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Math Help - Operations with Algebraic Vectors

  1. #1
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    Operations with Algebraic Vectors

    The question that I am having trouble with is:
    a) Find the point on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
    b) Find a point not on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).

    I'm unsure of where to even start, so if anyone can give me any hints or suggestions that would be much appreciated! Please and thank you!
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  2. #2
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    Re: Operations with Algebraic Vectors

    Quote Originally Posted by misiaizeska View Post
    The question that I am having trouble with is:
    a) Find the point on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
    b) Find a point not on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
    For part a):
    find the y-intercept of the plane through (1,0,2) with normal <2,-2,-3>.

    For part b), find any other point on that same plane.
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  3. #3
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    Re: Operations with Algebraic Vectors

    Hello, misiaizeska!

    a) Find the point P on the y-axis that is equidistant from the points A(2,-1,1) and B(0,1,3)..

    The word "equidistant" should suggest the Distance Formula . . .

    Let point P be (0,y,0)

    We have: . \begin{Bmatrix}\overline{PA} &=& \sqrt{(0-2)^2 + (y+1)^2 + (0-1)^2} &=& \sqrt{y^2 + 2y - 6} \\ \overline{PB} &=& \sqrt{(0-0)^2 + (y-1)^2 + (0-30^2} &=& \sqrt{y^2-2y+10} \end{Bmatrix}


    Then: . \sqrt{y^2+2y+6} \:=\:\sqrt{y^2-2y+10}

    . . . . . . . y^2+2y + 6 \:=\:y^2 - 2y + 10

    . . . . . . . . 4y \:=\:4 \quad\Rightarrow\quad y \:=\:1


    \text{Therefore: }\:P(0,1,0)
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