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Dot product, cross product

  1. Jun 7, 2005 #1
    i need help with the following:
    note that the big dots represents the dot product


    1. suppose that [tex]a \bullet b = c \bullet b [/tex] for all vectors [tex]\overrightarrow{b} [/tex]. show that [tex]\overrightarrow{a} = \overrightarrow{c} [/tex].

    i suppose i can't simply divide out the b, right? anyway, i tried writing out the components of each vector - for example a=(a1, a2, a3), b=(b1, b2, b3), c=(c1, c2, c3). i got as far as:

    (b1, b2, b3) [tex]\bullet[/tex](a1-c1, a2-c2, a3-c3) = 0

    but i don't know how to simplify this further to eliminate the b. any help is appreciated.
     
  2. jcsd
  3. Jun 7, 2005 #2
    cross product

    my second question is:

    2. suppose that [tex]\overrightarrow{a} X \overrightarrow{b} = \overrightarrow{c} X \overrightarrow{b} [/tex] for all vectors [tex]\overrightarrow{b}. [/tex] how are [tex]\overrightarrow{a} [/tex] and [tex]\overrightarrow{c} [/tex] related?.

    i wrote the vectors in terms of its components, took the cross product, and i got as far as:

    (b3(a2-c1) + b2(a3-c3), b1(a3-c3) + b3(a1-c1), b2(a1-c1) + b1(a2-c2)) = 0

    again, i'm not sure how to simplify this further or what conclusion to draw about the relationship between a and c. any help is greatly appreciated.
     
  4. Jun 7, 2005 #3
    For the first, did you know that [itex] \vec{a} \bullet \vec{b} = |a||b|cos(\theta) [/itex]? That should be all you need.

    There is a similar identity for the cross product.
     
  5. Jun 7, 2005 #4
    i have considered that, but the angles between a and b, and c and b, are different (presumably). so i end up with

    |a| cos(t) = |c| cos (s)

    ...which doesn't seem to help very much.
     
  6. Jun 7, 2005 #5

    Curious3141

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    For the dot product, here's a hint. Consider the cases where [tex]\vec{b}[/tex] is a unit vector first in the direction of [tex]\vec{a}[/tex] then in the direction of [tex]\vec{c}[/tex]. Draw up some equations and see what you can conclude about the angle between vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex], then what that implies about the relationship between their magnitudes.
     
  7. Jun 7, 2005 #6

    Curious3141

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    Here's the solution for the dot product :

    Let the angle between vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] be [tex]\theta[/tex].

    Let [tex]\vec{b}[/tex] be a unit vector in the same direction as [tex]\vec{a}[/tex]

    Then [tex]\vec{a} \bullet \vec{b} = |\vec{a}|[/tex]

    [tex]\vec{c} \bullet \vec{b} = |\vec{c}|\cos\theta[/tex]

    These two are equal, so we have :

    [tex]|\vec{a}| = |\vec{c}|\cos\theta[/tex] --eqn(1)

    Now let [tex]\vec{b}[/tex] be a unit vector in the direction of [tex]\vec{c}[/tex]

    Working through it in the same way gives :

    [tex]|\vec{c}| = |\vec{a}|\cos\theta[/tex] --eqn(2)

    Solving those we get :

    [tex]|\vec{c}|\cos^2{\theta} = |\vec{c}|[/tex]

    Either [tex]|\vec{c}| = |\vec{a}| = 0[/tex], meaning both of them are null vectors (in which case they're equal in any case),

    or [tex]\cos^2{\theta} = 1[/tex]

    giving [tex]\cos{\theta} = 1[/tex] (the negative root is inadmissible because if you look at equations 1 and 2, the magnitudes are positive, so the cosine term can't be negative).

    So we have [tex]\theta = 0[/tex], and putting that back into either equation 1 or 2, we get [tex]|\vec{a}| = |\vec{c}|[/tex].

    So the two vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] are of the same magnitude and oriented in the same direction. By definition, therefore, [tex]\vec{a} = \vec{c}[/tex] (QED)
     
  8. Jun 7, 2005 #7

    Galileo

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    Write the equation in a different form:

    [tex](\vec a -\vec c)\times \vec b=\vec 0[/tex]
    for any vector [itex]\vec b[/itex].

    (You can do the same thing with the dot product also).
    What can you infer about [itex]\vec a-\vec c [/itex] from this?
     
  9. Jun 7, 2005 #8

    dextercioby

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    The first exercise is incorrect.

    [tex] \vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b} [/tex]

    Using the distributivity,one gets

    [tex] \left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0 [/tex]

    Which has the infinite set of solutions:

    [tex] \vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\} [/tex].

    The "V" is a linear space and the [itex] \vec{d}[/itex] and [itex] \vec{b} [/itex] are arbitrary vectors.

    So,please,think deeper before giving erroneous advice.

    Daniel.
     
  10. Jun 7, 2005 #9

    Galileo

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    The exercise is to show that [itex]\vec a \cdot \vec b=\vec c\cdot \vec b\forall \vec b \Rightarrow \vec a = \vec c[/itex].
    This is clearly true, you just have to pick a clever value for b, say b=a-c
     
    Last edited: Jun 7, 2005
  11. Jun 7, 2005 #10

    dextercioby

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    That's not true.That difference is not zero,but an arbitrary vector.Zero vector is not arbitrary.

    Daniel.
     
  12. Jun 7, 2005 #11

    Galileo

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    Are you sure you have the same understanding of the problem as I do?
    [itex]\vec b[/itex] can be any vector.
     
  13. Jun 7, 2005 #12

    dextercioby

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    Yes.

    HINT:[tex]\vec{c}\cdot\vec{b}=\vec{c}\cdot\left(\vec{b}+\vec{c}\times \vec{d}\right) [/tex]

    ,where \vec{d} is an arbitrary vector...:wink:

    Daniel.
     
  14. Jun 7, 2005 #13
    here is how I would handle the two dimentional case:

    a * b = (ax + ay)(bx + by) = ax*bx + ay*by

    c * b = (cx + cy)(bx + by) = cx*bx + cy*by

    Two vectors are equal only if their corresponding components are equal. So if a * b = c *b then we have:

    ax*bx = cx*bx

    ay*by=cy*by

    Which shows that a and c are equal.
     
  15. Jun 7, 2005 #14

    arildno

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    You are wrong, Daniel.
    [tex]\vec{a},\vec{c}[/tex] are fixed vectors, and the [tex]\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}[/tex] is supposed to hold for ALL [tex]\vec{b}[/tex]
    This is equivalent to:
    [tex](\vec{a}-\vec{c})\cdot\vec{b}=0[/tex]
    In particular, this must hold for the choice [tex]\vec{b}=\vec{a}-\vec{c}[/tex] from which follows [tex]\vec{a}=\vec{c}[/tex]
     
    Last edited: Jun 7, 2005
  16. Jun 7, 2005 #15

    dextercioby

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    That's wrong

    [tex] \vec{a}\cdot\vec{b} [/tex] is a scalar

    and hence

    [tex] a_{x}b_{x}+a_{y}b_{y}=c_{x}b_{x}+c_{y}b_{y} \nRightarrow \left\{\begin{array}{c}a_{x}b_{x}=c_{x}b_{x}\\a_{y}b_{y}=c_{y}b_{y} \end{array}\right [/tex]

    Okay??

    Daniel.
     
  17. Jun 7, 2005 #16

    dextercioby

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    I'm not wrong,my solution does not exclude the 0 vector.That "d" is arbitrary,just as "b",so it can be the 0 vector.

    Daniel.
     
  18. Jun 7, 2005 #17

    arildno

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    Whatever are you talking about???
    [tex]\vec{a},\vec{c}[/tex] are FIXED vectors; their difference do not change if [tex]\vec{b}[/tex] changes.
     
  19. Jun 7, 2005 #18

    dextercioby

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    Who said "a" and "c" are fixed?

    Daniel.
     
    Last edited: Jun 7, 2005
  20. Jun 7, 2005 #19

    arildno

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    It says for all b here.
     
  21. Jun 7, 2005 #20

    dextercioby

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    Yes,and my solution exploits the equality of the 2 scalar products.

    Daniel.
     
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