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Help me determine a vector that's perpendicular (with work)

  1. Sep 7, 2015 #1
    Problem: The vectors A and B together define a plane surface. Determine a vector of magnitude 3 that is perpendicular to that surface.

    What I am thinking is Vector A is on the x-axis and B is on the y-axis. Therefore, the vector that I must determine, runs along the z-axis which is 90 degrees to that plain.

    So, can I say that the answer is: The vector that must be determined is 90 degrees z-axis of x-axis with a magnitude of 3?

    Sorry if what I am doing or saying is not even remotely close to the valid answer.

    Can you please tell me how to approach this problem, maybe even go through it?

    Here is a picture of what I think the Determined Vector of magnitude 3 is:
    Picture URL: http://tinypic.com/r/25u16jd/8
    http://tinypic.com/r/25u16jd/8
     
  2. jcsd
  3. Sep 7, 2015 #2

    fzero

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    Have you studied the cross product of vectors?
     
  4. Sep 8, 2015 #3
    Two vectors are perpendicular if their dot product is zero.
     
  5. Sep 8, 2015 #4
    No I haven't, I am still learning the basics. I will spend the rest of tonight looking over it.
     
  6. Sep 8, 2015 #5
    So |A|*|B|*cos(theta) = 0
    cos(theta) = 0 / (|A|*|B|)
    theta = arccos(0)
    theta = 90 degress
    I can say that the VECTOR that is perpendicular to the plane is 90 degrees counterclockwise and have its magnitude = 3?
     
  7. Sep 8, 2015 #6
    I think you should rather look at cross product
     
  8. Sep 8, 2015 #7
    LOL, yeah I am, thank you.
     
  9. Sep 8, 2015 #8
  10. Sep 8, 2015 #9
    So I am not sure, but this is what I got.

    First I find the cross product of A and B, since it will give me a vector that is perpendicular to both A and B.

    A x B = C

    And now that I have C in vector components, but from here I am lost. Do I multiply it by 3 since it wants a "vector with a magnitude of 3"?

    so 3C = 3*(# i + #j + #k) is my answer?

    Can you tell me if this is wrong and explain to me the correct way?
     
  11. Sep 8, 2015 #10
    So I am not sure, but this is what I got.

    First I find the cross product of A and B, since it will give me a vector that is perpendicular to both A and B.

    A x B = C

    And now that I have C in vector components, but from here I am lost. Do I multiply it by 3 since it wants a "vector with a magnitude of 3"?


    so 3C = 3*[#i + #j + #k] is my answer?

    Can you tell me if this is wrong and explain to me the correct way?
     
  12. Sep 8, 2015 #11

    fzero

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    You should compute the magnitude of ##\mathbf{C} = \mathbf{A}\times \mathbf{B}##. You can either multiply ##\mathbf{C}## by an appropriate number to obtain a vector with magnitude 3 or derive a unit vector ##\hat{\mathbf{C}}## as an intermediate step and then multply that.
     
  13. Sep 8, 2015 #12

    RUber

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    This is just about right. However, as Fzero just said, if the magnitude of C is not 1, then 3*C will not have magnitude 3, it will have magnitude 3|C|.
     
  14. Sep 8, 2015 #13
    So for the second method that you mentioned you mean like this?

    A: 4i + 3j - 2k
    B: 2i - 3j - 2k

    AxB = C = i -12j - 18k


    ##\hat{\mathbf{C}}## = C / |C|

    3*##\hat{\mathbf{C}}## = 3 * [-12i - 18k / sqrt(468)]
     
  15. Sep 8, 2015 #14

    fzero

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    Yes, that's the general idea. But with ##\mathbf{A}## and ##\mathbf{B}## as given I find
    $$\mathbf{A}\times\mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & -2 \\ 2 & -3 & -2 \end{vmatrix} $$
    gives a slightly different result from yours. So recheck your calculation before working out ##\hat{\mathbf{C}}##.
     
  16. Sep 8, 2015 #15

    RUber

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    Where did your i component go?
     
  17. Sep 8, 2015 #16
    Sorry for line "AxB = C = i -12j - 18k", I meant to say AxB = C = 0i -12j - 18k.

    I forgot to go back and fix it after realizing my Num Lock was off.
     
  18. Sep 8, 2015 #17
    I am not going to lie, I find physics and things associated with the course to be very hard and that I am struggling. Therefore, thank you everyone for you guys' help.
     
  19. Sep 8, 2015 #18

    RUber

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    I don't think you are handling the negatives properly.
    I also get a different result.
     
  20. Sep 8, 2015 #19
    You are right, I am going to redo that.
     
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