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Vectors - show that the lines intersect

  1. Apr 25, 2006 #1

    smn

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    Hi, i'm currently revising for a maths exam and i'm stuck on the following question:

    Show that the lines:

    r = (i+j+k) + s(i+2j+3k)

    r = (4i+6j+5k) + t(2i+3j+k)

    Intersect.

    My work so far:

    Let (i+j+k) + s(i+2j+3k) = (4i+6j+5k) + t(2i+3j+k)

    So (i) 1+s = 4+2t

    (j) 1+2s = 6+3t

    (k) 1+3s = 5+t

    I'm unsure where to go from here, any help would be appreciated.

    Regards

    smn
     
  2. jcsd
  3. Apr 25, 2006 #2

    Hootenanny

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    You have to find values for s and t and show that they satisfy all three vector component equations (i,j & k)

    ~H
     
  4. Apr 25, 2006 #3

    robphy

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    So, you have three equations that must be satisfied for an intersection.
    Is there a pair of values (s,t) that satisfies all three equations simultaneously?

    [Try using, say, equation (i) with equation (j), then the result with (k) etc...] Once you determine a pair (s,t), check that it satisfies each equation.
     
  5. Apr 25, 2006 #4
    Well, if they intersect, then they must by necessity have a point in common. You started on that above. If they have a point in common, what can you say about possible solutions to the system of equations you found?
     
  6. Apr 25, 2006 #5

    smn

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    Thanks for the prompt replys.

    I realise that you have to solve for s and t and these values should equal if the lines intersect.

    I was unsure what to do next with the 3 equations in order to solve for s and t.
    I'm now going to try using simultaneous equations, as mentioned, to try and solve for s and t.

    Regards

    smn
     
  7. Apr 25, 2006 #6

    Hootenanny

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    That's the way to go!

    ~H
     
  8. Apr 25, 2006 #7

    HallsofIvy

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    You have two variables s and t. You should be able solve two of the equations for them. Do those two values also satisfy the third equation?
     
  9. Apr 25, 2006 #8

    smn

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    Yes, i worked out that s=1 and t= -1. I then sub'd these values into the 3rd equation and it satisfied this also ( 4=4).

    Thanks for all your help

    Regards

    Sam
     
  10. Apr 25, 2006 #9

    robphy

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    Now, the follow-up question...
    can you show that there exists an intersection WITHOUT solving explicitly for s and t?
     
  11. Apr 25, 2006 #10
    Using pre-calculus methods? Only way I can think of is to show they are coplanar and not parallel.
     
  12. Apr 25, 2006 #11

    robphy

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    Are use of the dot- and cross-product operations considered pre-calculus?
    Note that the OP has already written lines in parametric vector form:
    [tex]\vec A=\vec A_0 + s\vec U [/tex]
    [tex]\vec B=\vec B_0 + t\vec V [/tex]
    which is already somewhat advanced by introductory standards.
     
  13. Apr 25, 2006 #12
    It wan't when I was in HS, but that was back in the late 70s.
     
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