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Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+cd.

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    [tex]
    a^{2} + b^{2} = c^{2} + d^{2} = 1
    [/tex]
    [tex]ac + bd = 0
    [/tex]
    3. 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):

    [tex]

    (a + c)^{2} + (b + d)^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ac + bd) = 1 + 1 = 2
    [/tex]
    [tex]
    (a + d)(b + c) = ab + ac + bd + cd = ab + 0 + cd = ab + cd

    [/tex]

    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.
     
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    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!
     
  4. Feb 6, 2010 #3
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    I tried that approach earlier and obtained
    [tex]
    c^{2}(2-\frac{1}{b^{2}}) = 1
    [/tex]
    I was unable to manipulate it any further into something meaningful.
    It feels like I am missing something from this approach as well.
     
  5. Feb 6, 2010 #4
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    Hmm, I will double-check my own calculation but I ended up with b^2 = c^2 and a^2 = d^2.
     
  6. Feb 6, 2010 #5
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    If a = -bd/c, then

    [tex]a^2 + b^2 = b^2 d^2 / c^2 + b^2 = 1[/tex]

    so

    [tex]d^2 = c^2 (1-b^2) /b^2[/tex]

    [tex]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} [/tex]

    Hence [tex]b^2 = c^2 [/tex].
     
  7. Feb 6, 2010 #6
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    OH, I see my mistake. I had
    [tex]a^2 = -\left(\frac{bd}{c}\right)^2[/tex]
    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!
     
    Last edited: Feb 6, 2010
  8. Feb 6, 2010 #7
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

    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
     
  9. Feb 6, 2010 #8
    Re: Suppose a,b,c,d in Reals s.t. a^2 + b^2 = c^2 + d^2 = 1 and ac+bd=0. What is ab+c

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