Let $a^2+b^2=1,c^2+d^2=1$ and $ac+bd=0.$ Prove that $a^2+c^2=1, b^2+d^2=1$ and $ab+cd=0.$

Question

Let $a^2+b^2=1,c^2+d^2=1$ and $ac+bd=0.$ Prove that $a^2+c^2=1, b^2+d^2=1$ and $ab+cd=0.$

My solution goes like this:

We have, $$(a^2+b^2)(c^2+d^2)=1=(ac+bd)^2+1\implies a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2c^2+b^2d^2+2abcd+1\implies (ad-bc)^2=1\implies ad-bc=\pm 1.$$ Also, we have, $ac+bd=0\implies ac=-bd\implies a=\frac{-bd}{c}.$ Again, from previous result, we had, $ad-bc=\pm 1.$ Now, substituting, tha value of $a=\frac{-bd}{c},$ in this, equation, we get, $$\frac{-bd^2}{c}-bc=\pm1\implies -bd^2-bc^2=\pm c\implies -b(d^2+c^2)=\pm c\implies -b=\pm c\implies b=\mp c.$$ So, $a^2+b^2=1\implies a^2+c^2=1.$ Also, since, $a^2+c^2=1$ and $c^2+d^2=1$ ( given), so, we can say, $a^2-d^2=0\implies d=\pm a.$ We also have, $c^2+d^2=1\implies b^2+d^2=1.$ Now, if $ab+cd=a(\mp c)+c(\pm a)=0.$ This completes the solution so as to say, we have, $a^2+c^2=1, b^2+d^2=1$ and $ab+cd=0.$

However, we claimed, $a=\frac{-bd}{c},$ is valid, which, is under the assumption $c\neq 0.$ But, if $c=0$ then $d=\pm 1$ and since $ac+bd=0\implies bd=0$ and hence, $b=0$ so, $a=\pm1.$ Substituting these values, i.e $a=d=\pm1$ and $b=c=0,$ we see, that $a^2+c^2=1,$ $b^2+d^2=1$ and $ab+cd=1$ are automatically satisfied.

Is the above solution valid? If not, where is it going wrong?

Answer 1

Your solution is correct. Another possibility is to consider the matrix$$M=\begin{bmatrix}a&c\\b&d\end{bmatrix}$$and to observe that\begin{align}\left\{\begin{array}{l}a^2+b^2=1\\c^2+d^2=1\\ac+bd=0\end{array}\right.&\iff M^TM=\operatorname{Id}_2\\&\iff MM^T=\operatorname{Id}_2\\&\iff\left\{\begin{array}{l}a^2+c^2=1\\b^2+d^2=1\\ab+cd=0.\end{array}\right.\end{align}

Answer 2

HINT: from $a^2+b^2=1$ and $c^2+d^2=1$ you can write $$ a=\cos u, b=\sin u\qquad\text{and}\qquad c=\cos v, d=\sin v $$ for some $u$, $v\in\mathbb{R}$. Now $ac+bd=0$ simply means that $u-v=\frac{\pi}2+k\pi$ for some $k\in\mathbb{Z}$ (e.g. write the matrices corresponding to rotation by $u$ and $-v$ and multiply them).

Then, use trigonometry.

Answer 3

An other way : $$\{x^2+y^2=1\mid x,y\in\mathbb R\}=\{(\cos\theta ,\sin\theta )\mid \theta \in [0,2\pi]\}.$$

Therefore, there are $\theta \in [0,2\pi]$ and $\varphi \in [0,2\pi]$ s.t. $$(a,b)=(\cos\theta ,\sin\theta )\quad \text{and}\quad (c,d)=(\cos\varphi ,\sin\varphi ).$$

Using the condition $ac+bd=0$ gives $$\cos(\theta -\varphi )=\cos(\theta )\cos(\varphi )+\sin(\theta )\sin(\varphi )=0,$$

and thus $$\theta -\varphi =\frac{\pi}{2}+k\pi,\quad k\in\mathbb Z.\tag{C}$$

Finally, $$a^2+c^2=\cos^2(\theta )+\cos^2(\varphi ).$$ The condition $(C)$ yields $$\cos(\theta )=\sin(\varphi +k\pi)=\pm\sin(\varphi ),$$

and thus $$a^2+c^2=\sin^2(\varphi)+\cos^2(\varphi )=1.$$

The rest goes through the same.

Answer 4

Your solution's rectified version :

We've :
$a^2 + b^2 = 1 ...(1)$
$c^d + d^2 = 1 ...(2)$
$ac + bd = 0 ...(3)$

Multiplying (1) and (2) and then from (3) , we got $ad - bc = ±1$ ...(4).

Now, case 1: when $ad - bc$ = 1

Then, multiplying (3) by $c$ and (4) by $d$ and adding , we get : $a(c^2 + d^2) = d$ => $a = d$

Then from (3) $ab + cd = 0$ and from (1) $a^2 + c^2 = 1$

Similarly obtain the same thing from another case where $ad - bc = -1$ .

Answer 5

Your solution is on the right track, but the division by $c$ can be avoided. Here is a bit cleaner attack along similar lines.

Assuming $a,b,c,d\in\mathbb{R}$, note that $$ \overbrace{\left(a^2+b^2\right)}^1\overbrace{\left(c^2+d^2\right)}^1=\overbrace{(ac+bd)^2}^0+\color{#C00}{\overbrace{\color{#000}{(ad-bc)^2}}^1}\tag1 $$ If $ad-bc=1$, $$ \begin{align} (a-d)^2+(b+c)^2 &=\overbrace{\left(a^2+b^2\right)}^1+\overbrace{\left(c^2+d^2\right)}^1-\overbrace{2(ad-bc)\vphantom{a^2}}^2\tag{2a}\\ &=0\tag{2b}\\[1pt] (c,d)&=(-b,a)\tag{2c} \end{align} $$ Else $ad-bc=-1$, $$ \begin{align} (a+d)^2+(b-c)^2 &=\overbrace{\left(a^2+b^2\right)}^1+\overbrace{\left(c^2+d^2\right)}^1+\overbrace{2(ad-bc)\vphantom{a^2}}^{-2}\tag{3a}\\ &=0\tag{3b}\\[1pt] (c,d)&=(b,-a)\tag{3c} \end{align} $$ In either case $\text{(2c)}$ or $\text{(3c)}$, we have $b^2=c^2$; therefore, $$ \overbrace{\,a^2+c^2\,}^{a^2+b^2}=\overbrace{\,b^2+d^2\,}^{c^2+d^2}=1\tag4 $$ and since $c=\pm b$, $$ \begin{align} ab+cd &=\pm(ac+bd)\tag{5a}\\ &=0\tag{5b} \end{align} $$