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richyw Group Title

Show that the line of intersection of the planes \[x+2y-z=2\]and\[3x+2y+2z=7\] is parallel to the line \[\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle\] Find the equation of the plane determined by these two lines.

  • 2 years ago
  • 2 years ago

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  1. richyw Group Title
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    I can do the first part, the trouble I am having is finding the plane. How can I do this?

    • 2 years ago
  2. helder_edwin Group Title
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    first, solve the system \[ \large \begin{cases} x+2y-z=2\\ 3x+2y+2z=7 \end{cases} \]

    • 2 years ago
  3. TuringTest Group Title
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    I would do it a bit differently I would cross the normal vectors of the two planes. The resulting vector should point in the same direction as the line.

    • 2 years ago
  4. richyw Group Title
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    they are parallel lines

    • 2 years ago
  5. TuringTest Group Title
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    what are parallel line?

    • 2 years ago
  6. TuringTest Group Title
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    lines*

    • 2 years ago
  7. richyw Group Title
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    I need the plane defined by the two lines.

    • 2 years ago
  8. TuringTest Group Title
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    you are given the planes

    • 2 years ago
  9. TuringTest Group Title
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    and you are only given one line

    • 2 years ago
  10. richyw Group Title
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    right and I have already found the direction of the line defined by the intersection of the two planes, and shown that it is perpendicular to the given line.

    • 2 years ago
  11. richyw Group Title
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    now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.

    • 2 years ago
  12. TuringTest Group Title
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    it's supposed to be parallel to the line, I am getting confused

    • 2 years ago
  13. richyw Group Title
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    it is. but two parallel lines can still define a plane.

    • 2 years ago
  14. helder_edwin Group Title
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    doing what @TuringTest sugested u get the vector \[ \large (6,-5,-4) \] which the direction of the given line.

    • 2 years ago
  15. TuringTest Group Title
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    I'll let @helder_edwin take it then, I am a bit confused as to what you're asking

    • 2 years ago
  16. TuringTest Group Title
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    ^right, that's what the Q seems to be asking

    • 2 years ago
  17. richyw Group Title
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    so what I did was find a point on the line of intersection which I did by setting z=0 and got the point (5/2,-1/4, 0)

    • 2 years ago
  18. helder_edwin Group Title
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    for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)

    • 2 years ago
  19. TuringTest Group Title
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    indeed

    • 2 years ago
  20. richyw Group Title
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    why can they not? I see no reason why they can't.

    • 2 years ago
  21. helder_edwin Group Title
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    a plane is two-dimensional whareas a line is one-dimensional

    • 2 years ago
  22. TuringTest Group Title
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    you need a point and a vector to define a plane

    • 2 years ago
  23. helder_edwin Group Title
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    but the point cannot belong to the line.

    • 2 years ago
  24. TuringTest Group Title
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    right, I mean a point in the plane and a normal vector|dw:1349383884636:dw|

    • 2 years ago
  25. richyw Group Title
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    but I have two lines. so I could get a displacment vector from one line and a point on the other line?

    • 2 years ago
  26. TuringTest Group Title
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    vectors can be moved, so having two does not define any point

    • 2 years ago
  27. richyw Group Title
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    but these are not vectors. These are lines!

    • 2 years ago
  28. richyw Group Title
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    I have one line, and If I could find the other, then I would have TWO LINES. I cannot visualize how there is more than one unique plane determined by these two lines

    • 2 years ago
  29. TuringTest Group Title
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    but you can't get another line because these vectors are linearly dependent

    • 2 years ago
  30. richyw Group Title
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    ok. what I need to do is find the equation of the line formed by the intersection of the two planes. This line has the direction (6,-5,-4), now I just need a point along that line. So I can just solve for where those planes equal eachother right?

    • 2 years ago
  31. richyw Group Title
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    doing that I find that the point (5/2,-1/4,0) is on the intersection of the two planes.

    • 2 years ago
  32. TuringTest Group Title
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    you just need one point of intersection, then use the vector you get from crossing the two normal vectors of the plane to write the equation of the line

    • 2 years ago
  33. richyw Group Title
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    now I have a point and a direction, meaning I have a line. How do I find this line?

    • 2 years ago
  34. TuringTest Group Title
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    use r(t)=P+tv where P is a point of intersection of the planes and v is the vector that points in that direction

    • 2 years ago
  35. richyw Group Title
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    ok I have done this and get \[\vec{r}(t)=\langle 5/2+6t, -1/4-5t,-4t \rangle\]

    • 2 years ago
  36. TuringTest Group Title
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    looks reasonable enough

    • 2 years ago
  37. TuringTest Group Title
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    oh I see, so now you want the plane between this line and the given one, eh?

    • 2 years ago
  38. richyw Group Title
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    ok now I have another line givin in the initial question.\[\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle\]

    • 2 years ago
  39. richyw Group Title
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    so two lines. need this plane!

    • 2 years ago
  40. TuringTest Group Title
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    take the given point P1=(1,3,2) and the point you just found P2=(5/2,-1/4,0) and make a vector u=P1-P2 then cross that with the vector for the line of intersection uXv=n (n is the normal vector of the plane you are looking for) then you can use n(dot)P2=n(dot)P where P=(x,y,z) to find the equation of that plane

    • 2 years ago
  41. richyw Group Title
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    alright cool thanks a lot! I got super confused when helder_edwin said "two parallel lines CANNOT define a plane. u need two independent directions (vectors)". So just to confirm I can get a second direction vector by subtracting a point on line 1 from a point on line two, correct?

    • 2 years ago
  42. helder_edwin Group Title
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    yes. and by doing that u got the second direction u needed.

    • 2 years ago
  43. TuringTest Group Title
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    yes @richyw that is right. Sorry for the laps in communication earlier, I was thinking we were just talking about two parallel vectors

    • 2 years ago
  44. TuringTest Group Title
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    ... and you're welcome :)

    • 2 years ago
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