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Vectors help?

  1. Sep 19, 2003 #1


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    [SOLVED] Vectors help?

    The vector 2 + j2 is the same as what?

    I know that j2 is -1 so the answer would be -2 for the magnitude but what about the angle in degrees? I don't comprehend the problem can anyone help make sense outta this for me, plz!
    Dx :wink:
  2. jcsd
  3. Sep 19, 2003 #2
    A negative magnitude?

    I'm sorry but I didnt understand your question very well.
  4. Sep 19, 2003 #3
    I don’t follow what you’re doing at all. That problem represents a point in a two-dimensional plane, and it is found in reference to both a horizontal and vertical axis. The first number is how far you move along the horizontal axis and the second is how far you move along the vertical. Because both numbers are positive, not negative, you will be going to the right 2 units and upward 2 units, placing the point in quadrant number one. Because the real part (2) and the imaginary part (j2) are of equal magnitude this is going to form a square, and the angle you will always get with a square is 45 degrees. But, the math goes like this;

    C = 2 + j2
    C = [squ][(2)2 + (2)2] = the magnitude you need to find.


    [the] = tan-1(2/2) = the angle you need to find.

    In the equation to find the angle, where the 2/2 is located, the number that goes in the numerator will be the imaginary j value and the denominator will be the real value. This would have been easier to see if the values hadn’t been the same.

    Hope that helps,
    Good luck.

    I should expand a little about the square always having 45 degrees; If the point resides in quadrant one it is 45, in quad-2 it is 90 + 45, and so on.
    Last edited by a moderator: Sep 19, 2003
  5. Sep 21, 2003 #4


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    Sorry to have confused everybody. I was trying to make sense outa this problem but didn't realize I confused you guys along with myslef. :wink:
    I got it Boulderhead, THANKS!
    Dx :wink:
  6. Sep 21, 2003 #5
    You're welcome!
    Do you have a good calculator?
  7. Sep 21, 2003 #6


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    Its not the latest and greatest but me and my TI-83 have been though good times n bad. :wink:
  8. Sep 21, 2003 #7


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    One thing that is confusing to me is your statement "I know that j2 is -1". This is a mathematics forum and mathematicians use i for the imaginary unit. Only electricians use "jmaginary" numbers!

    Assuming that that is what you mean (and that you mean "^2" or squared rather than "j2") then 2- i^2= 2-(-1)= 3.

    The "degrees" would be 0 since this is a real number.
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