If $ a^2 + b^2=1,c^2+d^2=1,ac+bd=0 $ , find $ab+cd $?

Question

I have tried everything except trigonometry.We have not still started doing trigonometry in classes. I don't know how am I supposed to solve this,since I put 5 hours and nothing led to solution.

Answer 1

If $b\ne 0$ then $d=-{ac\over b}$ so $$ c^2+{a^2c^2\over b^2}=1\implies c^2=b^2$$

Now $$ab+cd = ab-{ac^2\over b} = {a(b^2-c^2)\over b}=0$$

Now try if $b=0$ by yourself.

Answer 2

$ab+cd=0$ because \begin{align*} cd &=cd(a^2+b^2)\\ &=a(ac)d+(bd)bc\\ &=a(-bd)d+(-ac)bc\\ &=-ab(d^2+c^2)\\ &=-ab. \end{align*}

Answer 3

Let $$u=\begin{pmatrix} a \\ b\end{pmatrix}.$$ The set of vectors orthogonal to $u$ is $$\Bigl\{\begin{pmatrix} bt \\ -at\end{pmatrix}:t\in\mathbb{R}\Bigr\}.\tag{$1$}$$ Only two vectors in this set have unit length: $$\begin{pmatrix} \pm b \\ \mp a\end{pmatrix}.$$ So $\begin{pmatrix} c \\ d\end{pmatrix}$ must be one of these two, and $ab+cd=0$ follows wither either choice from direct calculation.

To see why $(1)$ holds, we can just check that the solution set to the matrix equation $$\begin{pmatrix} a & b\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} = (0)$$ is $$ \Bigl\{\begin{pmatrix} bt \\ -at\end{pmatrix}:t\in\mathbb{R}\Bigr\}.$$