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Parametric vector equation

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    The position vectors of two points A,B of a line are a = < 2 ,1 ,7> : b = < 1 , 4 , -1 >
    Find the parametric vector equation for any point on the line AB using λ as the parameter.


    2. Relevant equations

    In general x = a + λb
    Where b is the unit vector of b.


    3. The attempt at a solution
    b = b/ |b|
    So b = <1/ 3√2,4/ 3√2,-1/ 3√2>
    so we have
    x = < 2 , 1, 7> + λ<1/ 3√2,4/ 3√2,-1/ 3√2>


    This correct? Please do not give me the answer but I prefer a hint. Thank you .
     
  2. jcsd
  3. May 8, 2013 #2

    Mark44

    Staff: Mentor

    The vector b in this equation is not the same as the position vector b above. You need a vector with the same direction as your line.
     
  4. May 8, 2013 #3
    The vector b in this equation is not the same as the position vector b above. You need a vector with the same direction as your line.

    I don't follow you dude. So b is not the same because I made b into a unit vector. So it is lambda times the unit vector b
    I don't quite follow you. Thanks :)
     
  5. May 8, 2013 #4

    Mark44

    Staff: Mentor

    You have two points on your line: A( 2 ,1 ,7) and B( 1 , 4 , -1). As a vector, <1, 4, -1> is NOT in the same direction as the line from A to B. You need to find that vector; i.e. AB.
     
  6. May 8, 2013 #5
    Mark44,

    OK, So a = < 2 , 1 ,7 > and b = < 1 , 4 , -1 >

    AB = a - b = < 2 , 1 ,7 > - < 1 , 4 , -1 > = < 1 , - 3, 2 >
    Turn AB into a unit vector. So you have < 1 / sqrt(14), -3/ sqrt(14) , 2/ sqrt(14) >
    Then any point x is represented by x = < 2, 1, 7 > + (lambda) < 1 / sqrt(14), -3/ sqrt(14) , 2/ sqrt(14) >

    Good?Thanks
     
  7. May 8, 2013 #6

    Mark44

    Staff: Mentor

    Yes, that's right.
     
  8. May 8, 2013 #7
    Thanks Mark44.
     
  9. May 8, 2013 #8
    So if I wanted this in Cartesian coordinates then x , y , z = 2 , 1 , 7 + λ ( 1/√14, - 3/√14, 2/ √14)
    I'm sorry pretend they are written as column's
    Then we have
    x = 2 + λ(1/√14)
    y = 1 + λ(- 3/√14)
    z = 7 + λ(2/ √14)

    Eliminate lambda and I got.

    √14(x - 2) = -(y -1)(√14)/ 3 = ( z - 7)(√14)/2
    Correct?
     
  10. May 8, 2013 #9

    Mark44

    Staff: Mentor

    Looks fine.
     
  11. May 8, 2013 #10
    Kind of ugly equation don't you think? Thanks
     
  12. May 10, 2013 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Actually, the vector from a to b is b-a, not a-b, if that matters.
     
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