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

Vector Question

  1. Oct 28, 2006 #1
    Hello I was doing some study for a maths test involving vectors when I came across this question:

    For non zero vectors show that: |a-b|=|a+b| if and only if a and b are perpendicular.
    Deduce that a parallelogram is a rectangle if and only if its diagonals are equal in length.

    I did the first question using dot products:

    Therefore cosX=0
    x=90, 270, 480.....etc.
    Thus vector a and vector b must be perpendicular for|a-b|=|a+b| to be valid.

    I'm not sure if this way is correct though....

    Could someone please check if the stuff that I've done above is right? Also, could I get some help doing the second part of teh question? Thank you it will be much apprechiated.
  2. jcsd
  3. Oct 28, 2006 #2
    1. [tex] A-B = (A_{x} - B_{x}, A_{y}-B_{y}) [/tex] and

    [tex] A+B = (A_{x} + B_{x}, A_{y}+B_{y}) [/tex]

    If [tex] A [/tex] and [tex] B [/tex] are perpendicular, then [tex] A \bullet B = 0 [/tex] or [tex] A_{x}B_{x} + A_{y}B_{y} = 0 [/tex]

    You also know that [tex] (A_{x}-B_{x})^{2} +(A_{y}-B_{y})^{2} = (A_{x}+B_{x})^{2} +(A_{y}+B_{y})^{2} [/tex]

    So how would you go from there?

    For the second question use the parallelogram law (i.e. proof of the commutative law of vector addition)
    Last edited: Oct 28, 2006
  4. Oct 28, 2006 #3
    Ah, so you expand and simplify:
    [tex] (A_{x}-B_{x})^{2} +(A_{y}-B_{y})^{2} = (A_{x}+B_{x})^{2} +(A_{y}+B_{y})^{2} [/tex], which eventually equals:
    [tex] A_{x}B_{x} + A_{y}B_{y} = 0 [/tex]
    Thus proving that A is perpendicular to B, and that |a-b|=|a+b|is only true under this circumstance. (I'm still a little unsure about the formal wording at the end of the proof)

    The second question about the parallelogram, is basically asking to prove the first question. What does the word "deduce" ask you to do?
  5. Oct 28, 2006 #4
    So draw a parallelogram and let |a-b| be one diagonal and |a+b| be another diagonal. You know that they are equal if a and b are perpendicular. Thus we have a rectangle. QED
    Last edited: Oct 28, 2006
  6. Oct 28, 2006 #5
    I understand now. Thank you for your help and quick response.
  7. Oct 28, 2006 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Use words. Maths should not be presented as a series of symbols without any link between them. It will do you good and your teachers. There is also no need to invoke cos at all: just leave it as a.b.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook