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Thread: Coordinates of intersection

  1. #1
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    Coordinates of intersection

    Hey,

    I have a couple of questions I've been doing online which have left me a little puzzled. The first one, I'm not really sure how to go about. I think a lot of that comes down to having not had a lot of experience with parametric forms.

    I'll just post screenshots of where I'm up to on them, as they'll probably explain better where I'm up to.

    This is the one I'm having the most trouble with
    Coordinates of intersection-screenshot_5.jpg




    This one, I've got most of the questions, but the last one is leaving me a little confused.
    Coordinates of intersection-screenshot_6.jpg



    Any help would be amazing. Thanks
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  2. #2
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    Re: Coordinates of intersection

    Given the equations x- 5=y- 3= z/2 we can set all those common values equal to t: x- 5= t, y- 3= t, and z/2= t so we have the parametric equations x= t+ 5, y= t+ 2, and z= 2t. Putting those into the equation of the plane, 2x- 2y+ 2z= -4, 2(t+ 5)- 2(y+ 2)+ 2(2t)= 2t+ 10- 2t- 4+ 4t= 4t+ 6= -4. Solve that for t, then put it into the equations for x, y, and z.
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  3. #3
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    Re: Coordinates of intersection

    Quote Originally Posted by HallsofIvy View Post
    Given the equations x- 5=y- 3= z/2
    That should, of course, be y- 2, not y- 3.

    we can set all those common values equal to t: x- 5= t, y- 3= t
    And this should be y- 2= t. I had caught that error below.

    , and z/2= t so we have the parametric equations x= t+ 5, y= t+ 2, and z= 2t. Putting those into the equation of the plane, 2x- 2y+ 2z= -4, 2(t+ 5)- 2(y+ 2)+ 2(2t)= 2t+ 10- 2t- 4+ 4t= 4t+ 6= -4. Solve that for t, then put it into the equations for x, y, and z.
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