1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proveing Perpendicular of two vectors by applying Pythagoras rule

  1. Jun 30, 2011 #1
    Given that a=(a1,a2,a3) and b=(b1,b2,b3) by applying the Pythagoras rule, Prove that a1b1+a2b2+a3b3=0 if a and b perpendicular



    The Scalar product a.b = |a||b|CosQ -------------(1)



    if two vectors are perpendicular; Q=90degrees
    then CosQ=0;
    from (1)

    a.b=0
    (a1,a2,a3).(b1,b2,b3)=0
    a1b1+a2b2+a3b3=0

    is this a correct solution for the above Question?
    please can someone give me an opinion! :confused:
     
    Last edited: Jun 30, 2011
  2. jcsd
  3. Jun 30, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    The proof is correct, but they are asking for a proof by Pythagorean rule, which this isn't.

    What would the Pythagorean rule be for a triangle with vectors a and b as its sides?
     
  4. Jun 30, 2011 #3
    according to the Pythagorean rule;
    for a Right triangle
    |a|2+|b|2=|c|2

    can you give me a hint or something to solve my question properly, :smile:
     
  5. Jun 30, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    Can you relate c to a and b?

    (I think that is a hint. :wink:)
     
  6. Jun 30, 2011 #5
    vectors are realy hard lesson to me, so please show me how to prove that equation, I dont know where to start:confused: please help me!!
     
  7. Jun 30, 2011 #6

    I like Serena

    User Avatar
    Homework Helper

    All right.

    c is the vector that starts from the end of vector a, and ends at the end of vector b.
    This means that c = b - a.
    Do you know this type of vector algebra?

    So you have |a|2 + |b|2 = |b - a|2

    Fill in the definition of the norm and work out the algebra?
     
  8. Jun 30, 2011 #7
    Thank you so much I like Serena, finally I was able to proved that:smile:, can you explain how did you get the equation |a|2+|b|2=|b-a|2 more?:smile:
     
  9. Jun 30, 2011 #8

    I like Serena

    User Avatar
    Homework Helper

  10. Jun 30, 2011 #9
    I know about addition and subtraction of vectors, what i wanted to understand is how did you get c=b-a instead of c=a+b,
    can you explain little bit more, what did you mean by,
    "c is the vector that starts from the end of vector a, and ends at the end of vector b.
    This means that c = b - a" ?
     
  11. Jun 30, 2011 #10

    I like Serena

    User Avatar
    Homework Helper

    You can use c=a+b as well.
    On the wiki link in my previous post you can see that you'll get a different triangle, but that does not matter for Pythagoras, since the angle is 90 degrees.


    That is exactly what it means.

    As opposed to c = a + b, where c is the vector that begins at the beginning of a, and ends at the end of b (the head-to-tail construction of vector addition).
    In this case vector b would be shifted so its tail is at the head of a.
     
  12. Jun 30, 2011 #11
    Thank you very much I like Serena. I got the idea, you are a really good helper:smile: hope to catch you again,:smile: bye...
     
  13. Jun 30, 2011 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    a,b vectors determine a parallelogram, and one diagonal is c1=a+b, the other is c2=b-a, see attachment. It is all the same which one you select. You get Pythagoras equation only in case of a1b1+a2b2+a3b3=0

    ehild
     

    Attached Files:

  14. Jun 30, 2011 #13
    wow... thank you ehild:) your attachment really helpful for me, now I got a more clear idea about |c| = |b - a|.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook