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Help with vector proof

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  1. May 1, 2016 #1
    1. The problem statement, all variables and given/known data
    help.png

    2. Relevant equations

    let the point where the line through B and X intersects with AC be P

    3. The attempt at a solution

    I know that

    ACdotBP = 0

    AC = AD+DC

    BP = PC+ CD

    Therefore (AD+DC)dot(PC+CD)=0

    I also know that:

    ECdotCE = 0

    BCdotDA=0

    However I am stuck on showing that the dot product of ACdotBP is equal to 0

    If anyone can help me it would be much appreciated

    thanks
     
  2. jcsd
  3. May 1, 2016 #2

    Samy_A

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    How did you get BP = PC+ CD?
    And isn't AC⋅BP = 0 what you have to prove?
     
  4. May 1, 2016 #3
    thanks[/QUOTE]
    Sorry that should say BP=BC+CP

    And yes I am trying to prove AC⋅BP = 0

    I am just unsure on the way that I should go about it
     
  5. May 1, 2016 #4

    Samy_A

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    You know that AX⋅BC=0, CX⋅BA=0.

    What you could do is pick an origin O (that could be any point, doesn't really matter), and express the vectors in terms of this origin.
    For example: AX = OX - OA, BC = OC - OB, and similarly for the other vectors.
    Now express AX⋅BC=0, CX⋅BA=0 in these terms, and compare the two expressions that you get.

    Notice that you have to prove that BX⋅CA = 0.
     
  6. May 1, 2016 #5
    When I substitute AX = OX - OA, BC = OC - OB and BA = BO+OA, CX=CA+AB

    I get (OX-OA)⋅(OC-OB)=0 and (CA+AE)⋅(BO+OA)=0

    What would I do from here would I have to rearrange the equations?
     
  7. May 1, 2016 #6

    Samy_A

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    Actually, when you rewrite AX⋅BC=0 and CX⋅BA=0 in terms of vectors originating in O, what you get is:
    AX⋅BC=0 indeed becomes (OX-OA)⋅(OC-OB)=0
    But CX⋅BA=0 becomes (OX-OC)⋅(OA-OB)=0.

    Now use the distributive property of the dot product to compute these two expressions.
    Write the results one under the other and look at them for a moment. Combining them in the correct way will lead to the hoped for result.
     
  8. May 1, 2016 #7
    I am not sure if this is the distributive property of the dot product but this is what I got:

    OXOC+OAOB=0
    OXOA+OCOB=0

    How would i got about combining these?
     
  9. May 1, 2016 #8

    Samy_A

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    The distributive property of the dot product is:
    (a + b)⋅c = a⋅c + b⋅c (where a,b and c are vectors).
    From this it follows, for example, that (a + b)⋅(c + d) = a⋅c + b⋅c + a⋅d + b⋅d,
    or (a - b)⋅(c - d) = a⋅c - b⋅c - a⋅d + b⋅d (where a,b c and d are vectors).

    Apply this correctly to:
    (OX-OA)⋅(OC-OB)=0
    (OX-OC)⋅(OA-OB)=0
     
  10. May 1, 2016 #9
    (OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

    (OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB

    How would I combine this now?
     
  11. May 1, 2016 #10

    Samy_A

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    You left out that both expressions are 0.

    Did you notice that two terms appear in both expressions? That should give you a clue about the next step: how to combine the two expressions.


    Remember that the final goal is to prove that BX⋅CA = 0.
     
  12. May 1, 2016 #11
    (OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB=0

    (OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=0

    OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

    OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

    now how would I prove that BX⋅CA = 0?
     
  13. May 1, 2016 #12

    Samy_A

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    The distributive property of the dot product is still valid.
    Two terms contain OX, two terms contain OB ...
     
    Last edited: May 1, 2016
  14. May 1, 2016 #13
    OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

    OX⋅OA-OX⋅OC=OA⋅OB+OC⋅OB

    OX(OA-OC)=(OA+OC)OB

    How would I prove it now?
     
  15. May 2, 2016 #14

    Samy_A

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    You have a sign error: what you should get is OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

    Now, look back at post #4: for any vector YZ, YZ = OZ - OY.
     
  16. May 2, 2016 #15
    BX⋅CA=0

    BX=OX-OB

    CA=OA-OC

    (OX-OB)⋅(OA-OC) = 0

    OX⋅OA-OC⋅OX-OB⋅OA+OB⋅OC=0

    OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

    Therefore: OX⋅OA=OA⋅OB-OC⋅OB+OX⋅OC

    OA⋅OB-OC⋅OB+OX⋅OC-OC⋅OX-OB⋅OA+OB⋅OC=0

    0=0

    Is this now correct?

    Thanks
     
  17. May 2, 2016 #16

    Samy_A

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    I'm not sure I follow the logic here. BX⋅CA = 0 is what you have to prove.

    You had OX⋅OA - OX⋅OC = OA⋅OB - OC⋅OB.
    That can be written as OX⋅(OA-OC) = OB⋅(OA - OC)
    That becomes OX⋅CA = OB⋅CA.
    (All these steps actually appear in your previous post.)

    From that you easily get the value of BX⋅CA.
     
  18. May 2, 2016 #17
    OX⋅CA-OB⋅CA. =0

    (OX-OB)⋅CA=0

    BX=(OX-OB)

    therefore

    BX⋅CA=0

    Is this now correct?
     
  19. May 2, 2016 #18

    Samy_A

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    Yes.
     
  20. May 2, 2016 #19
    Thanks for your help I really appreciate it!!
     
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