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Math Help - Vectors Question

  1. #1
    Newbie SeerAlpha's Avatar
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    Vectors Question

    Hi I'm stuck on the following questions:

    a) Find an equation to describe the set of points equidistant from A(2, -1, 3) and B(1, 2, -3).
    b) Find the coordinates of two points that are equidistant from A and B.

    Please show the steps and thanks for your help!
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  2. #2
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    Re: Vectors Question

    (1) Write the (square of) distance from (x, y, z) to A(2, -1, 3).
    (2) Write the (square of) distance from (x, y, z) to B(1, 2, -3).
    (3) Equate (1) and (2).
    Thanks from SeerAlpha
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  3. #3
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    Re: Vectors Question

    Quote Originally Posted by SeerAlpha View Post
    Hi I'm stuck on the following questions:

    a) Find an equation to describe the set of points equidistant from A(2, -1, 3) and B(1, 2, -3).
    b) Find the coordinates of two points that are equidistant from A and B.
    Here is a second way. Write the equation of the plane that contains the midpoint of \overline{AB} that has normal \overrightarrow {AB} .
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  4. #4
    Newbie SeerAlpha's Avatar
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    Re: Vectors Question

    The book's answer is x-3y+6z = 0 where P(x,y,z) is the point but I didn't get that as an answer... so did you get the book's answer or a different one? I probably missed something
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  5. #5
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    Re: Vectors Question

    What did you get and how did you get it? What is the point halfway between A and B? What is the vector \vec{AB}?
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  6. #6
    Newbie SeerAlpha's Avatar
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    Re: Vectors Question

    I did PA = (2, -1, 3) - (x,y,z)
    = (2-x, -1-y, 3-z)

    PB = (1,2,-3) - (x,y,z)
    = (1-x, 2-y, -3-z)

    |PA| = |PB|

    sqrt[(2-x)^2 + (-1-y)^2 + (3-z)^2] = sqrt[(1-x)^2 + (2-y)^2 + (-3-z)^2]
    4 - 4x + x^2 + 1 + 2y + y^2 + 9 - 6z + z^2 = 1 - 2x + x^2 + 4 - 4y + y^2 + 9 +6z + z^2
    -2x + 6y - 12z = 0

    (sorry i'm not sure how to add the direction arrows)
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  7. #7
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    Re: Vectors Question

    Quote Originally Posted by SeerAlpha View Post
    I did PA = (2, -1, 3) - (x,y,z)
    = (2-x, -1-y, 3-z)

    PB = (1,2,-3) - (x,y,z)
    = (1-x, 2-y, -3-z)

    |PA| = |PB|

    sqrt[(2-x)^2 + (-1-y)^2 + (3-z)^2] = sqrt[(1-x)^2 + (2-y)^2 + (-3-z)^2]
    4 - 4x + x^2 + 1 + 2y + y^2 + 9 - 6z + z^2 = 1 - 2x + x^2 + 4 - 4y + y^2 + 9 +6z + z^2
    -2x + 6y - 12z = 0
    That is that correct answer.
    -2x+6y-12=0 is x-3y+6z=0.
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  8. #8
    Newbie SeerAlpha's Avatar
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    Re: Vectors Question

    ... Doh! Should of compared the two answers more carefully. Hmm I'm going to try part b but can you guys give me a hint on how to start it?
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  9. #9
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    Re: Vectors Question

    Oh and if possible I have one last question that I really need help with... Please see the attached image for the question and diagram. Vectors Question-15-d.png
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  10. #10
    Newbie SeerAlpha's Avatar
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    Re: Vectors Question

    Ok I got part b just need some guidance on the diagram question.
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  11. #11
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    Re: Vectors Question

    It is better to start a new thread.

    You have coordinates of all vectors. Use the fact that \overrightarrow{OP} \cdot\overrightarrow {AE}=|\overrightarrow{OP} |\cdot|\overrightarrow{AE}|\cos\alpha where \alpha is the required angle, \overrightarrow{OP} \cdot\overrightarrow{AE} is the dot product and |\overrightarrow{OP} | is the length of \overrightarrow{OP}, i.e., \sqrt{\overrightarrow{OP}\cdot\overrightarrow{OP}}.
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  12. #12
    Newbie SeerAlpha's Avatar
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    Re: Vectors Question

    I think I solved it. You've all been a great help!
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