Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conditions of dot product

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Under what conditions are the following true?

    a) [tex]|\mathbf{a}\cdot \mathbf{b}| =|\mathbf{a}||\mathbf{b}|[/tex]

    b) [tex](\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2[/tex]

    2. Relevant equations

    None.

    3. The attempt at a solution

    a) I dont understand why the answer is [tex]\mathbf{a} = \lambda \mathbf{b}[/tex]. Surely it would be equivalent all the time?

    b) Also looks equivalent all the time to me.
     
  2. jcsd
  3. Sep 12, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Are you aware that the dot product of any two perpendicular vectors is 0? It isn't likely that [itex]|\mathbf{u}||\mathbf{v}|= 0[/itex] in that case is it?

    a) One definition of dot product is [itex]\mathbf{a}\cdot\mathbf{b}= |\mathbf{a}||\mathbf{b}| cos(\theta)[/itex] where [itex]\theta[/itex] is the angle between the two vectors. For what [itex]\theta[/itex] is [itex]cos(\theta)= 1[/itex]?

    b) [tex](\mathbf{a}+ \mathbf{b})\cdot(\mathbf{a}- \mathbf{b})= \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{a}- \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{b}[/tex]

    Now, is [itex]\mathbf{x}\cdot\mathbf{x}= |\mathbf{x}|^2[/itex]? Is [itex]\mathbf{x}\cdot\mathbf{y}= \mathbf{y}\cdot\mathbf{x}[/itex]?
     
  4. Sep 13, 2010 #3
    Yeah I am aware. For a) I would say it could only be equivalent if the dot product doesnt equal 0, but I dont get why the answer is [tex]\mathbf{a} = \lambda \mathbf{b}[/tex]

    b) I agree as x^2 gives you the absolute value and x * y is the same as y * x, but how does this apply to b?
     
  5. Sep 13, 2010 #4

    Mark44

    Staff: Mentor

    No. What if the dot product were 1/2? Would that make the two sides of the equation equal? The two sides of that equation are equal only if the dot product is 1. That's different from what you said.

    Look at what HallsOfIvy wrote about the coordinate-free definition of the dot product.
     
  6. Sep 14, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    The dot product not equaling 0 is not relevant. Answer my question: For what [itex]\theta[/itex] is [itex]cos(\theta)= 1[/itex]?

    Then you agree that [itex](a+ b)\cdot(a- b)= a\cdot a+ a\cdot b- b\cdot a- b\cdot b[/itex] (I miswrote that as "[itex]+ b\cdot b[/itex] before)
    [itex]= |a|^2+ a\cdot b- a\cdot b- |b|^2[/itex]. What does that reduce to (for all a and b)?
     
  7. Sep 14, 2010 #6
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook