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Find parametric equation at point, and parallel to planes?

  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Find parametric equation of the line through the point ##(4,0,-4)## that is parallel to the planes ##x-8y+7z=0## and ##4x+3y-z+4=0##.

    2. Relevant equations

    ## \vec r = \vec r_0 + t\vec v ##
    (for orthogonal vectors) ## v \bullet w = 0 ##


    3. The attempt at a solution

    So I started out develeping the equation of the line through the point and came up with

    ## L = \lbrack 4,0,-4 \rbrack + t\lbrack a,b,c \rbrack ##

    Now I needed to find the direction vector a, b, and c.

    Since the line is parallel to the 2 planes, the direction vector should be orthogonal to both of the planes' normal vectors.

    ## \lbrack a,b,c \rbrack \bullet \lbrack 1,-8,7 \rbrack = 0##

    ## \lbrack a,b,c \rbrack \bullet \lbrack 4,3,-1 \rbrack = 0 ##

    So, now I have 2 equations with the 3 variables, and I went to solve them in terms of c.

    ## a - 8b + 7c = 0 ## and ## 4a + 3b - c = 0 ##

    ## b = \frac {29c} {35} ##
    ## a = \frac {-216c} {35} ##

    I wasn't really sure what to do at this point, as these answers seemed pretty obscure, and didn't know what to plug in for C, besides maybe c = 35. But when I do that, it doesn't match any of the multiple choice responses.

    But the problem comes with pretty defined answers to pick from, multiple choice:

    ##a) x = 4 + t, y = -8t, z = -4 + 7t ##
    ##b) x = 4 - t, y = 8t, z = -4 - 7t ##
    ##c) x = 4 - 13t, y = 29t, z = -4 + 35t ##
    ##d) x = 4 + 13t, y = -29t, z = -4 - 35 ##
    ##e) x = 4 + 4t, y = 3t, z = -4 - t ##
     
  2. jcsd
  3. Jan 24, 2016 #2

    LCKurtz

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    I didn't check your arithmetic, but you can probably pick c arbitrarily since many different length vectors have the same direction. But wouldn't it be easier to just note the direction you want is in the direction of the cross product of your two normals?
     
  4. Jan 24, 2016 #3

    Mark44

    Staff: Mentor

    Your value for a is incorrect, which you can see by substituting back into your equations.

    I found it much easier just to take the cross product of the normals to the two planes.
     
  5. Jan 24, 2016 #4
    How do you mean? Do you just mean literally taking the cross product of the normals and doing it that way? How would you make sure it goes through the point then?
     
  6. Jan 24, 2016 #5

    LCKurtz

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    You have the point given. You just need the direction vector to finish the equation of the line. That's what the cross product gives.
     
  7. Jan 24, 2016 #6
    Okay, I get it! Thank you!
     
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