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Clarification on a dot product

  1. Sep 20, 2007 #1
    This technically a homework question, but needed for homework and understanding for homework to come. Just hope to get it cleared up, thanks again!

    1. This formula is given:

    r[tex]_{1}[/tex]s[tex]_{1}[/tex] + r[tex]_{2}[/tex]s[tex]_{s}[/tex]+ r[tex]_{e}[/tex]s[tex]^{3}[/tex] = [tex]\sum[/tex]r[tex]_{n}[/tex]s[tex]_{n}[/tex] (with the limits etc. not too important).

    Then, in respect to scalar products, the magnitude of any vector is denoted by l r l or by Pythoagora's theorem: square root of[r(1)[tex]^{2}[/tex] + r(2)[tex]^{2}[/tex] + r(3)[tex]^{2}[/tex]]
    (couldn't find square root in latex)

    and that THAT is the same as square root of [r . r]
    This last step I do not follow...
     
  2. jcsd
  3. Sep 20, 2007 #2

    Astronuc

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    Staff: Mentor

    Last edited: Sep 20, 2007
  4. Sep 20, 2007 #3
    I know that it's true, (in reading my text) but how does 3 squared terms boil down to r^2?
     
  5. Sep 20, 2007 #4

    Astronuc

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    Staff: Mentor

    This is a vector operation.

    One squares the corresponding vector components represented in each of three dimensions, which are orthogonal in the Cartesian system.

    The magnitude r is given as the sqrt of the sum of the squares, i.e. r = sqrt (r12 + r22 + r32), so

    r2 = r12 + r22 + r32
     
    Last edited: Sep 21, 2007
  6. Sep 20, 2007 #5
    See [tex] \vec{r}\cdot\vec{r}[/tex] is nothing but [tex]\sca{r}\sca{r}\cos\theta[/tex]. But [tex]\theta[/tex] = 0 and so [tex]\cos\theta[/tex] = 1.

    Hence [tex] \vec{r}\cdot\vec{r}[/tex] = [tex]^{}r^2[/tex]

    Now a projection of a vector on an axis is known as the component of the vector on that axis. So take a cartesian system and draw any arbritary vector and drop perpendiculars on the x y and z axis. Now try using pythagoras theorem for this and you will see that how

    [tex]\sca{r^2}[/tex] = [tex]\sqrt{r^2_x + r^2_y + r^2_z}[/tex]
     
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