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Equations of lines

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Find equations for a line perpendicular to both of these lines.



    2. Relevant equations
    (x/3)=(y/2)=(z/2) and (x/5)=(y/3)=(z-4)/2

    3. The attempt at a solution

    i dont know how to start???I know the two lines are skew, if i take the cross product will it be perp. to both lines??Can i take the cross product of two lines that lie in different planes?
     
    Last edited: Oct 5, 2007
  2. jcsd
  3. Oct 5, 2007 #2

    Dick

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    You don't take the cross product of the lines. You take the cross product of the direction vectors of the lines. The result is a direction vector for the perpendicular line.
     
  4. Oct 5, 2007 #3
    how do i find the direction vectors?
     
  5. Oct 5, 2007 #4

    Dick

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    Do you have a text book? Isn't it covered in there?
     
  6. Oct 5, 2007 #5
    No!!My book leaves a lot of details out, it expects us to know certain calculus stuff since its a post calculus course.I just dont remember direction vectors but have studied them in the past.Ill see what i can find n google.thnx anyway.
     
  7. Oct 5, 2007 #6

    Dick

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    Then you may need another book to keep on hand. ax=by=cz has direction vector (1/a,1/b,1/c).
     
  8. Oct 5, 2007 #7
    Hmm, i did the cross product of the direction vectors and got -2i+4J-k, but the questions asks to find the equations, i got a vector.The answer in the book is .5x-52/7=-.25y+52/21=z-208/21, they surely used another method.
     
  9. Oct 5, 2007 #8

    Dick

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    No. Look at the direction vector of the line they give as a solution. It's (1/.5,1/(-.25),1/1) which is (2,-4,1). You got the direction vector right. Now it looks like they want you to fix the constants by requiring that the perpendicular intersect the other two lines.
     
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