Solve ab+cd Given a^2+b^2=c^2+d^2=1 and ac+bd=0

[Kizaru]

Homework Statement

Suppose that a, b, c, d are real numbers such that a^2 + b^2 = c^2 + d^2 = 1 and ac + bd = 0. What is ab + cd?

Homework Equations

$$a^{2} + b^{2} = c^{2} + d^{2} = 1$$
$$ac + bd = 0$$

The Attempt at a Solution

Clearly, ac = -bd. I know the solution is 0, but I am having trouble proving or deriving it.
A few things to note (some may be useless, but this is an involved problem and I'm getting my hands dirty):

$$(a + c)^{2} + (b + d)^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ac + bd) = 1 + 1 = 2$$
$$(a + d)(b + c) = ab + ac + bd + cd = ab + 0 + cd = ab + cd$$

And there are many other expressions like these, but not sure how to piece them together. The first equation tells me that the hypotenuse of a rectangle with sides (a+c) and (b+d) is sqrt(2).
The second equation tells me the area of a rectangle with sides (a+d)(b+c) = ab+cd, assuming that one value is < 0. I know there is something else I'm missing, just not sure what it is.

Any help would be greatly appreciated.

 

[owlpride]

I really think there should be an elegant solution. For the time being, here's a brute force approach: You have three equations in four unknowns. You can solve for three of the variables in terms of the fourth, and use that to conclude that ab + cd = 0.

If one of the numbers is zero, it is pretty straight forward to show that ab + cd = 0. So let's assume that all variables are nonzero. Then you can use ac + bd = 0 to solve for a in terms of the other variables: a = -bd/c. Plug that into a^2 + b^2 = 1 to get d in terms of b and c. Then plug that into c^2 + d^2 = 1 and see what you get. I don't want to spoil the surprise!

 

[Kizaru]

I tried that approach earlier and obtained
$$c^{2}(2-\frac{1}{b^{2}}) = 1$$
I was unable to manipulate it any further into something meaningful.
It feels like I am missing something from this approach as well.

 

[owlpride]

Hmm, I will double-check my own calculation but I ended up with b^2 = c^2 and a^2 = d^2.

 

[owlpride]

If a = -bd/c, then

$$a^2 + b^2 = b^2 d^2 / c^2 + b^2 = 1$$

so

$$d^2 = c^2 (1-b^2) /b^2$$

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

Hence $$b^2 = c^2$$.

 

[Kizaru]

If a = -bd/c, then

$$a^2 + b^2 = b^2 d^2 / c^2 + b^2 = 1$$

so

$$d^2 = c^2 (1-b^2) /b^2$$

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

Hence $$b^2 = c^2$$.

OH, I see my mistake. I had
$$a^2 = -\left(\frac{bd}{c}\right)^2$$
Which is why I had the 2 inside the parentheses.

And from there we get a^2 = d^2 as well, and since one must be negative to satisfy ac+bd= 0 blah blah ab+cd = 0.

Thank you very much!

 

[boboYO]

interpret it as perpendicular vectors with norm 1:a=sin(t)
b=cos(t)

c=sin(t+pi/2)=cos(t)
d=cos(t+pi/2)=-sin(t)

ab+cd=0

 

[Kizaru]

boboYO, that is a very interesting solution. Thank you!