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Math Help - Vector calculus question

  1. #1
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    Vector calculus question

    Hi - I'm totally stuck with this question: how to interpret it and tackle it. Any advice woiuld be greatly received!! We've not covered anything like this in classes...

    Let

     A = \left( x_{A},y_{A},z_{A} \right)
     B = \left( x_{B},y_{B},z_{B} \right)

    be two given distinct points in the Euclidean space. By finding the cartesian equation, descibe the surface representing the location of points M which are solutions of the equation

     \vec{AM}.\vec{MB} = 0




    Thanks, FH
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  2. #2
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    Re: Vector calculus question

    It is easy
    in the plane this equation represents a circle.
    in the 3d it will be a sphere. Put M(x,y,z) find the vectors AM and MB and get their scalar product....the rest is easy.
    Last edited by MINOANMAN; March 13th 2013 at 12:56 PM.
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  3. #3
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    Re: Vector calculus question

    OK. So ive taken the scalar prodcut and get:
     \left( X_{M}.X_{A}-X_{B}.X_{A}-X_{M}.X_{M} \right)+\left( Y_{M}.Y_{A}-Y_{B}.Y_{A}-Y_{M}.Y_{M} \right)+\left( Z_{M}.Z_{A}-Z_{B}.Z_{A}-Z_{M}.Z_{M} \right)

    But what next?
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  4. #4
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    Re: Vector calculus question

    No, that's incorrect. Taking the ordinates of point M to be (x, y, z), the vectors AM and BM are (x_a- x, y_a- y, z_a- z) and (x- x_b, y- y_b, z- b_z) respectively. The dot product is (x_a- x)(x- x_b)+ (y_a- y)(y- y_b)+ (z_a- z)(z- z_b)= -x_ax_b- (x_a+ x_b)x- x^2- y_ay_b- (y_a+ y_b)y- y^2- z_az_b- (z_a+ z_b)z- z^2= 0.
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  5. #5
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    Re: Vector calculus question

    Quote Originally Posted by HallsofIvy View Post
    No, that's incorrect. Taking the ordinates of point M to be (x, y, z), the vectors AM and BM are (x_a- x, y_a- y, z_a- z) and (x- x_b, y- y_b, z- b_z) respectively. The dot product is (x_a- x)(x- x_b)+ (y_a- y)(y- y_b)+ (z_a- z)(z- z_b)= -x_ax_b- (x_a+ x_b)x- x^2- y_ay_b- (y_a+ y_b)y- y^2- z_az_b- (z_a+ z_b)z- z^2= 0.
    How do you get the sign to be negative for the terms - (x_a+ x_b)x, - (y_a+ y_b)y & -(z_a+ z_b)z ?? Thats not how the above parenthesis expands?
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  6. #6
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    Re: Vector calculus question

    Hello, FelixHelix!

    Let: \begin{Bmatrix}A &=& \left( x_a,y_a,z_a\right) \\ B &=& \left(x_b,y_b,z_b\right)\end{Bmatrix} be two given distinct points in Euclidean space.

    By finding the cartesian equation, descibe the surface representing the location
    of points M which are solutions of the equation:  \overrightarrow{AM}\cdot\overrightarrow{BM}\,=\,0

    Let M \,=\,(x,y,z)

    Then: . \begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}

    We have: . \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle\cdot \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle  \;=\;0

    . . (x-x_a)(x-x_b) + (y-y_a)(y-y_b) + (z-z_a)(z-z_b) \;=\;0

    . . x^2 - (x_a\!+\!x_b)x + x_ax_b + y^2 - (y_a\!+\!y_b)y + y_ay_b + z^2 - (z_a\!+\!z_b)z + z_az_b \;=\;0

    . . x^2 - (x_a\!+\!x_b)x + y^2 - (y_a\!+\!y_b)y + z^2 - (z_a\!+\!z_b)z \;=\;-x_ax_b - y_ay_b - z_az_b


    Complete the square:

    . . x^2 - (x_a+x_b)x + (\tfrac{x_a+x_b}{2})^2 + y^2 - (y_a+b_b)y + (\tfrac{y_a+y_b}{2})^2 + z^2 - (z_a+z_b)z + (\tfrac{z_a+z_b}{2})^2

    . . . . . .  \;=\;(\tfrac{x_a+x_b}{2})^2 + (\tfrac{y_a+y_b}{2})^2 + (\tfrac{z_a+z_b}{2})^2 - x_ax_b - y_ay_b - z_az_b


    . . \left(x-\tfrac{x_a+x_b}{2}\right)^2 + \left(y - \tfrac{y_a+y_b}{2}\right)^2 + \left(z - \tfrac{z_a+z_b}{2}\right)^2 \;=\;\left(\tfrac{x_a-x_b}{2}\right)^2 + \left(\tfrac{y_a-y_b}{2}\right)^2 + \left(\tfrac{z_a-z_b}{2}\right)^2


    The locus of M is a sphere with center C\left(\frac{x_a+x_b}{2},\,\frac{y_a+y_b}{2},\, \frac{z_a+z_b}{2}\right)

    . . and radius: r \:=\:\sqrt{\left(\frac{x_a-x_b}{2}\right)^2 + \left(\frac{y_a-y_b}{2}\right)^2 + \left(\frac{z_a-z_b}{2}\right)^2}

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  7. #7
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    Re: Vector calculus question

    Quote Originally Posted by Soroban View Post
    Hello, FelixHelix!Let M \,=\,(x,y,z)Then: . \begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}We have: . \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle\cdot \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle  \;=\;0. . (x-x_a)(x-x_b) + (y-y_a)(y-y_b) + (z-z_a)(z-z_b) \;=\;0. . x^2 - (x_a\!+\!x_b)x + x_ax_b + y^2 - (y_a\!+\!y_b)y + y_ay_b + z^2 - (z_a\!+\!z_b)z + z_az_b \;=\;0. . x^2 - (x_a\!+\!x_b)x + y^2 - (y_a\!+\!y_b)y + z^2 - (z_a\!+\!z_b)z \;=\;-x_ax_b - y_ay_b - z_az_bComplete the square:. . x^2 - (x_a+x_b)x + (\tfrac{x_a+x_b}{2})^2 + y^2 - (y_a+b_b)y + (\tfrac{y_a+y_b}{2})^2 + z^2 - (z_a+z_b)z + (\tfrac{z_a+z_b}{2})^2 . . . . . .  \;=\;(\tfrac{x_a+x_b}{2})^2 + (\tfrac{y_a+y_b}{2})^2 + (\tfrac{z_a+z_b}{2})^2 - x_ax_b - y_ay_b - z_az_b. . \left(x-\tfrac{x_a+x_b}{2}\right)^2 + \left(y - \tfrac{y_a+y_b}{2}\right)^2 + \left(z - \tfrac{z_a+z_b}{2}\right)^2 \;=\;\left(\tfrac{x_a-x_b}{2}\right)^2 + \left(\tfrac{y_a-y_b}{2}\right)^2 + \left(\tfrac{z_a-z_b}{2}\right)^2The locus of M is a sphere with center C\left(\frac{x_a+x_b}{2},\,\frac{y_a+y_b}{2},\, \frac{z_a+z_b}{2}\right). . and radius: r \:=\:\sqrt{\left(\frac{x_a-x_b}{2}\right)^2 + \left(\frac{y_a-y_b}{2}\right)^2 + \left(\frac{z_a-z_b}{2}\right)^2}
    Then: . \begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}the vector isnt BM but MB. Will this matter?
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