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How do I check my solution?

  1. May 10, 2015 #1
    Would somebody be so kind as to tell how I can check my solution to this problem(s) below:

    Find parametric equations for the line through the points P=(-2,0,3) and Q=(3,5,2).

    More specifically, I feel that I should be able to substitute some value for t or (x,y,z) to confirm my solutions. Thanks!
     
  2. jcsd
  3. May 10, 2015 #2

    mfb

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    I moved your thread to our homework section.

    What did you get as solution?
    You should be able to plug in some values of t into your solution to get your two points.
     
  4. May 11, 2015 #3

    HallsofIvy

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    If you have parametric equations, take any two values for the parameter and check if the points you get are on the same lie as the two given points. Depending upon how you find the parametric equations, it should be easy to choose the parameters so you get the given points. Of course, a line is determined by two points so you only need to check two.
     
  5. May 11, 2015 #4
    Ok, I'm about to reveal my ignorance but how do I determine what the parameter values are?
     
  6. May 11, 2015 #5
    My solution is r(t)=< 5t-2, 5t , 3-5t >
     
  7. May 11, 2015 #6

    Mark44

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    Is there some value of the parameter t so that r(t) = (-2, 0, 3)? Is there another value of t so that r(t) = (3, 5, 2)? If the answers to these questions aren't obvious by inspection, set up an equation with <5t - 2, 5t, 3 - 5t> on one side, and either of the given points on the other side, and solve for t.
     
  8. May 13, 2015 #7
    Gottca, thanks for the help! I'm undergoing a self study program and some things that should be obvious aren't at times.
     
  9. May 13, 2015 #8

    Ray Vickson

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    Basically, that is what "through" means in this type of problem When you say the line passes through the points P=(-2,0,3) and Q=(3,5,2), this means that for some value of t you will have <5t - 2, 5t, 3 - 5t> = <-2,0,3>, while for some other value of t you will have <5t - 2, 5t, 3 - 5t> = <3,5,2>. Componentwise, <a,b,c> = <d,e,f> means a = d, b = e and c = f.

    BTW: another (easier?) way is to note that the line through ##P = \langle -2,0,3 \rangle## and ##Q = \langle 3,5,2 \rangle## has the form
    [tex] (1-t) P + t Q = (1-t) \langle -2,0,3 \rangle + t \langle 3,5,2 \rangle [/tex]
    For ## 0 \leq t \leq 1## the points are on the line-segment joining P and Q (with point = P when t = 0 and point = Q when t = 1). For t < 0 the points are on the line "before" P (that is, on the side away from Q), while for t > 1 they are on the line "after" Q (that is, on the side away from P).
     
  10. May 17, 2015 #9

    HallsofIvy

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    If you are referring to something like "[itex]3x- 7\le 2[/itex] so [itex]3x\le 9[/itex], [itex]x\le 3[/itex], you can't do that for functions that are not "one to one".
     
  11. May 17, 2015 #10

    mfb

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    You can always simplify ##3x- 7\le 2## to ##x \le 3## (assuming real numbers).
    And all the functions that might come up here are injective.
     
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