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Math Help - Proof

  1. #1
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    Arrow Proof

    Hi,
    This is my first post on the forum, but I have been reading this forum for at least 2 years now. It has always been very useful to read some of the posts you guys make.

    However, now, it seems like I can't solve my problem simply by reading the forum...I would need a little extra help.

    Here is how it goes:

    (Note: Let bold letters be vectors)
    I know that u=(u1,u2,u3), v=(v1,v2,v3) and w=(w1,w2,w3)
    Let A= \begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end  {array}

    I have to prove that:
    (Note: Let u,v and w, be vectors)
    det(A) \leq \parallel u \parallel \parallel v \parallel  \parallel w \parallel


    I tried to replace the norm of u times norm of v times norm of w by its equivalent using their values (u1,u2,u3), (v1,v2,v3), (w1,w2,w3).

    Hence, \parallel u \parallel = \sqrt{u \cdot u} ...


    But then, all I got was a huge equation and the strange impression that I am not even close from doing the right thing...

    Anyone could help me head in the right direction ?

    Thanks a lot !
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by elitethut View Post
    Hi,
    This is my first post on the forum, but I have been reading this forum for at least 2 years now. It has always been very useful to read some of the posts you guys make.

    However, now, it seems like I can't solve my problem simply by reading the forum...I would need a little extra help.

    Here is how it goes:

    (Note: Let bold letters be vectors)
    I know that u=(u1,u2,u3), v=(v1,v2,v3) and w=(w1,w2,w3)
    Let A= \begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end  {array}

    I have to prove that:
    (Note: Let u,v and w, be vectors)
    det(A) \leq \parallel u \parallel \parallel v \parallel \parallel w \parallel


    I tried to replace the norm of u times norm of v times norm of w by its equivalent using their values (u1,u2,u3), (v1,v2,v3), (w1,w2,w3).

    Hence, \parallel u \parallel = \sqrt{u \cdot u} ...


    But then, all I got was a huge equation and the strange impression that I am not even close from doing the right thing...

    Anyone could help me head in the right direction ?

    Thanks a lot !
    The scalar triple product should do the trick.

    \vec{a} \cdot (\vec{b} \times \vec{c})=\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

    Now just use the geometric defintions of the dot and cross product
    \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}

    where \vec{n} is a unit vector normal to the plane formed by b and c ||n||=1

    and \vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)

    and remember the sine and cosine are always less than or equal to 1.
    Last edited by TheEmptySet; October 27th 2009 at 02:00 PM. Reason: editied for clairity
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  3. #3
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    Hmmm, I'm not sure I understand the role that \vec{n} has to play in this.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by elitethut View Post
    Hmmm, I'm not sure I understand the role that \vec{n} has to play in this.
    When you cross two vectors you get a vector that is normal(perpendicular) to the plane spanned by the two vectors.

    The n is just giving you the direction that the resulting vector is going.
    The ||b|||c||\sin(\theta)is the length(magnitude) of the resulting vector.

    This is needed becuase the dot product is a binary operation on two vectors, not a vector and magnitude. If we didn't have the unit normal the dot product would not make sense.

    I hope this helps.

    I have editied the above post for clarity as well.
    Last edited by TheEmptySet; October 27th 2009 at 02:01 PM. Reason: To make the argument clearer
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  5. #5
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    Ok... So

    <br />
\vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}<br />
    and
    <br />
\vec{a}=||a||\cos(\theta) \vec{n}<br />

    Therefore,
    <br />
\vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)<br />

    Is that right?
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  6. #6
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    Quote Originally Posted by elitethut View Post
    Ok... So

    <br />
\vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}<br />
    and
    <br />
\vec{a}=||a||\cos(\theta) \vec{n}<br />

    Therefore,
    <br />
\vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)<br />

    Is that right?
    <br />
\vec{a}=||a||\cos(\theta) \vec{n}<br />

    Is incorrect. a is just the vector a.
    Now:
    d\cdot e=||d||||e||cos(\theta) where \theta is the angle beteween d and e.

    Therefore:
    a\cdot(b\times c)=||a||\Bigl(||\bigl(||b|||c||\sin(\theta) \vec{n}<br />
\bigr)||\Bigr)\cos(\theta_2)=||a||||b||||c||\sin(\  theta)\cos(\theta_2)||n||

    Now n is a unit vector so:

    ||a||||b||||c||\sin(\theta)\cos(\theta_2)||n||=||a  ||||b||||c||\sin(\theta)\cos(\theta_2)


    Be careful, don't confuse \theta and \theta_2 together.
    The one inside sin is the angle between b and c, and the one inside cos is the angle between a and the resulting vector of (bxc).
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  7. #7
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    Ok,

    So basically,

     ||a||||b||||c||\sin(\theta)\cos(\theta_2)<br />
    is equal to Det(A)...but how does this prove that the det(A) is smaller than <br />
||a|||b|||c||<br />

    Edit: Oh, is it simply that \sin(\theta) and \cos(\theta_2) will always give values bellow 1 (or equal to one), and therefore the value of  ||a||||b||||c|| times 2 values that are equal or smaller to 1 will always give a smaller or equal value than  ||a||||b||||c||
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  8. #8
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    Quote Originally Posted by elitethut View Post
    Edit: Oh, is it simply that \sin(\theta) and \cos(\theta_2) will always give values bellow 1 (or equal to one), and therefore the value of  ||a||||b||||c|| times 2 values that are equal or smaller to 1 will always give a smaller or equal value than  ||a||||b||||c||
    Yes, that is correct.
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  9. #9
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    Ok, I understand everything except one thing.

    Isn't the formula supposed to be
    ||\vec{b} \times \vec{c}||=||b|||c||\sin(\theta) \vec{n}

    and not :

    \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n} ???
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  10. #10
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    Quote Originally Posted by elitethut View Post
    Ok, I understand everything except one thing.

    Isn't the formula supposed to be
    ||\vec{b} \times \vec{c}||=||b|||c||\sin(\theta) \vec{n}

    and not :

    \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n} ???
    \vec{b} \times \vec{c} is a vector while ||\vec{b} \times \vec{c}|| is a length of the vector.

    The length of the vector that is the result of a cross product \vec{b} \times \vec c is
    ||\vec{b} \times \vec c||=||b|||c||\sin(\theta),
    but the vector it self is ||b|||c||\sin(\theta) \vec{n}

    Read again what TheEmptySet posted above:
    When you cross two vectors you get a vector that is normal(perpendicular) to the plane spanned by the two vectors.

    The n is just giving you the direction that the resulting vector is going.
    The ||b|||c||\sin(\theta)is the length(magnitude) of the resulting vector.

    This is needed becuase the dot product is a binary operation on two vectors, not a vector and magnitude. If we didn't have the unit normal the dot product would not make sense.

    I hope this helps.

    I have editied the above post for clarity as well.
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