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Math Help - vector analysis zero dot product proof

  1. #1
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    vector analysis zero dot product proof

    My professor gave the following theorem today:

    Let A and B be vectors. Then A * B =0 if and only if A and B are perpendicular.

    Now, in the proof for this my professor gave the following reasoning in the case that A or B are the zero vector, in the "iff direction" of if 0 => perpendicular:

    |A| = 0 or |B| = 0 => A=0 or B=0 => Give A are direction perp. To B or give B a direction perp to A => A is perp. To B.

    I understand that we can attribute any direction to the zero vector, but does anyone else feel it is a bit sketchy to "choose" A to be perp. to B in order to prove that A is perp to B?

    In other words.. the z ero vector's direction is anything. I don't have to "choose" it to be perpendicular to anything for the dot product to be zero. Therefore, as the theorem reads,"if and only if A and B are perpendicular" is a bit... not "wrong" but at the same time, could be more accurate.
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  2. #2
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    Re: vector analysis zero dot product proof

    Quote Originally Posted by OneMileCrash View Post
    My professor gave the following theorem today:
    Let A and B be vectors. Then A * B =0 if and only if A and B are perpendicular.
    Now, in the proof for this my professor gave the following reasoning in the case that A or B are the zero vector, in the "iff direction" of if 0 => perpendicular: |A| = 0 or |B| = 0 => A=0 or B=0 => Give A are direction perp. To B or give B a direction perp to A => A is perp. To B. I understand that we can attribute any direction to the zero vector, but does anyone else feel it is a bit sketchy to "choose" A to be perp. to B in order to prove that A is perp to B? In other words.. the z ero vector's direction is anything. I don't have to "choose" it to be perpendicular to anything for the dot product to be zero. Therefore, as the theorem reads,"if and only if A and B are perpendicular" is a bit... not "wrong" but at the same time, could be more accurate.
    Without knowing what text/set of notes your professor follows, it is very hard to comment.
    But most Vector Calculus textbooks give that theorem as: Two non-zero vectors are perpendicular if and only if their scalar product(dot product) is zero. At the same time it is usual to define an angle between two non-zero vectors. Moreover, the scalar products is defined as A\cdot B=\|A\|\|B\|\cos(\theta), where \theta is the angle between vectors A~\&~B. Thus the question you have does not even come up.
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  3. #3
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    Re: vector analysis zero dot product proof

    My question is about the proof given, not the dot product.

    I just feel like if I wrote, in proving A is perpendicular to B, "Choose A to be perpendicular to B. Therefore A is perpendicular to B" the professor who taught me how to write proofs would chokeslam me.
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  4. #4
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    Re: vector analysis zero dot product proof

    Quote Originally Posted by OneMileCrash View Post
    My question is about the proof given, not the dot product.
    I just feel like if I wrote, in proving A is perpendicular to B, "Choose A to be perpendicular to B. Therefore A is perpendicular to B" the professor who taught me how to write proofs would chokeslam me.
    Did you read my reply carefully? I think you missed the point.
    There is no reason to even consider a zero vector in that theorem. I don't understand why your professor would do that.
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  5. #5
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    Re: vector analysis zero dot product proof

    Ah, I see. I actually, after making my point suggested putting "nonzero" in the theorem.
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