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Find the values of k so that lines are perpendicular using symetric equations

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Line 1: x-3/3k+1=Y+6/2=Z+3/2K
    Line 2: x+7/3=y+8/-2k=z+9/-3

    2. Relevant equations
    Cross product and dot product

    3. The attempt at a solution
    vector equation for line 1: (x,y,z)=(4,-4,-3)+k(3,0,2)

    vector equation for line 2: (x,y,z)=(-4,-8,-12)+k(0,-2,0)

    The product of (3,0,2)and (0,-2,0) is zero so they are perpendicular but i don't know how to find k :confused:
  2. jcsd
  3. Apr 27, 2009 #2


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    a) Use parentheses: you mean, I think (x-3)/(3k+1)= (y+6)/2= (z+ 3)/2k
    b) Don't use y and Y or k and K for the same thing. Those are different symbols and typically mean different values.

    Setting each of those equal to t, (x-3)/(3k+1)= t so x- 3= (3k+1)t or x= 3+ (3k+1)t, (y+6)/2= t so y+ 6= 2t or y= -6+ 2t, and (z+3)/2k= t so z+ 3= 2kt or z= -3+ 2kt

    (x+7)/3= t so x+ 7= 3t or x= -7+ 3t, (y+8)/(-2k)= t so y+8= -2kt or y= -8- 2kt, and (z+9)/(-3)= t so z+ 9= -3t or z= -9- 3t.

    No, that is not an equation for line 1. For one thing, k is given in the symmetric equation: one value of k corresponds to one line so without another parameter, this would be just a single point, not a line.

    Same comment.

    You don't have equations for the lines. As I said above your vector equation should be
    (x, y, z)= (-7, -8, -3)+ t(3k+1, 2, 2k)t. The parameter t determines the point, k is fixed for a line.

    For line 2, (x, y, z)= (-7, -8, -9)+ t(3, -2k, -3).

    In order that the lines be perpendicular you must have (3k+1, 2, 2k).(3, -2k, -3)= 9k+ 3- 4k- 6k= -k+ 3= 0.
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