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richyw

  • 2 years ago

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.

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

  3. TuringTest
    • 2 years ago
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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.

  4. richyw
    • 2 years ago
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    they are parallel lines

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

  6. TuringTest
    • 2 years ago
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    lines*

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

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

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

  10. richyw
    • 2 years ago
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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.

  11. richyw
    • 2 years ago
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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.

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

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

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

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

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

  17. richyw
    • 2 years ago
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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)

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

  19. TuringTest
    • 2 years ago
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    indeed

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

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

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

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

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

  25. richyw
    • 2 years ago
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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?

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

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

  28. richyw
    • 2 years ago
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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

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

  30. richyw
    • 2 years ago
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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?

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

  32. TuringTest
    • 2 years ago
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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

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

  34. TuringTest
    • 2 years ago
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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

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

  36. TuringTest
    • 2 years ago
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    looks reasonable enough

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

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

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

  40. TuringTest
    • 2 years ago
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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

  41. richyw
    • 2 years ago
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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?

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

  43. TuringTest
    • 2 years ago
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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

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

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