$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0$, compute $ab+cd$

Question

Consider that $\{a,b,c,d\}\subset \Bbb R$ and it is known that $$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0.$$

Compute the value of $ab+cd$. (Edit: originally written with a typo as $ab+bd$),

This is from a list of training problems used within the preparation for the "Cone-Sul" math olympics. I tried to solve using newton identities, but without success. Any help will be appreciated. Sorry if this is a duplicate.

Answer 1

$$(ab+cd)^2=(ab+cd)^2-(ac+bd)^2=a^2b^2+c^2d^2-a^2c^2-b^2d^2=$$ $$=(a^2-d^2)(b^2-c^2)=-(b^2-c^2)^2.$$ Thus, $$ab+cd=0.$$

Answer 2

If $ a^2 + b^2 = 1$, then $\exists \theta \in \mathbb{R}/ \cos \theta = a \text{ and } \sin \theta = b$.

Similarly, $ \exists \phi \in \mathbb{R}/ \cos \phi = c \text{ and } \sin \phi = d$.

Then, you can translate $ac+bd$ and $ab+bd$.

Answer 3

One approach is to write the $\{a, b, c, d\}$ in terms of sines and cosines and utilise the angle sum formulae.

Thus write \begin{align} a &=\sin x \\ b &=\cos x\\ d&=\sin y \\ c &=\cos y \end{align}

and computing $ac+bd=\sin(x+y)$, also since $\cos(x+y)=bc-ad$ we obtain $$a^2+b^2=1,c^2+d^2=1\implies (ac+bd)^2+(ad-bc)^2=1$$

Alternatively, one could infer form complex numbers that $|z|=|w|=1\implies |zw|=1$. We can show more generally that the complex norm is multiplicative, of course.

Answer 4

Original: There is not information to compute $ab+bd$. Here are some examples that satisfy your conditions with different $ab+bd$: $$ \begin{array}{cccc|c} a & b & c & d & ab+bd \\ \hline 1 & 0 & 1 & 0 & 0 \\ 0.8 & 0.6 & -0.6 & 0.8 & 0.96 \\ 0.8 & -0.6 & 0.6 & 0.8 & -0.96 \end{array} $$

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After correction to question: $ac+bd$ is the determinant of $$ \begin{pmatrix} a & -d \\ b & c \end{pmatrix} $$ where each column is a unit vector in $\mathbb R^2$ due to $a^2+b^2=c^2+d^2=1$. So the only way for this determinant to be $0$ is for the two columns to be either equal or each other's negative.

In the case $a=-d$, $b=c$ we have $ab+cd=ab-ab=0$.

In the case $a=d$, $b=-c$ we have $ab+cd=ab-ab=0$ again.

Answer 5

Let $a=\cos x,b=\sin x,c=\cos y,d=\sin y$. Then $$ ac+bd=\cos x\cos y+\sin x\sin y=\cos(x-y) $$ and hence $$ \cos(x-y)=0.$$ So $$ ab+cd=\cos x\sin x+\cos y\sin y=\frac12(\sin(2x)+\sin(2y))=\sin(x+y)\cos(x-y)=0$$

Answer 6

Since $$ac+bd=0\implies a=-\frac{bd}{c}; c\neq0$$

Then $$1=a^2+b^2=b^2[\frac{d^2}{c^2}+1]=b^2\cdot\frac{c^2+d^2}{c^2}=\frac{b^2}{c^2}$$

$$\implies c=\pm b\implies d=\mp a$$

The solution is given by

$$(a,b,c,d)=(a,b,\pm b,\mp a)$$

So $$ab+cd=ab+(\pm b)(\mp a)=ab-ab=0$$

(For $c=0$, the case is trivial.)

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Another solution

Let $a=mb$, $c=nd$.

$$ac+bd=0\implies mbnd+bd=0\implies bd=0 \text{ or }mn=-1$$

For $mn=-1$

$$a^2+b^2=b^2(m^2+1)=1\implies b^2=\frac{1}{m^2+1}$$

Similarly, $\displaystyle d^2=\frac{1}{n^2+1}$

Then

$$ab+cd=mb^2+nd^2=\frac{m}{m^2+1}+\frac{n}{n^2+1}=\frac{(m+n)(mn+1)}{(m^2+1)(n^2+1)}=0$$

Answer 7

One has $$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ac+bd)(ad+bc)=0.$$

Of course, the "right" proof of this is to note that the hypotheses mean that $$\pmatrix{a&b\\c&d}\pmatrix{a&c\\b&d}=\pmatrix{1&0\\0&1}$$ and since for square matrices, $AB=I$ implies $BA=I$ then $$\pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}=\pmatrix{1&0\\0&1}$$ so that $ab+cd=0$ (and $a^2+c^2=b^2+d^2=1$).