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Scalar Product Explanation

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data

    What surface is represented by r . a = conts. that is described if a is a vector of constant magnitude and direction from the origin and r is the position vector to the point P(x1, x2, x3) on the surface?


    3. The attempt at a solution

    I know that the dot product of two vectors is a scalar. But I don't exactly understand what that scalar is a representation of. I don't necessarily want to have my homework question answered for me, I just don't know how I am supposed to approach it because I can't visualize what the scalar product is telling me (if there is any visualization possible).
     
  2. jcsd
  3. Sep 10, 2007 #2
  4. Sep 11, 2007 #3
    I think it has. I at least have an answer that I feel comfortable writing down and turning in. Thank you for the link :)
     
  5. Sep 11, 2007 #4

    learningphysics

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    expand the dot product... see if the equation seems familiar...
     
  6. Sep 11, 2007 #5
    Maybe I'm not too comfortable with my answer.

    I must be missing something obvious, I have that problem. Expanding the dot product of A . B = A1B1 + A2B2 + A3B3.

    The closest equation to that I can think of is the magnitude of a vector?
     
  7. Sep 11, 2007 #6

    learningphysics

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    instead of B, use R = (x,y,z).

    your equation is

    A.R = k

    what do you get as the equation after expanding A.R
     
  8. Sep 11, 2007 #7
    A.R = AxRx + AyRy + AzRz = k
     
  9. Sep 11, 2007 #8

    learningphysics

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    almost...
    calling A = (A1,A2,A3), and R = (x,y,z)

    A.R = A1x + A2y + A3z

    A.R = k

    A1x + A2y + A3z = k is the equation of the surface... does this equation seem familiar to you?
     
  10. Sep 11, 2007 #9
    It looks like the equation of a line in three dimensions
     
  11. Sep 11, 2007 #10
    My preferred way to think about the scalar product is as the length of one vector projected against the other. It's good to draw some pictures, and convince yourself that it's symmetric (it doesn't matter which one you choose to project and which to project against), and also understand the connection with planes. It's a very nice elementary exercise in geometry.
     
  12. Sep 11, 2007 #11

    learningphysics

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    Yes, this is a nice way to visualize it.
     
  13. Sep 11, 2007 #12

    learningphysics

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    No, it isn't. You might not have studied it if you haven't taken linear algebra... but it's the equation of a plane.

    I think genneth's idea is the best... to think of [tex]\frac{\vec{r}\cdot\vec{a}}{|a|}[/tex] as the scalar projection of r onto a... and think of all the points that give a constant scalar projection...
     
    Last edited: Sep 11, 2007
  14. Sep 11, 2007 #13
    My supervisor in my first year told us a simple four step procedure to solving any physics problem:

    1. Draw a diagram.
    2. Write some words.
    3. Think about it, hard.
    4. Write down the answer.

    and if it's too hard,

    3b. Write some equations.
    3c. Put the numbers in.

    Unfortunately, PF makes step 1 difficult -- it's actually one of my biggest gripes with physics and internet these days, as a single diagram often makes whole pages of misunderstanding disappear.
     
  15. Sep 11, 2007 #14
    Alright, I should've known that was the equation of a plane. My mind is growing numb.

    Okay, I think I am finally grasping it. I understand that one way of stating A . B is the projection of vector A onto vector B multiplied by the magnitude of vector B.

    So if you have A . B where vector A has magnitude of 1 and vector B has a magnitude of 2 and the angle between them is 30 degrees.

    scalar projection of A onto B is 1 . cos 30 which is [tex]\sqrt{3}[/tex]/2

    That multiplied by the magnitude of B gives you [tex]\sqrt{3}[/tex].

    So A . B = [tex]\sqrt{3}[/tex]

    And that describes the plane resulting from the projection of one vector onto the other?
     
  16. Sep 11, 2007 #15
    And thank you guys for being patient. Sometimes I think so hard about things that I miss the point altogether
     
  17. Sep 11, 2007 #16

    learningphysics

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    I think you have the idea. I messed up my definition of scalar projection in my previous post, but I've fixed it now.

    The equation is:
    [tex]\vec{r}\cdot\vec{a} = k[/tex]

    then if you divide both sides by |a|

    [tex]\frac{\vec{r}\cdot\vec{a}}{|\vec{a}|} = \frac{k}{|\vec{a}|}[/tex]

    Since a is a constant vector, the right side is just a new constant...

    So the points that satisfy this equation, are all the points such that the scalar projection of r onto a is constant...

    Draw a picture... start with 2D... you have [tex]\vec{a}[/tex] going through the origin, you have some point (x,y,z) so r is just (x-0,y-0,z-0) = (x,y,z)

    So you have r and a through the origin... draw the projection of r onto a... then find more points that give the same scalar projection... all those points should form a line that is perpendicular to a...

    But in 3D you get a plane that is perpendicular to a...
     
    Last edited: Sep 11, 2007
  18. Sep 11, 2007 #17
    I finally understand it. It's a miracle :) Thank you guys very much for your help with this.
     
  19. Sep 11, 2007 #18

    learningphysics

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    :smile: I like those steps. Yes, diagrams make things so much simpler... The process of drawing pictures... scanning pictures onto the computer is still too cumbersome... I suspect in the future we'll have easier ways to communicate visually on the internet...
     
  20. Sep 11, 2007 #19

    learningphysics

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    no prob. thanks to your posts and genneth's posts in this thread, I've also improved my understanding... :smile:
     
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