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Projection of line to plane

  1. May 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello! :smile:

    Find the equation of the projection of the line [tex]\frac{x}{4}=\frac{y-4}{3}=\frac{z+1}{-2}[/tex] of the plane x-y+3z+8=0.

    So the line projects its self on the plane...

    2. Relevant equations



    3. The attempt at a solution

    First I find equation of line which passes throught the line and it has vector "a" parallel to the normal vector of the plane...

    The equation of that line is: [tex]\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}[/tex]

    The equation of the line we need to find is probably: [tex]\frac{x-x_1}{4}=\frac{y-y_1}{3}=\frac{z-z_1}{-2}[/tex]

    So we just need to find the point...

    That point is [tex]M(\frac{1}{11},4-\frac{1}{11},-1+3\frac{1}{11})[/tex]

    But in my text book results I get tottaly different solution: [tex]\frac{x+9}{7}=\frac{y+1}{4}=\frac{z}{-1}[/tex]

    I don't know what is the problem. Please help.
     
    Last edited: May 19, 2008
  2. jcsd
  3. May 19, 2008 #2

    tiny-tim

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    Hello Theofilius! :smile:

    Call the line L, the normal to the plane N, and the projection P.

    Hint: what mathematical relationship is there between the three lines L N and P? :smile:
     
  4. May 19, 2008 #3
    I don't know, but recently I figured out that [tex]\vec{a} \circ \vec{n} \neq 0[/tex]. So, my way of solving the task is wrong...
     
  5. May 19, 2008 #4

    tiny-tim

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    (Well, we both knew that! :rolleyes: )

    Do you know how to draw a projection?

    If so, draw the diagram, and then:
     
  6. May 19, 2008 #5
    Like this? P is parallel to L
     
  7. May 19, 2008 #6

    tiny-tim

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    hmm … useful tip … you're looking for triangles and suchlike, so try to get everything to join up.

    Diagrams are supposed to help you … to stop you having to "leave things to the imagination".

    So draw it again, for yourself, with the line actually meeting the plane.

    And mark in any right-angles!

    Then what do you notice about L N and P? :smile:
     
  8. May 19, 2008 #7
    Here is something but I don't quite understand anything... Can you give me better diagram?
     
  9. May 19, 2008 #8

    tiny-tim

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    Hi Theofilius! :smile:

    Let's just describe it:

    Let the line L meet the plane at the point A.

    Let a normal to the plane meet the line L at B and the plane at C.

    Then ABC is a right-angled triangle, isn't it?

    L is AB, N is BC, P is CA.

    So what's the relationship? :smile:
     
  10. May 19, 2008 #9
    I understand now. So ABC is triangle. I don't know for what relationship you mean? Probably all lines are intersect each other... :smile:
     
  11. May 19, 2008 #10

    tiny-tim

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    ABC is a triangle, so all three lines are in the same plane.

    So the vector of the projection, CA, must be a linear combination of AB and BC, both of which are known. :smile:
     
  12. May 19, 2008 #11
    What should I do now? Should I find the intersection point of the plane and the line?
     
  13. May 19, 2008 #12

    HallsofIvy

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    You are not talking about just lines- you are talking about vectors. Vectors add and subtract.
     
  14. May 19, 2008 #13
    But I don't know those vectors...
     
  15. May 19, 2008 #14
    What are the coordinates of AB BC and AC?
     
  16. May 19, 2008 #15

    tiny-tim

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    Theofilius, follow HallsofIvy's hint:
    Take two fixed vectors, AB and BC.

    Make any linear combination of those vectors …

    what do you get? :smile:
     
  17. May 19, 2008 #16
    Can we just don't complicate? Lets make like this. Look at this picture... I will choose points from the line, and make 2 new normal lines out from the points. After that I will find the two points which are intersecting with the plane... Then I will use this formula:

    [tex]\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}[/tex]

    Does this make any sense?
     
  18. May 19, 2008 #17

    tiny-tim

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    Hi Theofilius! :smile:

    Yes, that will certainly work, once you've found those two points of intersection.

    Have a go! :smile:
     
  19. May 19, 2008 #18
    Unfortunately I don't get same solution as the text book results... I get for
    [tex]\vec{a}(\frac{57}{11}, \frac{20}{11} , \frac{17}{11})[/tex]
    and they get (7,4,-1)

    the vector a is the vector parallel to the line that we are finding...
     
  20. May 19, 2008 #19

    tiny-tim

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    Theofilius, how can we help by telling you where you've gone wrong, if you don't show us any working? :confused:
     
  21. May 20, 2008 #20
    The first point on the line L:

    A(0,4,-1) ;

    the second point on the line L:

    B(4,7,-3)

    The line that is passing throught A(0,4,-1) and is normal to the plane:

    [tex]\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}[/tex]

    The line that is passing throught B(4,7,-3) and is normal to the plane:

    [tex]\frac{x-4}{1}=\frac{y-7}{-1}=\frac{z+3}{3}[/tex]

    The interesection point with the line L and the plane:

    [tex](Aa_1+Ba_2+Ca_3)t+Ax_1+By_1+Cz_1+D=0[/tex]

    [tex]at+b=0[/tex]

    [tex]t=\frac{-b}{a}[/tex]

    The parametric equation of the line which is passing through A:

    [tex]x=t[/tex]

    [tex]y=4-t[/tex]

    [tex]z=-1+3t[/tex]

    [tex]a=1*1+(-1)(-1)+3*3=1+1+9=11[/tex]

    [tex]b=0*1+4*(-1)+(-1)*3+8=-4-3+8=1[/tex]

    So [tex]t=\frac{-1}{11}[/tex]

    Now we are substituting for t

    [tex]x=\frac{-1}{11}[/tex]

    [tex]y=4-\frac{-1}{11}=\frac{45}{11}[/tex]

    [tex]z=-1+3*\frac{-1}{11}=\frac{-14}{11}[/tex]

    The first point which is lying on the plane is:

    [tex]M_1(\frac{-1}{11},\frac{45}{11},\frac{-14}{11})[/tex]


    The parametric equation of the line which is passing throught B:

    [tex]x=4+t[/tex]

    [tex]y=7-t[/tex]

    [tex]z=-3+3t[/tex]

    a=11

    b=4-7-9=-12

    [tex]t=\frac{12}{11}[/tex]

    Now we are substituting for t:

    [tex]x=4+\frac{12}{11}=\frac{56}{11}[/tex]

    [tex]y=7-\frac{12}{11}=\frac{65}{11}[/tex]

    [tex]z=-3-\frac{12}{11}=\frac{-45}{11}[/tex]

    The second point which is lying on the plane is:

    [tex]M_2(\frac{56}{11},\frac{65}{11},\frac{-45}{11})[/tex]

    Now the equation of the projection line that we are looking for will be:

    [tex]\frac{x+\frac{1}{11}}{\frac{56}{11}+\frac{1}{11}}=\frac{y-\frac{45}{11}}{\frac{65}{11}-\frac{45}{11}}=\frac{z+\frac{14}{11}}{\frac{-45}{11}+\frac{14}{11}}[/tex]

    [tex]\frac{x+\frac{1}{11}}{\frac{57}{11}}=\frac{y-\frac{45}{11}}{\frac{20}{11}}=\frac{z+\frac{14}{11}}{\frac{-31}{11}}[/tex]

    How should I check if their answer is correct?
     
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