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I hate Vector Algebra

  1. Sep 7, 2004 #1
    I hate Vector Algebra!!

    I am having some trouble doing some vector algebra and any help or direction is greatly appreciated.

    In our lecture we drew a vector C(s) = A + s(B-A) which described a line passing through points A and B and found that the vector of min. length (shortest dist. from origin to line) was given by so = [A (dot) (A-B)]/[|B-A|2. The actual problem is to plug so into C(s) to show that C(so) = (sqrt(|A|2|B|2 - (A dot B)2))/(|B-A|).

    What I have so far consists of changing [A (dot) (A-B)]/[|B-A|2 to [-A (dot) (B-A)]/[|B-A|2 so I can square the (B-A) however, when I do that I do not know how to then multiply that by the dot product of A. I know that A dot (B-A) is A dot B minus A dot A (A squared) but I have no idea how to do A dot (B-A)2. Thanks for any help.
     
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  3. Sep 7, 2004 #2

    robphy

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    For a vector V, the quantity V2= ( V (dot) V ) is a scalar.
    Note that A (dot) ( (B-A)2 ) is not a legal operation.
     
  4. Sep 7, 2004 #3
    Sorry about my notation - I should have been more obvious about the A squared part being a scalar but thanks for telling me that A (dot) ( (B-A)2 ) is not a legal operation, now I know not to work in that direction.
     
  5. Sep 8, 2004 #4

    robphy

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    Maybe not the most elegant solution

    [tex]
    \def\vA {\vec A}
    \def\vB {\vec B}
    \def\vC {\vec C}
    \begin{align*}
    \vC
    &= \vA + s(\vB-\vA)\\
    C^2 &= A^2 + 2s\vA\cdot(\vB-\vA)+s^2(\vB-\vA)\cdot(\vB-\vA)\\
    \end{align*}
    [/tex]

    with
    [tex]
    \def\vA {\vec A}
    \def\vB {\vec B}
    \def\vC {\vec C}
    s_0=\frac{\vA\cdot(\vA-\vB)}{|B-A|^2}
    [/tex]

    [tex]
    \def\vA {\vec A}
    \def\vB {\vec B}
    \def\vC {\vec C}
    \def\vAB {(\vA\cdot\vB)}
    \begin{align*}
    C(s_0)^2
    &= A^2 + 2 \left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]\vA\cdot(\vB-\vA)+\left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]^2(\vB-\vA)\cdot(\vB-\vA)\\
    &= A^2 - 2 \left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]\vA\cdot(\vA-\vB)+ \frac{(\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\
    &= A^2 - \frac{(\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\
    &= \frac{A^2 (\vB-\vA)\cdot(\vB-\vA)- (\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\
    &= \frac{A^2 (B^2-2\vAB+A^2)- (A^2-\vAB)^2}{|B-A|^2}\\
    &= \frac{A^2 (B^2-2\vAB+A^2)- (A^4-2A^2\vAB +\vAB^2)}{|B-A|^2}\\
    &= \frac{A^2 B^2-2A^2\vAB+A^4- A^4+2A^2\vAB -\vAB^2}{|B-A|^2}\\
    &= \frac{A^2 B^2-\vAB^2}{|B-A|^2}\\
    \end{align*}
    [/tex]
     
  6. Sep 9, 2004 #5
    I think that's as elegenat as it can get. After about an hour two days ago, I finally came up with a solution that was just as long if not longer than yours. It seems as if it's just brute force is the route here. :bugeye: Thanks for all your help.
     
  7. Sep 9, 2004 #6

    robphy

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    If you allow yourself to use some trigonometric functions, it may have been easier to manipulate.
    Note that [itex]\vec A\cdot \vec B=AB\cos\theta[/itex] so that [itex]A^2B^2-(\vec A\cdot \vec B)^2=A^2B^2-A^2B^2\cos^2\theta=A^2B^2(1-\cos^2\theta)=A^2B^2\sin^2\theta[/itex].
     
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