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Show that a line is parallel to and below the plane

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Letting the line be x=0 , y = t and z = t

    Show that the line is parallel to and below the plane 5x - 3y + 3z = 1

    3. The attempt at a solution

    - First I assigned t = 1 and t = 4 to obtain two points of the line:
    P1 = (0,1,1) and P2 = (0,4,4)

    - Then I created a directional vector P1P2 = (0-0, 4-1 , 4-1) = (0,3,3)

    - We know that n = (5, -3, 3)

    - So if we prove that n and P1P2 are orthogonal (i.e.their dot product is 0), then THE PLANE AND LINE ARE PARALLEL.

    Please tell me if I made any mistake so far...

    - But how do you prove the line is BELOW the plane???

    Last edited: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2
    ... the next question asks for the same but this time ABOVE the plane :P

    Thanks in advance!
  4. Oct 20, 2011 #3


    Staff: Mentor

    Every point on the line has an x-coordinate of 0, and is contained in the y-z plane. Look at the intersection of your given plane with the y-z plane, which will be another line. Call this line L2. To show that the line is below the plane, every point on the line should have a z-value that is less than that of the corresponding point on L2.

    Similar idea for your follow-on question.
    Last edited: Oct 20, 2011
  5. Oct 20, 2011 #4
    Ok, so I need to find the line of intersection between the given plane and the yz plane.

    Hmmm, to do that we set the system:
    5x - 3y + 3z = 1
    y - z = 0


    Then solve it and obtain:
    y = z (or y = t and z = t)
    x = 1/5

    I don't understand the part about "every z-value must be less than that of the corresponding point on L2."

    thx again
  6. Oct 20, 2011 #5


    Staff: Mentor

    No, not at all. The line whose parametric equations you're working with does not intersect the plane, so the system above has no solution.

    Every point in the y-z plane (AKA, the plane x = 0) has an x-coordinate of 0, so set x = 0 in the equation of your plane.
    L is the line given parametrically. L2 is the line of intersection between the plane (x - 3y + 3z = 1) and the y-z plane.

    Once you figure out the equations of the two lines it should be fairly obvious which one is above, and which one is below.
  7. Oct 20, 2011 #6
    I'm kind of lost now :S I can't see it lol
  8. Oct 20, 2011 #7
    We know the given line and the given plane are parallel - i.e. they DON'T intersect, correct?
  9. Oct 20, 2011 #8


    Staff: Mentor


    Your parametric line (L) can be written as z = y, right? (And x = 0.)

    What is the equation of line L2, which is the intersection of the plane x - 3y + 3z = 1 in the plane x = 0? Substitute 0 for x in the equation of the plane. What do you get? This really isn't very complicated.
  10. Oct 20, 2011 #9
    yeah, I guess the answer will be obvious. I don't know what I'm missing here loll

    Well, if you substitute 0, you get -3y + 3z = 1 , which looks like the equation of a yz plane? but if y = z then you get nothing (0 = 1 ? No solution)
  11. Oct 20, 2011 #10


    Staff: Mentor

    In the yz plane, this is a line. If you solve for z, what do you get?

    The two lines are parallel, so of course they don't intersect. You seem to be forgetting what you're trying to show, that one of them is above the other.
  12. Oct 20, 2011 #11
    yeah, I'm getting all mixed up here...

    So, solving for z you'd get z = 1/3 + y
  13. Oct 20, 2011 #12


    Staff: Mentor

    OK, good.

    In the x = 0 plane, one line's equation is z = y. The other line's equation is z = y + 1/3. The latter line is in the plane.

    Which one is above?

    If you're having trouble seeing this, graph both lines. Label the axes with the y-axis being the horizontal axis and the z-axis being vertical axis.

    Are you trying to do this without having a drawing?
  14. Oct 20, 2011 #13
    Yeah, no drawing (until now), I was trying to get a numerical proof of this.

    So, I'm drawing the graph...

    It looks like the x=0, y=z line is JUST below the other one (by about 0.3333 units in the z-axis).

    I believe this is what you meant by "every z-value must be less than that of the corresponding point on L2"

    What confuses me is the fact that we are using the same x=0 and y=z values in the
    5x - 3y + 3z = 1 plane ... Are we just trying to set two very similar planes and lines?
  15. Oct 20, 2011 #14


    Staff: Mentor

    In this context, "above" and "below" are visual concepts.
    How about exactly 1/3.
    The title of this thread is "Show that a line is parallel to and below the plane"
    The line of the title is the one with the parametric equations. It lies entirely in the plane x = 0 (the y-z plane). "Below" suggests that we look at the z-values on the line and the z-values on the portion of the plane that is directly above the line (IOW, the intersection of the plane with the y-z plane).

    You have to work a lot harder to see this if you don't have a drawing.
  16. Oct 20, 2011 #15
    It's a lot clearer now! I'm self-teaching all this stuff, so you might guess why I'm having some trouble with it hehe

    By the way, do you know any good Linear algebra and vectors book? I've been using Elementary linear algebra (anton/rorre - 9th edition) but I don't think it prepares the student well for this kind of problems

    thanks a lot for your help and time
  17. Oct 21, 2011 #16


    Staff: Mentor

    The book you have is probably OK. I have one by just Anton, 5th Ed. I would guess that your problems stem from some lack of understanding of precalculus concepts. If you're self-teaching, take a look at the materials at khanacademy.org - I think that's the site.
  18. Oct 21, 2011 #17
    I'll take a look at that website... thanks for the advice!
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