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Math Help - Find the intersection of two lines in 3d space.

  1. #1
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    Find the intersection of two lines in 3d space.

    -2x-5y=3, -8y+4z=-7

    How do you go about finding (x,y,z)? I can't find this in my textbook.
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  2. #2
    A Plied Mathematician
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    Those are planes, not lines. Are you trying to find the equation of the line of intersection?
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  3. #3
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    You're right. I'm trying to find the intersection of two planes
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  4. #4
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    So, set up a system of two equations in three unknowns and start solving it. What do you get?
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  5. #5
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    First find the cross product of the two normals. That is the direction vector of your line.
    This point is on both planes: \left( {\frac{{ - 3}}{2},0,\frac{{ - 7}}{4}} \right).
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  6. #6
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    I would prefer Ackbeet's method. Since z does not appear in the first equation and x does not appear in the second, it is easy to solve the two equations for x and z in terms of y. Then you can use "y" as parameter. Once you have x= f(y) and z= g(y), your parametric equations for the line of intersection are x= f(t), y= t, z= g(t).
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  7. #7
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     \left\{\begin{array}{ccc}<br />
x&=-\displaystyle\frac{5y+3}{2} \\ \\<br />
z&=\displaystyle\frac{8y-7}{4}<br />
\end{array}\right.\rightarrow <br />
\left[\begin{array}{ccc}<br />
x\\y\\z<br />
\end{array}\right]=<br />
\left[\begin{array}{ccc}<br />
-\displaystyle\frac{5y+3}{2} \\ \\<br />
y \\ \\<br />
\displaystyle\frac{8y-7}{4}<br />
\end{array}\right]=<br />
\left[\begin{array}{ccc}<br />
-\displaystyle\frac{5}{2}y-\frac{3}{2} \\ \\<br />
y \\ \\<br />
\displaystyle\frac{8}{4}y-\frac{7}{4}<br />
\end{array}\right]=<br />
y\left[\begin{array}{ccc}<br />
-\displaystyle\frac{5}{2}\\ \\<br />
1 \\ \\ 2<br />
\end{array}\right]+\left[\begin{array}{ccc}<br />
-\displaystyle\frac{3}{2} \\ \\<br />
0 \\ \\<br />
-\displaystyle\frac{7}{4}<br />
\end{array}\right]<br />

    Line pass through Point (-\displaystyle\frac{3}{2},0,-\frac{7}{4}) with normal vector (-\displaystyle\frac{5}{2},1,2)^T
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  8. #8
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    Reply to math2009:

    That's fairly slick, too. I would say, though, that you have a direction vector there, not a normal vector.
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