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Math Help - Sorry 1 more plane problem

  1. #1
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    Sorry 1 more plane problem

    What is the line of intersection of the xy and xz planes?

    So far this is what I have:

    The planes I used are: xy=(3,4,0)+s(1,1,0)+t(2,3,0)
    and xz = (2,0,1)+p(1,1,0)+q(1,0,2)

    Now I need the scalar equations in the for Ax+By+Cz+D=0 so I found the cross product of the direction vectors to get the normals which were:
    n1=(0,0,1) and n2=(0,-1,0) where n=(A,B,C)

    Heres where my problem is, I dont see a way to plug these normals in to get a scalar equation. Did I do something wrong, or what do I do next?
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  2. #2
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    Quote Originally Posted by sugar_babee View Post
    What is the line of intersection of the xy and xz planes?
    I do not know what the difficulty is. Draw by hand the xy-plane and the xz-plane and you shall easily see that their intersection forms the x-axis.

    Are you asking how to do this problem using algebra?
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  3. #3
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    oooh yeah that's obvious to me now, but I do need the algebraic way because I think the point is to figure that out with the question rather than just knowing it
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  4. #4
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    Quote Originally Posted by sugar_babee View Post
    oooh yeah that's obvious to me now, but I do need the algebraic way because I think the point is to figure that out with the question rather than just knowing it
    The equation of the xy-plane is 0x+0y+1z=0.
    The equation of the xz-plane is 0x+1y+0z=0.

    I will use the following theorem that you learned.
    Theorem: Two planes distinct planes are either parallel (do not intersect) or interest in a line.

    Now, these two planes are distinct. Are they parallel? The answer is no because (0,0,0) is a point shared by both planes. So by the theorem they intersect in a straight line. But to find the equation of a straight line we need to points (two points determine a unique straight line). By inspection we see that (1,0,0) is another point.

    So we need to find the equation of a line passing through \{ (0,0,0),(1,0,0) \}.
    We need to first find a non-zero vector aligned with this line, and such a non-zero vector is \bold{v} = (1-0)\bold{i}+(0-0)\bold{j}+(0-0)\bold{k} = \bold{i}.
    Thus, the line is descirbed parametrically as,
    \left\{ \begin{array}{c} x = 1t+0 \\ y=0t+0\\ z=0t+0 \end{array} \right\}
    Which is the x-axis.
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