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Homework Help: Intersection of 3 planes

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Find value of k so that
    x+2y-z=0
    x+9y-5z=0
    kx-y+z=0
    intersect in a line

    2. Relevant equations



    3. The attempt at a solution
    multiply l1 by 5
    subtract l2 from 5l1
    end up with:
    4x+y=0

    subtract l1 from l2
    end up with:
    7y-4z=0

    i have no idea what to do from here, or even if what i did was correct.
    thanks for the help.
     
  2. jcsd
  3. Jan 21, 2009 #2

    rock.freak667

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    Rewrite it in maxtrix form and row-reduce until echelon form is obtained. For the planes to intersect in a line, then when you reduce in echelon form, the rank of the augmented matrix should be less than 3 (the number of variables)
     
  4. Jan 21, 2009 #3
    ahh thank you.

    i have another problem:

    Find the vector equation of the plane through A(1,-7,-2) and perpendicular to line (5,0,0)+t(2,0,7)

    i have parametric equations of the line
    x=5+t
    y=0
    z=7t

    what do i do from here?
     
  5. Jan 21, 2009 #4

    rock.freak667

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    If the line is perpendicular to the plane, what can you say about the normal to the plane and the line?
     
  6. Jan 21, 2009 #5
    line is normal to the plane?
     
  7. Jan 21, 2009 #6
    ok, i found took the vector from 5,0,0 from 1,-7,-2

    i made the z component of the vector 0

    then knowing that the dot product of the line and a vector of the plane is 0, i found the component of z of a vector of the plane which would make it perpendicular to the line

    is that possible?
     
  8. Jan 21, 2009 #7

    rock.freak667

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    Yes, so for a vector line in the for r=a+tu, where u is the direction of the line.

    In your example u=<2,0,7>, if the lines is perpendicular to the plane, then so is the direction.

    So if the direction (u) is normal to the plane, doesn't that mean that the vector u is parallel to the normal?
     
  9. Jan 21, 2009 #8
    absolutely
     
  10. Jan 21, 2009 #9

    rock.freak667

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    so if it is parellel to the plane then

    Q<2,0,7> would be a normal to this plane. Pick any non-zero value for Q, and you'll get the normal to the plane,N. When you get that, use the formule

    [tex](\vec{r}-\vec{r_0})\cdot \vec{N}=0[/tex]

    Where [itex]\vec{N}[/itex] is the normal vector and [itex]r_0[/itex] is a point on the plane (which you were given in the question).
     
    Last edited: Jan 21, 2009
  11. Jan 21, 2009 #10
    what is the other r in the question?
     
  12. Jan 21, 2009 #11

    rock.freak667

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    that is just <x,y,z>

    so that for the point <x0,y0,z0>

    [tex]\vec{r}- \vec{r_0} = <(x-x_0),(y-y_0),(z-z_0)>[/tex]
     
  13. Jan 21, 2009 #12
    thank you.

    i have another question:



    find a vector equation of the line through the point 4,5,5

    that meets the line: (x-11)/3=y+8=z-4

    i dont even know where to start on this one.

    perhaps i do the same thing as before?

    find the vector between 4,5,5 and the point on the line when t=0? (starting point?)
    then set one of the components to 0 and use the dot product formula?
     
    Last edited: Jan 21, 2009
  14. Jan 21, 2009 #13

    rock.freak667

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    For a vector line written in the form

    [tex]\frac{x-a}{p} = \frac{y-b}{q}= \frac{z-c}{r}[/tex]

    What does <a,b,c> represent and what does <p,q,r> represent?
     
  15. Jan 21, 2009 #14
    -a, -b, -c is a point
    p, q, r are components in the x, y, z axis
     
  16. Jan 21, 2009 #15

    rock.freak667

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    Good
    <p,q,r> would be the direction of the line.
     
  17. Jan 21, 2009 #16
    haha where do i go from there then? btw is my process that i edited in at 2:27 am correct?
     
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