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Resolve vector component

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Resolve the vector 6i+2j-2k into two vectors, one parallel and another perpendicular to i+j+k

    2. Relevant equations
    [tex]a\cdot b = 0 \; \text{for two perpendicular vectors}[/tex]

    [tex]a=\lambda \;b\;\text{for parallel vectors}[/tex] , [tex]\lambda = \text{parameter}[/tex]

    3. The attempt at a solution
    I have no idea to start. How to resolve one component of vector into two components ?

  2. jcsd
  3. Aug 23, 2009 #2


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    Have you learned about the geometrical meaning of dot product?
    I.e. if I drew v = 6i+2j-2k and w = i+j+k for you, could you explain to me how v · w appears in the picture?
  4. Aug 23, 2009 #3
    I don't think that he can give you geometrical interpretation because dot product is just a scalar and not vector.
    Although, he can give you the scalar projection of v onto w.

    And do you mean to resolve 6i+2j-2k = c + d ?

    If so, let c be the vector parallel to and d perpendicular to the vector (1,1,1) i.e i+j+k


    and do you know what d will equal to? What are the conditions of the task?

    Last edited: Aug 23, 2009
  5. Aug 23, 2009 #4


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    Okay, so a vector parallel to i+ j+ k must be [itex]\lambda i+ \lambda j+ \lambda k[/itex]. Suppose ai+ bj+ ck is the vector perpendicular to that. Then you have [itex](ai+ bj+ ck)\cdot(\lamba i+ \lambda j+ \lambda k)= a\lambda+ b\lambda+ c\lambda[/itex][itex]= \lambda(a+ b+ c)= 0[/itex] and [itex](ai+ bj+ ck)+ (\lambda i+ \lambda j+ \lambda k)[/itex][itex]= (a+\lambda)i+ (b+\lambda)j+ (c+ \lambda)k[/itex][itex]= 6i+ 2j- 2k[/itex]. That gives you four equations to solve for a, b, c, and [itex]\lambda[/itex].
  6. Aug 23, 2009 #5
    Hi CompuChip, Дьявол, and Mr. HallsofIvy

    I get it now. Sorry, but I have another simple question. What is the meaning of dot product ?

    It's easier for me to imagine cross product. If we cross two vectors, we will get third vector that is perpendicular to the previous two vectors.

    But, what about dot product ? If we dot 2 vectors, we get a numerical value, What does the numerical value represent?

  7. Aug 23, 2009 #6
    As I was taught, the dot product is where you only consider the part of the second vector being multiplied that is parallel to the first vector.
  8. Aug 24, 2009 #7
    http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Dot_Product.svg/300px-Dot_Product.png [Broken]

    And the dot product is A • B = |A| cos(θ) |B|

    |A| cos(θ) is the scalar projection of A onto B.

    So you got the part down there, just you multiply it with the magnitude of B.

    Last edited by a moderator: May 4, 2017
  9. Aug 24, 2009 #8
    Hi Дьявол and mg0stisha

    Wow, now I get the meaning of dot product. Thanks a lot to all of you !! (CompuChip, Дьявол, MR. HallsofIvy, mg0stisha) ^^
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