Help with simple dot product proof

  1. Here's what I got to prove where '.' is dot.

    A.B=A.C Then B=C True or false? If true, prove it in general terms, if false, provide a counter-example.

    Ok, I just need some body to comment on my little proof here, and any guidelines to make it more thorough or whatnot.
    I know that the dot product is commutative,
    A.(B+C)=A.B +A.C, but not sure if it really needs to be in my proof or not.

    Proof
    ------
    Say A.B=N and A.C=N (where N is a scalar number)
    so if N=N
    Then A.B=A.C
    If I cancel the A's, I get B=C.

    Is that a good way to approach that, or is there a better way of expressing it?
     
  2. jcsd
  3. You know that (A^-1)(A)=1 or the identity. Then
    (A^-1)(A)(B)=(A^-1)(A)C)
    with this we can multiply both sides and get
    1(B)=1(C) or B=C

    The raeson that (A^-1)(A)=1 is because (A^-1) is the inverse for A.
     
  4. What if A is the zero vector? Then A.B=A.C no matter what B and C are.

    And even if A <> 0 if you break A, B and C down into components, I think you will find that you can come up with other situations where A.(B-C) must equal 0 even though you know nothing about the values of B and C individually.

    Try it.
     
  5. HallsofIvy

    HallsofIvy 40,654
    Staff Emeritus
    Science Advisor

    No.

    In the first place, there is no "A-1" when you are talking about dot product. There is, start with, no "identity" since
    A.I= A would not make sense. A is a vector and the dot product of two vectors is a number, not a vector.

    You are not really using either commutative or distributive laws:
    you are using cancellation which is exactly what you are asked about: Is is true that when A.B= A.C, B MUST equal C. You cannot use what you are asked to prove.

    Here is a hint. Choose two vectors at right angles. Call them A and B. Now choose a third vector at right angles to A. Call it C.
    What are A.B and A.C. Does that answer your question?
     
  6. mathman

    mathman 6,566
    Science Advisor
    Gold Member

    a.b=a.c
    a.(b-c)=0
    Therefore a is perpendicular to b-c. This does not imply b=c.

    Example (3 space):
    a=(1,0,0)
    b=(x,u,v)
    c=(x,s,t)
    where x,u,v,s,t may assume any values.
     
    Last edited: Oct 3, 2003
  7. I thought this problem was going to be easy, but I keep on getting confused each time I come back here. Let's see if I get this straight mathman.

    Let's say that A and B are perpendicular to each other. Now another vector, C, is perpendicular to A and B.

    So A.B=0 and A.C=0, but this doesn't imply that B and C HAVE to equal each other?

    And one more thing.
    Example (3 space):
    a=(1,0,0)
    b=(x,u,v)
    c=(x,s,t)

    Just some clarification. Does x for vecter b and c have to be the same number?
     
  8. HallsofIvy

    HallsofIvy 40,654
    Staff Emeritus
    Science Advisor

    Take A= (1,0,0), B= (0,1,0), and C= (0,0,1). It can't get any simpler than that.


    You also say:
    "And one more thing.
    Example (3 space):
    a=(1,0,0)
    b=(x,u,v)
    c=(x,s,t)

    Just some clarification. Does x for vecter b and c have to be the same number?"

    I have absolutely no idea. Generally speaking we do NOT use the same letter to represent two different numbers, but what was the context?
     
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