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    plane1 has equation  r.(3i - 2j - k) = 5 and plane 2 has equation  r.(4i - j - 2k) = 5

    find the line of intersection

    working:

    Cartesian equation for 1):  3x - 2y - z = 5
    Cartesian equation for 2):  4x - y - 2z = 5

    multiplying the second one by 2 and taking away I get  5x - 3z = 5 so  z = \dfrac{5x - 5}{3}

    subbing this back into the first equation I get  4x - 6y = 10 rearranging getting  x = \dfrac{5 + 3y}{2} so  \dfrac{5 + 3y}{2} = x = \dfrac{3z +5}{5}

    so the line is  r = \frac{-5}{3}j - \frac{-5}{3}k + \lambda(i + \frac{5}{3}j + \frac{2}{3}j)

    however they got:  r = \frac{5}{2}j - \frac{5}{2}k + \lambda(\frac{3}{3}i + j + \frac{5}{2}k )

    am I wrong? I know there should be more than one answer, but how can I check that my answer is correct or not

    thank you
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    Found a mistake, I think? When you sub in your value of z into the first equation, you should get -(5x-5) aka a +5 on the LHS, making the constant 0 instead of 10.
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    (Original post by thorn0123)
    plane1 has equation  r.(3i - 2j - k) = 5 and plane 2 has equation  r.(4i - j - 2k) = 5

    find the line of intersection

    working:

    Cartesian equation for 1):  3x - 2y - z = 5
    Cartesian equation for 2):  4x - y - 2z = 5

    multiplying the second one by 2 and taking away I get  5x - 3z = 5 so  z = \dfrac{5x - 5}{3}

    subbing this back into the first equation I get  4x - 6y = 10 rearranging getting  x = \dfrac{5 + 3y}{2} so  \dfrac{5 + 3y}{2} = x = \dfrac{3z +5}{5}

    so the line is  r = \frac{-5}{3}j - \frac{-5}{3}k + \lambda(i + \frac{5}{3}j + \frac{2}{3}j)

    however they got:  r = \frac{5}{2}j - \frac{5}{2}k + \lambda(\frac{3}{3}i + j + \frac{5}{2}k )

    am I wrong? I know there should be more than one answer, but how can I check that my answer is correct or not

    thank you
    The answers aren't the same; you can tell by crossing one with the other (iff they're parallel and both non-zero, you will get 0).
    I don't get 4x-6y=10 when I substitute - although I just did that really quickly so I might have slipped.
    There's a nicer way to do this question, by the way: you have the normals to each plane, and you need to generate a line which is perpendicular to both normals.
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    (Original post by Smaug123)
    The answers aren't the same; you can tell by crossing one with the other (iff they're parallel and both non-zero, you will get 0).
    I don't get 4x-6y=10 when I substitute - although I just did that really quickly so I might have slipped.
    There's a nicer way to do this question, by the way: you have the normals to each plane, and you need to generate a line which is perpendicular to both normals.
    yes must be just an algebra slip... Done a few more questions using that method and every time I'd get the direction parallel to whatever the answer gives.

    Thanks for the new method (it's much quicker, and tidier).
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    (Original post by thorn0123)
    yes must be just an algebra slip... Done a few more questions using that method and every time I'd get the direction parallel to whatever the answer gives.

    Thanks for the new method (it's much quicker, and tidier).
    No problem my teacher laughed greatly at us when he set us this problem and we all did it the hard-but-obvious way…
 
 
 
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