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Line orthogonal to a plane with variable parameter

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    I have to find for which values of the real parameter ##b## the following plane is orthogonal to the following line:
    ##\pi : 5x + (2b - 1)y - (1 + 8b)z + 3 = 0##
    ##s : \begin{cases}
    x + z - 4 = 0 \\
    x - 3y + z + 2
    \end{cases}##

    2. Relevant equations


    3. The attempt at a solution
    For these two to be orthogonal, it means that the plane normal vector and the line directional vector must be parallel to each other, so linearly dependent. This also means that their scalar product is exactly the product of their norm, since the ##cos## of their angle would be ##1##. At this point I thought I just had to equal the two and find which ##b## satify this equation.
    ##\vec n_{\pi} * \vec v_s = ||n_{\pi}|| ||v_s||##
    Knowing that ##\vec n_{\pi} = (5, 2b - 1, -1 - 8b)## and ##\vec v_s = (3, 0, -3)##, with numbers we have:
    ##(5 * 3) + ((2b - 1) * 0) + ((-1 - 8b) * (-3)) = \sqrt{5^2 + (2b - 1)^2 + (-1 - 8b)^2} \sqrt{3^2 + 0^2 + (-3)^2}##
    After various calculations I end up with this:
    ##18 + 24b = \sqrt{68b^2 + 12b + 26} \sqrt{18}##
    And here I'm not sure anymore if I'm doing it right. Is the way I'm trying to do it correct? Do I have to use another method of finding out?
    The thing is that from the two vectors I noticed that if you put ##b = \frac{1}{2}## you get a parallel vector (##\vec n_{\pi} = (5, 0, -5)##)to the one of the line. But I've arrived at this only by looking at it, without a "real method", so I can't actually use this.
     
  2. jcsd
  3. Oct 6, 2016 #2

    Ssnow

    User Avatar
    Gold Member

    Hi, at first I suppose was ##x-3y+z+2=0## ... second in general two vectors ##v_{1},v_{2}## are parallel when ##v_{1}=\alpha v_{2}## for some ##\alpha## ...
    I suggest you to study this system with three equation and two value ## b,\alpha## to find ...

    Ssnow
     
  4. Oct 6, 2016 #3
    Yes, sorry, it's ## = 0##.
    But isn't ##b## the ##\alpha## I'm searching for?
    And how would the system look like? I have to multiply ##\alpha## to both the equations of the line?
     
  5. Oct 6, 2016 #4

    Ssnow

    User Avatar
    Gold Member

    I give another hint, if solving the system you can find ##b## and ##\alpha## as before then ##v_{1}## and ##v_{2}## can be parallel for this ##b##. The system in vectorial form is:

    ## (3,0,-3)=\alpha (5,2b-1,-1-8b)##
     
  6. Oct 6, 2016 #5
    Oh! Now I get it! This system:
    ##\begin{cases}
    5\alpha = 3 \\
    2b\alpha - \alpha = 0 \\
    -\alpha - 8b\alpha = -3
    \end{cases}##
    And I found what I had found before while noticing the two vectors:
    ##\begin{cases}
    \alpha = \frac{3}{5} \\
    b = \frac{1}{2} \\
    \end{cases}##
    Thank you very much!
     
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