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Math Help - Cross product and dot product relationships

  1. #1
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    Cross product and dot product relationships

    The question is the form of true or false
    If \underline a  \bullet \underline b  = \underline a  \bullet \underline c then \underline b  = \underline c

    The answer is given as false, I cannot understand why though. If the dot product of lhs and rhs equal each other and they share vector a in the calculation of the dot product how can vector b and c not equal each other ?

    and similarly for if \underline a  \times \underline b  = \underline a  \times \underline c then \underline b  = \underline c

    why does b not equal c in this case aswell?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Craka View Post
    The question is the form of true or false
    If \underline a  \bullet \underline b  = \underline a  \bullet \underline c then \underline b  = \underline c

    The answer is given as false, I cannot understand why though. If the dot product of lhs and rhs equal each other and they share vector a in the calculation of the dot product how can vector b and c not equal each other ?
    what if a is the zero vector?

    and similarly for if \underline a  \times \underline b  = \underline a  \times \underline c then \underline b  = \underline c

    why does b not equal c in this case aswell?
    if a \times b = a \times c then |a \times b | = |a \times c|. that means |a||b| \sin \theta_1 = |a||c| \sin \theta_2

    clearly if a is the zero vector, we have the same problem as we did with the dot product.

    if a is not the zero vector, then we can divide both sides by |a| to get


    |b| \sin \theta_1 = |c| \sin \theta_2

    does that mean |b| = |c|? if not, then clearly b \ne c
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  3. #3
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    Sorry I missed putting that part in that question, a is not a zero vector.

    Think I follow that with the cross product, and similarly I would do the same with the dot product though it would be in terms of cosine?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Craka View Post
    Sorry I missed putting that part in that question, a is not a zero vector.
    ok, then pick b such that it is perpendicular to a. so the dot product is zero. it means therefore, i can pick c to be any multiple of +/- b and i would still get the dot product to be zero (since multiples of b are parallel to be and would therefore also be perpendicular to a)

    thus, we found b and c that make the claim false, so it is false
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    is up to his old tricks again! Jhevon's Avatar
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    see here. look under "algebraic properties"
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