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Math Help - Cross Product Problem

  1. #1
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    Cross Product Problem

    Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).


    Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:


    1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?

    2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.

    Any ideas would be appreciated and I will definitely click the little thank you button for you
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  2. #2
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    Re: Cross Product Problem

    Quote Originally Posted by divinelogos View Post
    Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).


    Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:


    1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?


    2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.

    Correct.

    Any ideas would be appreciated and I will definitely click the little thank you button for you
    Hi divinelogos,

    If \underline{a}\mbox{ and }\underline{b} are two vectors the cross product is defined by,

    \underline{a}\times\underline{b}=\underline{\hat{n  }}|\underline{a}||\underline{b}|\sin\theta~;~\mbox  { where }\theta\mbox{ is the angle between }\underline{a}\mbox{ and }\underline{b}~;~\underline{\hat{n}}\mbox{ is the unit vector perpendicular to }\underline{a}\mbox{ and }\underline{b}

    The direction of \underline{\hat{n}} is given by the Right hand rule.

    Therefore, e_{1}\times e_{2}=\underline{\hat{n}}|e_{1}||e_{2}|\sin \frac{\pi}{6}=\frac{1}{2}\underline{\hat{n}}

    Since e_{3} is perpendicular to e_{1}\mbox{ and }e_{2};

    \underline{\hat{n}}=e_{3}

    \therefore e_{1}\times e_{2}=\frac{1}{2}e_{3}

    It should be noted that, we have to assume that \underline{\hat{n}}\neq -e_{3} to get the given answer.
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  3. #3
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    Re: Cross Product Problem

    Quote Originally Posted by Sudharaka View Post
    Hi divinelogos,

    If \underline{a}\mbox{ and }\underline{b} are two vectors the cross product is defined by,

    \underline{a}\times\underline{b}=\underline{\hat{n  }}|\underline{a}||\underline{b}|\sin\theta~;~\mbox  { where }\theta\mbox{ is the angle between }\underline{a}\mbox{ and }\underline{b}~;~\underline{\hat{n}}\mbox{ is the unit vector perpendicular to }\underline{a}\mbox{ and }\underline{b}

    The direction of \underline{\hat{n}} is given by the Right hand rule.

    Therefore, e_{1}\times e_{2}=\underline{\hat{n}}|e_{1}||e_{2}|\sin \frac{\pi}{6}=\frac{1}{2}\underline{\hat{n}}

    Since e_{3} is perpendicular to e_{1}\mbox{ and }e_{2};

    \underline{\hat{n}}=e_{3}

    \therefore e_{1}\times e_{2}=\frac{1}{2}e_{3}

    It should be noted that, we have to assume that \underline{\hat{n}}\neq -e_{3} to get the given answer.


    Thanks Sudharaka! I have one question about your solution though:

    Why can we assume n=e3 if n is a unit vector? Does "normalized basis" in the problem statement mean e1,e2, and e3 are unit vectors as well? If this is the case, your solution looks correct.

    Thanks again for your help
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  4. #4
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    Re: Cross Product Problem

    Quote Originally Posted by divinelogos View Post
    Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).
    This is not true. The length of the vector U\times V is |U||V| sin(\theta)


    Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:


    1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?

    2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.
    Yes, to "normalize" a vector means to divide by its length to get a vector in the same direction with length 1. A "normalized" basis is a basis of vectors, each of length 1. (An "orthonormal" basis would be a basis consisting of vectors,each of length 1, each at right angles to the others- but you are told that the distance between e1 and e2 so this is NOT an orthonormal vector.)

    The length of the vector e_1\times e_2 would be |e_1||e_2|sin(\theta)= (1)(1)(1/2)= 1/2. Since the cross product of two vectors is perpendicular to both, in the "right hand rule" direction, and we are given that e_3 is perpendiculat to both e_1 and e_2, e_1\times e_2 is either \frac{1}{2}e_3 or -\frac{1}{2}e_3 but we don't have enough information to decide which. That would depend upon in which of two possible directions, perpendicular to the plane determined by the two vectors, e_3. They are apparently assuming that the three vectors form a "right hand frame"- that is, that if you curl the fingers of your right hand from e_1 to e_2, your thumb would be pointing in the direction of e_3 but I don't see that explicitely stated.

    Any ideas would be appreciated and I will definitely click the little thank you button for you
    Last edited by mr fantastic; September 13th 2011 at 03:26 AM. Reason: Fixed tex tags.
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  5. #5
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    Re: Cross Product Problem

    Quote Originally Posted by HallsofIvy View Post
    This is not true. The length of the vector U\times V is |U||V| sin(\theta)
    I just missed the normal vector on the end of the definition... my bad.
    Last edited by mr fantastic; September 13th 2011 at 03:27 AM. Reason: Fixed quote tags.
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  6. #6
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    Re: Cross Product Problem

    I understand now how to get e1xe2, but what about e2xe3 and e1xe3?

    e1xe2 generates a vector perpendicular to the plane spanned by e1 and e2, but i'm not sure how they got 2e1-rad(3)e2?
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