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To prove that vectors (a + b) . (a - b) = a^2 + b^2 iff

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data

    The question says :
    Prove that
    [itex](\vec{a} + \vec{b}).(\vec{a} - \vec{b}) = \left | \vec{a} \right |^{2} + \left | \vec{b} \right |^{2}[/itex]
    if and only if [itex]\vec{a} \perp \vec{b}[/itex]

    2. Relevant equations

    This is known that
    [itex]\left | \vec{a} + \vec{b} \right | = \left | \vec{a} - \vec{b} \right |[/itex]
    if [itex]\vec{a} \perp \vec{b}[/itex]

    3. The attempt at a solution

    I tried substituting that and then using Cauchy–Schwarz inequality but somehow I can't open up the absolute brackets .
    Thanks in advance
     
  2. jcsd
  3. Jan 10, 2013 #2
    You can simply use the distributive property.

    EDIT: Are you sure that you have copied down the question correctly? I guess the RHS should be [tex]\left | \vec{a} \right |^{2} - \left | \vec{b} \right |^{2}[/tex]
     
    Last edited: Jan 10, 2013
  4. Jan 10, 2013 #3
    @Pranav The question is correct
     
  5. Jan 10, 2013 #4
    Substitute [itex]\vec{a}=\hat{i}[/itex] and [itex]\vec{b}=\hat{j}[/itex]. The relation doesn't satisfy this.
     
  6. Jan 10, 2013 #5
    That doesn't sound right: take the following example:

    [itex]\vec{a}=\left ( 1 , 0\right )[/itex]
    [itex]\vec{b}=\left ( 0 , -0.1\right )[/itex]

    [itex]\left ( \vec{a} + \vec{b} \right ) \cdot \left ( \vec{a} - \vec{b} \right ) = 0.99[/itex]

    but

    [itex] \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 = 1.01[/itex]

    J.
     
  7. Jan 10, 2013 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    As several people have told you now, the correct relation is
    [tex](\vec{a}+ \vec{b})\cdot(\vec{a}- \vec{b})= |\vec{a}|^2- |\vec{b}|^2[/tex]
    not what you wrote. You should have realised that from the usual numerical "difference of squares" formula.

    As to how to prove it, what "tools" do have to? You write as if you had never seen vectors before but that can't be true if you are expected to do a probolem like this!
    How are you working with vectors? In terms of components? Are you allowed to assume two or three dimensions? Or in terms of "length and direction"?

    How has the dot product been defined? As "[itex]<a_1, a_2, \cdot\cdot\cdot, a_n>\cdot<b_1, b_2, \cdot\cdot\cdot, b_n>= a_1b_1+ a_2b_2+ \cdot\cdot\cdot+ a_nb_n[/itex]? Or as [itex]|\vec{a}||\vec{b}|sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between [itex]\vec{a}|[/itex] and [itex]\vec{b}[/itex]?

    Either can be used but the proof depends on which you are using.
     
  8. Jan 10, 2013 #7

    D H

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    Staff Emeritus
    Science Advisor

    This is obviously false. Double check the question. It is probably asking you to prove that [itex](\vec a + \vec b) \cdot (\vec a + \vec b) = ||\vec a||^2 + ||\vec b||^2[/itex] iff [itex]\vec{a} \perp \vec{b}[/itex].
     
  9. Jan 10, 2013 #8

    mfb

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    2016 Award

    Staff: Mentor

    But this is true for all a,b.

    My guess: Show that
    [tex](\vec{a}+ \vec{b})\cdot(\vec{a}+ \vec{b})= |\vec{a}|^2+ |\vec{b}|^2[/tex]
    is true if and only if ##\vec{a} \perp \vec{b}##.

    Edit: Same minute :(
     
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