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Math Help - [SOLVED] Vector Calculus: Parallelogram Problem

  1. #1
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    [SOLVED] Vector Calculus: Parallelogram Problem

    Hello, I am stuck on a problem that at the conceptual level makes perfect sense. Here's the problem:

    Use vectors to show that the diagonals of a parallelogram have the same length if and only if the parallelogram is a rectangle. (Hint: let a and b be vectors along two sides of the parallelogram. Express vectors running along the diagonals in terms of a and b.)

    I have attached a picture. I can let:
    d1 = a + b
    d2 = b - a

    I know that I am trying to prove that:
    \|d1\| = \|d2\|
    When \cos\theta = 0 for the angle between a and b.

    Most of my work so far essentially amounts to me plugging in for the Angle Between Two Vectors formula and coming out with a dotted with b = 0. This occurs if either is the 0 vector or if they are perpendicular. However, this doesn't help me with the length of d1 and d2. I am not sure how I can relate the two together. Thanks for any assistance.
    Attached Thumbnails Attached Thumbnails [SOLVED] Vector Calculus: Parallelogram Problem-parallelogram.jpg  
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  2. #2
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    Suppose that \overrightarrow a ~\&~\overrightarrow b are the two adjacent vectors determining the parallelogram.
    The diagonals of the parallelogram are: \left( {\overrightarrow a  + \,\overrightarrow b } \right)\;\& \,\left( {\overrightarrow a  - \overrightarrow b } \right).
    Look at the following.
    \left( {\overrightarrow a  + \,\overrightarrow b } \right)\; \cdot \,\left( {\overrightarrow a  - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a}  - \overrightarrow b \overrightarrow b
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  3. #3
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    <br />
\left( {\overrightarrow a + \,\overrightarrow b } \right)\;<br />
\cdot \,\left( {\overrightarrow a - \overrightarrow b } \right) =<br />
\overrightarrow a \overrightarrow { \cdot a} - \overrightarrow b<br />
\overrightarrow b<br />

    I am not sure I understand the significance in dotting the diagonals together. The expression shows that the diagonals dotted together equal the length of a squared minus the length of b squared. Thanks for the help so far, I feel like I am missing something that should be very obvious to me.
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  4. #4
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    \|\vec{d_1}\|^2 = (\vec{a} +\vec{b}) \cdot (\vec{a} +\vec{b}) = \|\vec{a}\|^2+2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2
    \|\vec{d_2}\|^2 = (\vec{a} -\vec{b}) \cdot (\vec{a} -\vec{b}) = \|\vec{a}\|^2-2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2
     \|\vec{d_1}\|^2=\|\vec{d_2}\|^2\Leftrightarrow \vec{a} \cdot \vec{b}=0\Leftrightarrow \cos \theta=0
    Last edited by mr fantastic; September 18th 2009 at 08:27 AM. Reason: Restored original reply
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  5. #5
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    Luobo...thank you. That is what I was missing. That makes complete sense. I suppose was too tired to think clearly. I was hoping I could get a nudge, and that's what Plato tried, but I guess the roadblock in my mind just wouldn't allow me go past it. Thanks again!
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