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Math Help - Equation of a line that passes through a point.

  1. #1
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    Equation of a line that passes through a point.

    Find the equation of the line that passes through A(1,-4, 2) and is parallel to the intersection line of the two planes x - 2y + 3z - 1 = 0 and x - 4y+ 2z - 8 = 0.

    What I did:
    First I set the first and second equations to [1] and [2]:
    x - 2y + 3z - 1 = 0 [1]
    x - 4y+ 2z - 8 = 0 [2]
    I then multiply [1] by 2 and use elimination to get rid of the y variable for now:
    2x - 4y + 6z - 2 = 0
    x - 4y + 2z - 8 = 0
    ________________________
    x + 4z + 6 = 0 [3]

    I'll then let z = t to solve for x in equation [3]:
    x + 4t + 6 = 0
    x = -4t - 6
    Now I substitute z = t and x = -4t - 6 into equation [1] to solve for y:
    -4t - 6 - 2y + 3t - 1 = 0
    y = (-1/2)t - 7/2

    Now that I have the values of all the unknowns, I first express it in parametric form:
    x = -4t - 6
    y = (-1/2)t - 7/2
    z = t
    Knowing this, finally, the direction vector for the line that passes through A(1, -4, 2) can be expressed:
    (x,y,z) = (1, -4, 2) + t(-4, 1/2, 1)

    I just wanted to know, did I do this correctly? I feel as if I did something wrong. If I did, can you point where I went wrong? Thank you in advance.
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  2. #2
    MHF Contributor

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    Looks good to me!


    (Since the "direction vector", <-4, 1/2, 1>, shows only direction, its length is not important and any multiple of it would also work. If you really don't like fractions multiply it by 2: (x,y,z) = (1, -4, 2) + t(-8, 1, 2). That gives the same points, with this new t half what the old t was.)
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Looks good to me!


    (Since the "direction vector", <-4, 1/2, 1>, shows only direction, its length is not important and any multiple of it would also work. If you really don't like fractions multiply it by 2: (x,y,z) = (1, -4, 2) + t(-8, 1, 2). That gives the same points, with this new t half what the old t was.)
    Alrighty. Thanks HallsofIvy for all the help!
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