sin\alpha + \cos\beta \sin \beta $$
You are given that $\cos(\alpha-\beta) = mp + nq = 0$ and therefor $\alpha - \beta = \pm \pi/2$ (since $\cos(\pm \pi/2)=0$). Then $\alpha = \pi/2 + \beta$ and
Then
$$ 
mn + pq = \cos(\pi/2 + \beta) \sin(\pi/2 + \beta) + \cos\beta \sin \beta = \sin(-\beta) \cos(-\beta) + \cos\beta \sin \beta = -\sin\beta\cos\beta + \cos\beta \sin \beta = 0 $$
A: Think in terms of angles. Assume $m=\cos\theta$ and $n=\sin\theta$. We than have $p=\cos{(\pi/2+\theta)}=-\sin\theta$ and $q=\sin{(\pi/2+\theta)}=\cos\theta$ and therefore $mn+pq=0$
