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Equation of line?

  1. Sep 27, 2006 #1
    I think my coordinate geometry skills have turned rather rusty. :frown:
    Just help me out with this problem.

    Find the symmetric and parametric equation of a line passing through (3, 4, -1) and parallel to 2i - 3j + 6k.

    Here is my attempt at this:

    Since the line passes through (3,4,-1) the equation would be given by:
    (x - 3) + (y - 4) + (z +1) = 0

    But, this looks like the equation of a plane :bugeye: . I am all confused now. How do I relate this line to the vector parallel to it?
  2. jcsd
  3. Sep 27, 2006 #2


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    Staff Emeritus
    Science Advisor

    Yes, that's the equation of the plane through (3, 4, -1) and perpendicular to 2i- 3j+ 6k. You can't write a line in 3 dimensions in just one equation. Write it in parametric equations:
    x= x0+ at, y= y0+ bt, z= z0+ ct where (x0, y0, z0) is a point on the line and ai+ bj+ ck is a vector in the direction of the line.
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