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Mutual positions with varying paramenter

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    In the Euclidean space ##E^3## we have the following lines:
    ##r : \begin{cases}
    10bx + 4y - (b + 2) = 0 \\
    bx + z = 0
    \end{cases}##
    ##s : \begin{cases}
    4y - 10z + 1 = 0 \\
    x + 3y - 7z + 1= 0
    \end{cases}##
    Find the mutual position at the varying parameter ##b##.

    2. Relevant equations


    3. The attempt at a solution
    I tried both methods, with the matrix and the directional vectors and in both I find myself wrong I guess. (sorry for the long post but I really want to understand this)
    Matrix method:
    I have two matrices. One made of the full equations:
    ##(A|d) = \begin{pmatrix}
    10b & 4 & 0 & -b - 2 \\
    b & 0 & 1 & 0 \\
    0 & 4 & -10 & 1 \\
    1 & 3 & -7 & 1
    \end{pmatrix}##
    and the other one made of the same full equations minus the last column:
    ##A = \begin{pmatrix}
    10b & 4 & 0 \\
    b & 0 & 1 \\
    0 & 4 & -10 \\
    1 & 3 & -7
    \end{pmatrix}##
    Now:
    If the rank of the first is ##4## and the second is ##3##, then the two lines are askew.
    If the rank of the first is ##3## and the second is ##3##, then the two lines intersect in one point.
    If the rank of the first is ##3## and the second is ##2##, then the two lines are parallel.
    If the rank of the first is ##2## and the second is ##2##, then the two lines are equivalent.
    To know the rank of ##A## I'll take one of the squared submatrix of it and check the determinant. If the determinant is ##0## then I'll take another squared submatrix with one less row and one lest column.
    Taking the following submatrix
    ##A' = \begin{vmatrix}
    10b & 4 & 0 \\
    b & 0 & 1 \\
    0 & 4 & -10
    \end{vmatrix}##
    I have a determinant equal to ##0##, no matter what ##b## is. So I go deeper and check this other sub matrix:
    ##A' = \begin{vmatrix}
    10b & 4
    b & 0
    \end{vmatrix}##
    Which has a determinant equal to ##-4b##. It is different from ##0## only if ##b## is different from ##0##. So, we have that the rank of it is ##2## for ##b \neq 0##. This means that the other matrix must be either ##3## or ##2##.
    I'm going to write all the passages for the first matrix:
    ##(A|d) = \begin{vmatrix}
    10b & 4 & 0 & -b - 2 \\
    b & 0 & 1 & 0 \\
    0 & 4 & -10 & 1 \\
    1 & 3 & -7 & 1
    \end{vmatrix} =
    -b\begin{vmatrix}
    4 & 0 & -b - 2 \\
    4 & -10 & 1 \\
    3 & -7 & 1
    \end{vmatrix} -
    \begin{vmatrix}
    10b & 4 & -b - 2 \\
    0 & 4 & 1 \\
    1 & 3 & 1
    \end{vmatrix} =
    -b [-10(4 + 3b + 6) + 7(4 + 4b + 8)] - [4(10b + b + 2) - (30b - 4)] =
    -b [-100 - 30b + 84 + 28b] - [48b + 8 - 30b + 4] =
    -16b - 2b^2 + 18b + 12 = -2b^2 - 34b + 12##
    The thing is that this should be ##0##, right? Why isn't it? What did I do wrong?
    Directional vectors method:
    I brought the two lines into parametric form, having then:
    ##r : \begin{cases}
    x = -\frac{t}{b} \\
    y = 10t + b + 2 \\
    z = t
    \end{cases}##
    ##s : \begin{cases}
    x = -\frac{1}{2}t + 2 \\
    y = \frac{5}{2} - 1 \\
    z = t
    \end{cases}##
    So I have ##P_r (0, b + 2, 0), \vec v_r = (-\frac{1}{b}, 10, 1), P_s (2, -1, 0), \vec v_s = (-\frac{1}{2}, \frac{5}{2}, 1)##
    Putting all these in the following matrix and finding the determinant:
    ##\begin{vmatrix}
    0 - 2 & b + 2 - (-1) & 0 - 0 \\
    -\frac{1}{b} & 10 & 1 \\
    -\frac{1}{2} & \frac{5}{2} & 1
    \end{vmatrix} =
    \begin{vmatrix}
    - 2 & b + 3 & 0 \\
    -\frac{1}{b} & 10 & 1 \\
    -\frac{1}{2} & \frac{5}{2} & 1
    \end{vmatrix} = +5 - \frac{b}{2} - \frac{3}{2} - 20 + 1 + \frac{3}{b} = -\frac{31}{2} - \frac{b}{2} + \frac{3}{b} = -\frac{31 + b}{2} + \frac{3}{b}##
    Again, I find myself with something that should actually be ##0## independently of ##b##.
    What am I doing wrong?
     
  2. jcsd
  3. Oct 16, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Concerning the matrix: you ignored the last row in the determination of the rank, that does not work.
    I don't understand the steps that follow there, but the result is not the right determinant.
     
  4. Oct 16, 2016 #3
    It's not that I ignored it, it's that since I already found a determinant that is already ##0##, it's useless to calculate another squared submatrix containing the other missing row. So I directly went for the smaller submatrix of 2x2.

    I took the second row and did the cofactors with ##b## and ##1## (only these two since the other two numbers on the row are ##0##). Since they aren't on even positions, they are taken negative, so ##-b## and ##-1##. Same thing when the these two are multiplied to the other two determinants of the two matrices.

    EDIT: I just noticed a mistake in the signs. My bad. It's like on WolframAlpha.
     
  5. Oct 16, 2016 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    I don't understand the question. What is meant by "mutual position"?
     
  6. Oct 16, 2016 #5
    Meaning if they the two lines are parallel, intersecting or askew.
     
  7. Oct 16, 2016 #6

    mfb

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    2016 Award

    Staff: Mentor

    That only works if you start with square matrices and test possible all smaller matrices until you find one that has non-zero determinant, or until you tested all.
     
  8. Oct 16, 2016 #7
    So I have to test all the small square matrices of ##A## and find a non-zero determinant? Only if I don't find any I will try with the smaller ones?
     
  9. Oct 16, 2016 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If you find a 3x3 matrix in A with a non-zero determinant then A has rank 3. If all 3x3 matrices in A do not have rank 3 then A cannot have rank 3 and you have to dig deeper.
    It is usually easier to perform row/column operations that do not change the rank until the rank becomes obvious.
     
  10. Oct 20, 2016 #9
    Oh okay, thank you.
    The thing is that it feel easier for me to do it with the determinants than try randomly with row/column operations.

    P. S. Sorry for the late reply.
     
  11. Oct 20, 2016 #10

    mfb

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    2016 Award

    Staff: Mentor

    It is not random, it would be like the Gauss algorithm.
     
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