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Math Help - Simetric/Parametric Equation (Line in 3D Plane)

  1. #1
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    Simetric/Parametric Equation (Line in 3D Plane)

    Hello guys, I'm kinda of stuck in this question:

    "Find the simetric and parametric equation of the line that has the point (0,0,0) and is perpendicular (by intersection) to the line:

    (x-10)/4 = y/3 = z/2
    "
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  2. #2
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    Re: Simetric/Parametric Equation (Line in 3D Plane)

    First, you should be able to see that the line (x- 10)/4= y/3= z/2, which is the same as the line given by parametric equations, x= 10+ 4t, y= 3t, z=2t, has direction vector <4, 3, 2>. Further, any line through (0, 0, 0) to (a, b, c) can be written as x= as, y= bs, z= cs and has direction vector <a, b, c>. If the two lines are perpendicular, then their dot product, <4, 3, 2>.<a, b, c>= 4a+ 3b+ 2c= 0. Since the two lines intersect, we can take (a, b, c) to be a line on the given line and so must satisfy the equation: (a- 10)/4= b/3= c/2. That gives two equations for a, b, and c. Since we only want the equation of the line, and not a specific point, it is sufficient to solve for either of two unknows in terms of the other one, then choose a convenient value for it.
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    Re: Simetric/Parametric Equation (Line in 3D Plane)

    Thanks in advance for the help HallsofIvy

    But... look:

    I got this equation: 4a+3b+2c = 0

    And from (a- 10)/4= b/3= c/2 I get b = 3c/2

    Putting b in the first equation I got: a = -13c/8; b = 3c/2; c = c

    How can I advance?

    x = ta so => x = -13ct/8, even if I consider c as 1 it still gets wrong,

    The answer is: x = 13t y = -12t z = -8t
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    Re: Simetric/Parametric Equation (Line in 3D Plane)

    Guys, I'm really close to finish the question, please one last help:

    Question: Find the simetric and parametric equation of the line that has the point (0,0,0) and is perpendicular (by intersection) to the line:

     \frac{\(x-10)}{4} = \frac{\ y}{3} = \frac{\ z}{2}

    Attempt:

    1) The line in the question has the following direction vector: <4,3,2>
    2) If the line I want is perpendicular to it => <4,3,2>.<a,b,c> = 0 => 4a + 3b + 2c = 0
    3) Since the two lines intersect, we can use the equation of the line in question to a point in the equation I want:  \frac{\(a-10)}{4} = \frac{\ b}{3} = \frac{\ c}{2}
    4) From the equation of the step above, I find:  b = \frac{\ 3c}{2}
    5) And using the "b" above in the equation of the second step:  a = \frac{\ -13c}{8}; b = \frac{\ 3c}{2}; c = c

    And now, I dont see how can I advance...

    I mean, if I try: x = t*a = \frac{\ -13tc}{8} and consider c as 1 I get a wrong "a"

    Please?
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    Re: Simetric/Parametric Equation (Line in 3D Plane)

    Quote Originally Posted by frank1 View Post
    Guys, I'm really close to finish the question, please one last help:

    Question: Find the simetric and parametric equation of the line that has the point (0,0,0) and is perpendicular (by intersection) to the line:

     \frac{\(x-10)}{4} = \frac{\ y}{3} = \frac{\ z}{2}

    Attempt:

    1) The line in the question has the following direction vector: <4,3,2>
    2) If the line I want is perpendicular to it => <4,3,2>.<a,b,c> = 0 => 4a + 3b + 2c = 0
    3) Since the two lines intersect, we can use the equation of the line in question to a point in the equation I want:  \frac{\(a-10)}{4} = \frac{\ b}{3} = \frac{\ c}{2}
    4) From the equation of the step above, I find:  b = \frac{\ 3c}{2}
    5) And using the "b" above in the equation of the second step:  a = \frac{\ -13c}{8}; b = \frac{\ 3c}{2}; c = c
    You know, as I said before, that any line through the origin can be written in the form x= at, y= bt, z= ct. So this line is of the form x= -(13/8)ct, y= (3/2)ct, z= ct which you can write as x= -(13/8)s, y= (3/2)s, z= s with s= ct. That is your parametric equation. Or, if you don't like fractions, take c= 8 so that x= -13t, y= 12t, z= 8t.

    To write as "symmetric equations" solve each of those for t: t= -\frac{x}{13}= \frac{y}{12}= \frac{z}{8}

    And now, I dont see how can I advance...

    I mean, if I try: x = t*a = \frac{\ -13tc}{8} and consider c as 1 I get a wrong "a"

    Please?
    Thanks from MarkFL
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