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If c = A· E, then A - cE = 0.

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Let E be any unit vector, that is a vector of norm 1. Let c be the component
    of A along E. We saw that c = A·E. Then A - cE is perpendicular to E, and A = A - cE + cE
    Then A - cE is also perpendicular to cE ...

    2. Relevant equations
    3. The attempt at a solution

    I just cant seem to figure where they got A - cE = 0 from. Everyone just shows A - cE = 0 is true on its own instead of showing if c = A·E, then A - cE = 0* which is what I think the author means by "We saw that c = A·E. Then A - cE is perpendicular to E".

    Please, elaborate a little. Thanks.

    edit: * if c = A·E, then A - cE · E = 0
     
    Last edited: Mar 16, 2014
  2. jcsd
  3. Mar 16, 2014 #2
    Eh? Where does it say A - cE = 0? Are you trying to prove that A-cE is perpendicular to E?
     
  4. Mar 16, 2014 #3
    Oh yeah, A - cE . E = 0

    My bad.
     
  5. Mar 16, 2014 #4
    You mean (A-cE)·E=0, I suppose. Geometrically, if E is a unit vector, then is cE not the vector component of A along E? So what is A-cE?

    EDIT: heh, I wrote something silly, so I had to delete it
     
    Last edited: Mar 16, 2014
  6. Mar 16, 2014 #5
    My book has the picture of this situation. Geometrically it's clear that A-cE is a vector perpendicular to E and cE.

    A = A - cE + cE by parallelogram law. But why do we need A = A - cE + cE to see that A-cE is also perpendicular to cE when (A-cE)· E=0 implies (A-cE)· cE=0?

    E·E = 1.
     
  7. Mar 16, 2014 #6
    Yeah, that second part I wrote is clearly rubbish, so I deleted it. The scalar product of three vectors is nonsense, because the product of two is a scalar.

    A = A - cE + cE because -cE+cE=0. I think the point about A = A - cE + cE is that A = (A - cE) + cE, i.e. the vectors A, (A-cE) and cE form a right-angled triangle, with A as the hypotenuse. This is obvious geometrically, and I don't know what more is required.
     
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