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Review of vectors

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the unit vector that is parallel to the plane formed by [tex]\hat{x} + \hat{y} + \hat{z}[/tex], and 3{x} − 2{y} − 2{z}, and perpendicular to 2{x} + 2{y} − {z}

    2. Relevant equations
    The Cross product equation A x B = (A_1B_1) + (A_2B_2) +...+ (A_nB_n)


    3. The attempt at a solution
    I know two vectors define a plane- by definition. So do I choose a vector that when crossed with both {x} + {y} + {z} and 3{x} − 2{y} − 2{z} produces 0. And one that when dotted with 2{x} + 2{y} − {z} is 0? It seems there would be a more elegant way of doing this.

    1. The problem statement, all variables and given/known data
    If z = 0.1 -0.2j, find the following:
    z^X, sqrt(z), z^2, z^3
    for both the Real and Imaginary components.

    2. Relevant equations
    None, I can think of

    3. The attempt at a solution
    For the first part-
    (0.1 - 0.3j)^x :thats the real part, and the imaginary part is 0?
    And the others seem trivial. I don't think I am interpretting this problem correctly. It seems like these problems (finding the real and imaginary components for each assignment) seems straight-foward.


    Thanks,


    JL
     
    Last edited: Jul 6, 2009
  2. jcsd
  3. Jul 6, 2009 #2

    turin

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    The expression for your cross product in the relevant equations list is the expression for a dot product, not a cross product.

    I would work this problem as you suggest. Why do you think that it is inelegant? The use of invariants (e.g. dot product being invariant to rotation) is one of the more elegant approaches to physics problems.

    BTW: Does anyone else have a problem reading the OP? Maybe my browser is just not supporting the symbols.
     
  4. Jul 6, 2009 #3
    So to have my final vector be parallel to the plane (of the first two vectors), I have to do a guess and check kind of thing. Choose a vector (of my choice such that) when crossed with the initial two vectors has a result of 0. And that same vector [that I chose] when crossed with the last vector has a result of nonzero -means it is parallel to the first two vectors and perpendicular to the last?

    For some reason this forum wouldn't let me use the latex code [tex][/tex]. It kept scrambling my numbers into nonsense symbols. So I removed most of my latex code and that fixed it.

    As for the second question, I am not sure how to interpret it.


    Thanks Turin,


    JL

     
    Last edited: Jul 6, 2009
  5. Jul 6, 2009 #4

    turin

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    Think of your problem in terms of unknowns and equations. How many unknowns do you have? How many equations do you have? You can actually solve for the vector; no guessing is necessary. (Now does it sound more elegant?) Oh, one other thing that might be helpful: use a normal vector for the plane.

    BTW: try to avoid TeX, if it isn't too much trouble to do so. For example, you can use the symbols in my signature. There is also a post dedicated to supported symbols that don't require TeX. If you feel that you must, then the two tags are:

    [ t e x ] ... [ / t e x ]

    for line-broken TeX and [ i t e x ] ... [ / i t e x ] for inline TeX (without the spaces, of course).
     
  6. Jul 6, 2009 #5
    I found the normal vector to the plane, which is orthogonal to the first two vectors- but how does this help me ? Do you know what the second question is asking for?
     
  7. Jul 6, 2009 #6

    turin

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    They gave you one vector that they require to be perpendicular, so you know that the dot product with that one must vanish. You just found another vector that you know must also be perpendicular, by construction, so you know that the dot product with that one must vanish. You know the length of the vector (it is a unit vector). That's how many equations. You are (presumably) working in a 3-D vector space, so you must specify 3 vector coefficients. That's how many unknowns. Solve for the unknowns.

    For the second question, it looks like electrical engineering notation, where j is used instead of i for the sqrt of -1. I think they just want you to express those expressions without using z (i.e. the only variable should be x), and probably as simply as possible. I think that your guess is incorrect. I suggest polar notation. BTW, I think that the second question is not so much a vector question as it is a complex numbers question. But, then again, complex numbers are sort of like (2-D) vectors with some extra properties.

    For future reference: one question per thread, please.
     
  8. Jul 6, 2009 #7

    tiny-tim

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    Hi turin! :smile:

    use [noparse][noparse], then you can write [tex] … [/tex][/noparse] :wink:
     
  9. Jul 6, 2009 #8
    I am not fully sure what you mean. Should I have three equations to put into a matix-system of equations- to solve for the unknowns?
     
  10. Jul 6, 2009 #9
    So I took the cross product betweeen the following two vectors (first two vectors):
    x + y + z and 3x - 2y -2z,
    which resulted in the normal vector, I called n = 0{x} - 5{y} -5{z}.

    Now If I take the cross product between n and 2x + 2y - z (third vector), I should get a vector that is parallel to the first two, yet orthogonal to the third, which I will call vector q = -15{x} + 9{y} - 10{z} as my final answer.

    Is that correct?

    thanks,

    JL
     
    Last edited: Jul 6, 2009
  11. Jul 6, 2009 #10

    turin

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    You should be able to check your first cross product easily enough. Hint: take dot products of the result with the first two vectors. What should you get? What do you get?
     
    Last edited: Jul 6, 2009
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