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Cartesian and vector equations

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    The position vectors of two points a and b on a line are < 2,1,7> and < 1,4,-1>
    Find the parametric vector equation for any point on the line.
    Find the linear Cartesian equation of the line using x, y, z as coordinates.

    2. Relevant equations

    For the parametric eq.

    x = a + λu where u is a unit vector

    3. The attempt at a solution

    b -a = < -1, 3, -8 > , call it u
    Norm of b -a = √(-1)^2 + (3)^2 + -(-8)^2
    unit vector u = <-1/√74, 3/ √74 , -8 /√74>

    x = < 2, 1, 7 > + λ<-1/√74, 3/ √74 , -8 /√74>


    For the Cartesian equation I have.

    x = 2 + λ(-1/√74)

    y = 1 + λ(3/ √74)

    z = 7 + λ(-8 /√74)


    How is she looking?
    I have a question if it says linear Cartesian equation does that mean I have to eliminate lambda and write it all together or it is fine to leave as I have?

    Thanks
     
  2. jcsd
  3. May 20, 2013 #2

    CAF123

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    Gold Member

    You can solve each of the the Cartesian equations you have for λ and hence put the equations into 'symmetric' form.
     
  4. May 20, 2013 #3
    OK. Does it look correct otherwise? Thanks.
     
  5. May 20, 2013 #4

    Mark44

    Staff: Mentor

  6. May 20, 2013 #5
    It is the same but different. I have mad mistakes in the old one. So I redid it. Is it correct?
     
  7. May 20, 2013 #6

    CAF123

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    Gold Member

    I believe so
     
  8. May 20, 2013 #7
    Thanks, sorry for the re- post.
     
  9. May 20, 2013 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    When t= 0, x= 2, y= 1, z= 7 which is (2, 1, 7), one of the two given points.
    when [itex]\sqrt{74}[/itex], x= 2- 1= 1, y= 1+ 3= 4, z= 7- 8= -1 which is (1, 4, -1), the other point. Since all the calculations are linear, this is a straight line and since it contains the two points that define the line, it is the correct line.

    But I am puzzled at your use of [itex]\sqrt{74}[/itex]. You don''t have to use a unit vector. You could just as well use x= 2- t, y= 1+ 3t, z= y- 8t, avoiding the square root.




    Yes, eliminate [itex]\lambda[/itex]. The simplest way to do that is to solve each equation for [itex]\lambda[/itex]: [itex]\lambda= -\sqrt{74}(x- 2)[/itex], [itex]\lambda= \sqrt{74}(y- 1)/3[/itex], and [itex]\lambda= \sqrt{74}(z- 7)/(-8)[/itex]. Since those are all equal to [itex]\lambda[/itex] they are equal to each other:
    [tex]\frac{x- 2}{-1}= \frac{y- 1}{3}= \frac{z- 7}{-8}[/tex]
    Notice that the [itex]\sqrt{74}[/itex] terms all cancel.
     
  10. May 20, 2013 #9
    I don't know just used it. The unit vector I mean. THX DOOODE!
     
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