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Finding components of a vector

  1. Jul 5, 2011 #1
    hello! please someone help me,:smile: here is my question.

    Find the components of d=(3,5,7) along the directions of u, v and w
    consider: u=1/3(2,2,-1) v=1/3(2,-1,2) w=1/3(-1,2,2)

    I don't know where to start, I need some ideas to solve this
    thanx:smile:
     
  2. jcsd
  3. Jul 6, 2011 #2

    ehild

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    Start with the definition of "component of a vector". Think of scalar product.

    ehild
     
  4. Jul 6, 2011 #3
    please tell me how to apply scalar product, I know what is scalar product, but I don't know how to use it for find components
     
  5. Jul 6, 2011 #4

    ehild

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    You can write up a vector as a linear combination of the base vectors. a=x1u+x2v+x3w. x1, x2, x3 are the components of vector a. Notice that u,v,w are orthonormal. What do you get if multiplying the equation with one of them?

    ehild
     
  6. Jul 6, 2011 #5
    I don't understand, with which do I need to multiply??:confused: can you show me how to do that?
     
  7. Jul 6, 2011 #6

    ehild

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    Multiply both sides of the equation d=x1u+x2v+x3w by u:
    du=x1uu+x2vu+x3wu.



    ehild
     
  8. Jul 6, 2011 #7
    How do I suppose to calculate du? by using the scalar product??
     
  9. Jul 6, 2011 #8
    after solving du=x1uu+x2vu+x3wu, I got 27 for x1 is this correct ehild?
     
  10. Jul 6, 2011 #9

    ehild

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    No, how did you get it?

    ehild
     
  11. Jul 6, 2011 #10
    I used scalar product to solve it,

    du=x1uu+x2vu+x3wu
    (3,5,7)1/3(2,-2,-1)=x11/9(2,-2,-1)(2,-2,-1)+x21/3(2,-1,2)1/3(2,2,-1)+x31/3(-1,2,2)1/3(2,2,-1)
     
  12. Jul 6, 2011 #11

    ehild

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    What did you get for the individual products du, uu, vu, wu?

    ehild
     
  13. Jul 6, 2011 #12
    I'm sorry, what did you mean by individual products, is it du,x1uu,x2uv and x3uw,if so x2uv and x3uw become zero,:redface:
     
  14. Jul 6, 2011 #13

    ehild

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    Yes, but how much is uu?

    ehild
     
  15. Jul 6, 2011 #14
    oops I've made a mistake when solving :smile:, is UU=1, then I got 3 for x1 :smile:
     
  16. Jul 6, 2011 #15

    ehild

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    Very good!!!! Now do the same (without mistake) with v and w.

    Note: this method works for orthogonal u, v ,w vectors only. In general, you can write a linear system of equations for x1,x2,x3 and solve it.

    ehild
     
  17. Jul 6, 2011 #16
    Oh...thank you very much ehild, I really appreciate your help:smile:, I have another problem, can you explain me little bit about mutually perpendicular unit vectors:smile:
     
  18. Jul 6, 2011 #17

    ehild

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    You had such ones just now. u, v, w are mutually perpendicular if the scalar product of any two of them is zero, and a vector is unit vector if its modulus (or magnitude) is 1. You get it by multiplying the vector by itself and taking the square root.
    You can make an unit vector of any vector by dividing all components by the modulus.

    Can I help something more?

    ehild
     
  19. Jul 6, 2011 #18
    I found 3, 5 and 7 respectively for the components x1,x2 and x3. but I have a doubt about these values,

    I had to Find the components of d=(3,5,7) along the directions of u, v and w
    consider: u=1/3(2,2,-1) v=1/3(2,-1,2) w=1/3(-1,2,2), finally I got 3,5,7. this is really confusing me:confused:
     
  20. Jul 6, 2011 #19

    ehild

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    Your result means that by decomposing the vector d to three vectors parallel to u,v, w, these component vectors are
    3u, 5v and 7 w, that is: d=3u+5v+7w. Check.

    ehild
     
  21. Jul 7, 2011 #20
    Thanx ehild, then what is my final answer would be?
     
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