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Math Help - Equations of Planes 2

  1. #1
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    Equations of Planes 2

    For what value of k will the following set of planes intersect in a line?

    x-2y-z=0
    x+9y-5z=0
    kx-y+z=0
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  2. #2
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    Hint: To find the line of intersection of the planes x-2y-z=0 and x+9y-5z=0, use the Addition Method to find the projected line on the xy-plane:

    \begin{aligned}<br />
5x-10y-5z &= 0 \;\;\;\;\;\;\;\;\;\;\mbox{(multiply first equation by 5)}\\<br />
x+9y-5z &= 0 \\<br />
4x-19y &= 0 \;\;\;\;\;\;\;\;\;\;\mbox {(subtract second from first)}\\<br />
y &= \frac{4}{19}x<br />
\end{aligned}

    We can then find the z-coordinate of the line of intersection:

    z=x-2y=\frac{19}{19}x-2\cdot\frac{4}{19}x=\frac{11}{19}x.

    The equations of the line of intersection are therefore:

    \begin{aligned}<br />
y&=\frac{4}{19}x\\<br />
z&=\frac{11}{19}x.<br />
\end{aligned}

    Now, all that remains is to find the value of k that will result in these two equations of intersection for all three original equations.
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  3. #3
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    so k=-7/19?
    how would i check that it is correct?
    Last edited by skeske1234; September 12th 2009 at 07:41 AM.
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  4. #4
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    can anyone please verify this? scott H?
    Last edited by skeske1234; September 13th 2009 at 07:16 AM.
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  5. #5
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    can someone verify if k=-7/19 seems right
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  6. #6
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    Quote Originally Posted by skeske1234 View Post
    so k=-7/19?
    how would i check that it is correct?
    You can check it by putting it into the original problem and seeing what happens!

    If k= -7/19, then the equations become x-2y-z=0. x+9y-5z=0. and -(7/19)x-y+z=0 which we can rewrite as -7x- 19y+ 19z= 0. How many solutions does that have? If we subtract the first equation from the second, we eliminate x: (x+9y-5z)- (x-2y-z)= 11y- 4z= 0. If we multiply the first equation by 7 and add to the third, (7x-14y-7z)+(-7x-19y+19z)= -33y+ 12z= 0 or, dividing by 3, -11y+ 4z= 0, exactly the same equation as before! If z= (11/4)y ten the first equation becomes x- 2y- (11/4)y= 0 so x= (2+11/4)y= (33/4)y. That is any point on the line x= (33/4)t, y= t, z= (11/4)t lies on all three planes.
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