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Math Help - Finding equation of a plane

  1. #1
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    Finding equation of a plane

    Hi

    I'm sitting a quiz for maths and i'm having trouble with this question...

    If anyone could help I would be very much appreciated!


    So the question asks to...

    Find the equation of the plane containing the line

    x = (0, 0, 1) + t (0, -1, 0) and the point (1, 1, 0).




    Here is my attempt:

    This line passes through the pt P(0, 0, 1)

    So we need a second vector which spans the plane together with the direction vector of the line.

    If Q(1, 1, 0) then the second vector is:

    PQ = (1, 1, 0) - (0, 0, 1) = (1, 1, -1)


    So the equation of the plane is:
    r = (0, 0, 1) + t (0,-1, 0) + s (1, 1, -1)


    But I can't seem to get my head around which answer this comes to...
    A. z - x = 1
    B. z + x = 1
    C. z - x = - 1
    D. z + x = - 1
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  2. #2
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    Remember that a plane can be defined by just a point and its normal: \vec{n} \cdot \vec{p} = 0

    or a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0 where (a,b,c) is your normal and (x_{0}, y_{0}, z_{0}) is a point.

    We have the point (0, 0, 1) and we have the direction vector of the line (0, -1, 0). We have to find a vector perpendicular to this (i.e. the normal). This is simple as all we have to do is find a vector such that: (0,-1,0) \cdot (a,b,c) = 0 (remember, two vectors perpendicular to each other will give a dot product of 0).

    So expand a bit: 0a - b + 0c = 0. Any solution will work. ex. (1, 0, 1). (In this case, b has to be equal to 0 since there's no other variable that can cancel it out).

    So, plugging it all into a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0, we get ...
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  3. #3
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    Hello, caelestis!

    Find the equation of the plane containing the line:
    . . x \:=\:(0, 0, 1) + t (0,\text{-}1, 0) and the point (1, 1, 0).

    To write the equation of a plane, we need a point (x_1,y_1,z_1)
    . . and its normal direction: \vec{n}\:=\:\langle a,b,c\rangle
    Then: . a(x-x_1) + b(y-y_1) + c(z-z_1) \;=\;0



    We have three points:
    . . t = 0:\;P_1 \:=\:(0,0,1)
    . . t = 1:\;P_2\:=\:(0,\text{-}1,1)
    . . .and: . P_3 \:=\:(1,1,0)

    Vector \overrightarrow{v_1} \:=\:\overrightarrow{P_1P_2} \:=\:\langle 0,\text{-}1,0\rangle is in the plane.

    Vector \overrightarrow{v_2} \;=\;\overrightarrow{P_2P_3} \:=\: \langle 1,2,\text{-}1\rangle is in the plane.


    The normal vector is their cross product:

    . . \vec{n} \;=\;\overrightarrow{v_1}\times \overrightarrow{v_2} \;=\;\left|\begin{array}{ccc} i & j & k \\ 0 & \text{-}1 & 0 \\ 1 & 2 & \text{-}1 \end{array}\right| \;=\;i(1) - j(0) + k(1) \;=\;i + k \;=\;\langle 1,0,1\rangle


    We have: . P_1 = (0,0,1),\;\vec{n} = \langle 1,0,1\rangle

    The equation of the plane is: . 1(x-0) + 0(y-0) + 1(z-1) \:=\:0


    Therefore: . {\color{blue}z + x \:=\:1\quad\text{answer B}}

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