3. Let $a^{2}+b^{2}=1, c^{2}+d^{2}=1$ and $a c+b d=0$. Determine the value of the expression $a b+c d$?
Solution. First method. From $a^{2}+b^{2}=1$, it follows that $a$ and $b$ cannot both be zero simultaneously. Without loss of generality, we can assume that $a \neq 0$. Then $c=-\frac{b d}{a}$. We have

$$
1=c^{2}+d^{2}=\frac{b^{2} d^{2}}{a^{2}}+d^{2}=\frac{d^{2}\left(a^{2}+b^{2}\right)}{a^{2}}=\frac{d^{2}}{a^{2}}
$$

From which it follows $a^{2}=d^{2}$. Therefore,

$$
a b+c d=a b-\frac{b d^{2}}{a}=\frac{b}{a}\left(a^{2}-d^{2}\right)=0
$$

Second method. If we multiply both sides of the equation $a c+b d=0$ by $a b+c d$ sequentially, we get

$$
\begin{aligned}
& 0=(a c+b d)(a d+b c) \\
& 0=a^{2} c d+a b c^{2}+a b d^{2}+b^{2} c d \\
& 0=a b\left(c^{2}+d^{2}\right)+c d\left(a^{2}+b^{2}\right) \\
& 0=a b+c d
\end{aligned}
$$