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Math Help - trigonometry vectors in the plane

  1. #1
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    trigonometry vectors in the plane

    I am checking to see if I understand this problem correctly.

    Find the magnitude of the vector u = QP--> with Q = (6, 6) and P= (-3, 8)

    The arrow indicates the directed line segment.

    -3 - 6 = -9

    8 - 6 = 2

    ||v|| = sqrt( (-9)^2 + 2^2 ) = sqrt(85) = 9.2195
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  2. #2
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    Dear robasc, it's okay.
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  3. #3
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    Thank you for your answer. My book gives very brief explanations on exercises and that stinks!
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  4. #4
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    Original Problem:
    Find the magnitude of the vector u = QP--> with Q = (6, 6) and P= (-3, 8)

    The arrow indicates the directed line segment.

    -3 - 6 = -9

    8 - 6 = 2

    ||v|| = sqrt( (-9)^2 + 2^2 ) = sqrt(85) = 9.2195
    I have just one question pertaining to the problem above.

    Assume the direction of QP is going the opposite direction such as PQ-->

    Then I would find the components like this:

    6 - (-3) = 9

    6 - 8 = -2

    is that correct?
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  5. #5
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    Yes, or you can multiply -1.
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