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Equation of line parallel to another passing through a point

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Write the parametric and cartesian equations of the line passing through ##P = (-\frac{3}{10}, 0, \frac{1}{10})## and parallel to
    $$r : \begin{cases}
    10x + 4y - 3 = 0 \\
    x + z = 0
    \end{cases}$$

    2. Relevant equations


    3. The attempt at a solution
    Since the two lines have to be parallel, that means that the two directional vectors are linearly dependent. So I first calculate the directional vector of ##r## as follow:
    ##\vec v_r = \begin{vmatrix}
    \hat i & \hat j & \hat k \\
    10 & 4 & 0 \\
    1 & 0 & 1
    \end{vmatrix} = (4, -10, -4)##
    Now that I have the directional vector, I know that the directional vector of the line we are searching for (let's call it ##u##) has to be linearly dependent to it, so something like this: ##\vec v_u = \alpha(4, -10, -4) = (4\alpha, -10\alpha, -4\alpha)##
    With this I have already an overview of the parametric form of ##u## using ##\vec v_u## and ##P##, which is:
    $$u : \begin{cases}
    x = 4\alpha - \frac{3}{10}t \\
    y = -10\alpha \\
    z = -4\alpha + \frac{1}{10}t
    \end{cases}$$
    I was thinking about substituting this three equations into ##r## and find ##\alpha## but I stopped because I was doing it not knowing why. If it's the right way, why should I substitute them? If not, then, what should I do after arriving at this point?
     
  2. jcsd
  3. Nov 13, 2016 #2

    LCKurtz

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    You just need a single direction vector. You could divide out a ##2## to make it simpler: ##\vec D = \langle 2,-5,-2\rangle##. You don't need the ##\alpha##.

    Remember the form for a straight line: ##\langle x,y,z\rangle = \vec P_0 + t\vec D##, where ##\vec P_0## is the position vector to a point on the line and ##\vec D## is any direction vector. Recheck your equation.
     
  4. Nov 13, 2016 #3
    Oh sorry, I messed up the equation.
    So it would be like:
    $$u : \begin{cases}
    x = -\frac{3}{10} + 2t \\
    y = -5t \\
    z = \frac{1}{10} -2t
    \end{cases}$$
     
  5. Nov 13, 2016 #4

    Mark44

    Staff: Mentor

    You can (and should) check it yourself. Does the line you found have the right direction (i.e., is it parallel to the given line)? Is there a value of t so that it goes through the given point? If you can answer yes to both questions, then you know you have the right equation.
     
  6. Nov 16, 2016 #5
    The fact that has the right direction can be already seen from the beginning, since it's where I started. They are linearly dependent so they are surely parallel.
    About the value of ##t##, I can see that only ##t = 0## would give me the point again. Is it correct?
    By the way, the Cartesian form would be:
    $$\begin{cases}
    x + \frac{2}{5}y + \frac{3}{10} = 0 \\
    - \frac{2}{5}y + z - \frac{1}{10} = 0
    \end{cases}$$
     
  7. Nov 16, 2016 #6

    PeroK

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    If you express your Cartesian solution as equations in ##x, y## and ##x, z## then would expect the coefficients to be the same as in the original equations?
     
  8. Nov 16, 2016 #7
    I put ##t = \frac{y}{-5}## so I guess it's normal that the coefficients remained the same. But in general I actually don't know if the coefficients can be the same as in the parametric form or not.
     
  9. Nov 16, 2016 #8

    PeroK

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    The first equation you were given was ##10x +4y -3 =0##

    The first equation you got for the parallel line after multiplication was ##10x +4y +3=0##

    Is that a surprise or coincidence?
     
  10. Nov 16, 2016 #9
    Oh, it's because they are parallel? So only the coefficient changed?
     
  11. Nov 16, 2016 #10

    PeroK

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    If the lines are parallel, then ##x,y,z## change in the same proportion to each other, so only the constant term changes.

    For the other equation you should get ##x+z=## some constant.
     
  12. Nov 16, 2016 #11

    Mark44

    Staff: Mentor

    If two vectors are linearly dependent, then each one is a nonzero multiple of the other. The situation is not as obvious if you have three linearly dependent vectors, where none of them is necessarily a multiple of any of the other two.
     
  13. Nov 18, 2016 #12
    So I did do it wrong. Then I have to put ##t = \frac{x}{2} + \frac{3}{20}##. Substituting this I would get:
    $$\begin{cases}
    \frac{5}{2}x + y +\frac{3}{4} = 0 \\
    x + z + \frac{1}{5} = 0
    \end{cases}$$
    Which has the same equations but with different constants.

    You mean in case there were three parallel lines? Shouldn't they all be linearly dependent in order for them to be parallel? Like, having only one parameter as a difference between them. For example: ##\vec v = (a, b, c)##, ##\vec u = \alpha(a, b, c)##, and ##\vec w = \beta(a, b, c)##
     
  14. Nov 18, 2016 #13

    Mark44

    Staff: Mentor

    Well, I wasn't talking about three parallel lines. If two (or more) vectors are parallel, then it's pretty obvious that they are linearly dependent.
    It's possible for three vectors to be linearly dependent where none is a multiple of any of the others. All they have to do is lie in the same plane. For example, u = <1, 0>, v = <0, 1>, w = <1, 1>.
     
  15. Nov 18, 2016 #14
    Yes, yes, I already knew that. Putting them that way, they would still be parallel though, right?
     
  16. Nov 18, 2016 #15

    Mark44

    Staff: Mentor

    No, they are not parallel. Why would you think that? All three point in different directions.

    If two vectors are parallel, each one is a nonzero multiple of the other, which means that they are necessarily linearly dependent. My point was that if you have three vectors, they can be linearly dependent, but no two of them have to be parallel.
     
  17. Nov 18, 2016 #16
    Oh okay. Because I was getting confused about what I had said before.
     
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