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Math Help - Lines and Planes in space Question?

  1. #1
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    Unhappy Lines and Planes in space Question?

    Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

    It is problem # 114 in CALCULUS 8th edition by larson, hostetler and edwards, secction 11.5 page 809. It will be in section 10.5 in the 7th edition.

    Show that the plane 2x-y-3z=4 is parallel to the line
    x= -2+2t
    y= -1+4t
    z= 4

    and find the distance between them.

    I tried using the techniques given in the book but it's not for a plane and line.

    Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

    Thank you guys! Really desperate...
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  2. #2
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    Show that the plane 2x-y-3z=4 is parallel to the line
    x= -2+2t
    y= -1+4t
    z= 4
    and find the distance between them.


    First, note that the line you give in parametric form can alternatively be written r = (-2, -1, 4) + t(2, 4, 0).

    The vector (2, -1, -3) is normal or perpendicular to the plane 2x - y - 3z = 4.

    The direction vector for the given line is (2, 4, 0).

    To show that the line is parallel to the plane, show that the line's direction vector (2, 4, 0) is perpendicular to the plane's normal vector (2, -1, -3). That is, show that the (scalar) dot product of (2, 4, 0) and (2, -1, -3) is 0.

    The dot product of (2, -1, -3) and (2, 4, 0) = 4 + (-4) + 0 = 0. Niiiiice.

    Good luck!
    -Andy
    Last edited by abender; July 28th 2008 at 12:45 PM.
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  3. #3
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    Since the line r = (-2, -1, 4) + t(2, 4, 0) is parallel to the plane 2x-y-3z=4, the distance between any point on r to the plane is the same. So, we can simply let t=0 and get a point on our line. The point, P on the line r when t=0 is P(-2, -1, 4). So, our problem now becomes finding the distance between a point and plane. We use the distance formula as follows:

    d = |(2*-2)(-1*-1)(-3*-4) - 4| / sqrt[(2^2)+(-1^2)+(-3^2)]
    = |(-4*1*12) - 4| / sqrt(4+1+9)
    = 52 / sqrt(14)
    = [52 sqrt(14)] / 14
    = [26 sqrt(14)] / 7
    = (26/7) sqrt(14)

    Good luck! If you need need more questions worked out/explained/explicated, just post em or pm me.

    -Andy
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  4. #4
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    thank you for the response!

    Thank you for the response, i understand this problem now that someone actually explained it to me (imagine if my teacher even knew what he was talking about??) I do have one more question that i don't understand AT ALL. The post is located here:
    http://www.mathhelpforum.com/math-he...tml#post170860

    Thanks
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