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KatClaire Group Title

Find the equation of the line of intersection of the planes with the equations -3x+4y+z=1 and 5x-3z=5

  • one year ago
  • one year ago

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  1. KatClaire Group Title
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    Please help me work it out step by step :(

    • one year ago
  2. cherio12 Group Title
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    alright so you will have to use the cross product of the two normal vectors

    • one year ago
  3. KatClaire Group Title
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    I did and I got -12i-4j-20k

    • one year ago
  4. cherio12 Group Title
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    hm...how'd u get yours?

    • one year ago
  5. cherio12 Group Title
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    wait...i got it

    • one year ago
  6. amistre64 Group Title
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    practice makes perfect, good job so far cherio ;)

    • one year ago
  7. cherio12 Group Title
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    haha..gotta love calc =D..and ya i got -12i-4j-20k after i set it up right

    • one year ago
  8. KatClaire Group Title
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    :P

    • one year ago
  9. cherio12 Group Title
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    so this is your direction vector i believe

    • one year ago
  10. cherio12 Group Title
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    now you need to to find a point that lie on this line

    • one year ago
  11. cherio12 Group Title
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    you can do this by taking your two planes and setting x,y, or z to zero and finding a point

    • one year ago
  12. KatClaire Group Title
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    okay so if x=0 I get y=2/3 and z=-5/3 , right?

    • one year ago
  13. cherio12 Group Title
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    yes, so for the next part you have to make a vector with that point (this will be dotted with the direction vector from before)

    • one year ago
  14. cherio12 Group Title
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    so typically for a line you just select the (x,y,z) to be your point (if anything doesn't make sense let me know) so you get (x-0, y-(2/3), z+(5/3)) as that vector

    • one year ago
  15. cherio12 Group Title
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    now i dot it with any scalar of the direction vector and get L=(x, y-(2/3), z+(5/3))* t(-12,-4,-20)

    • one year ago
  16. KatClaire Group Title
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    Any reason why you do (x-0,y-(2/3),z+(5/3)? and you can't simplify that any more so that would be the answer?

    • one year ago
  17. cherio12 Group Title
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    well if you want to write in parametric or (something that starts with a s...symmetric?) you can simplify it a bit more..but as far at why x,y,z...um..i'll try to draw a picture.

    • one year ago
  18. amistre64 Group Title
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    in case i missed it, since the second plane has no y value, wouldnt it be simper to make y=0 in the first to solve for x,z ?

    • one year ago
  19. cherio12 Group Title
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    |dw:1359565267353:dw|

    • one year ago
  20. cherio12 Group Title
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    i just set x=0, so i didn't have to do substitution. other way works too

    • one year ago
  21. cherio12 Group Title
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    okay..not to try to explain the picture...

    • one year ago
  22. amistre64 Group Title
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    i get a common point of (-2,0,-5) :) but any common point will suffice

    • one year ago
  23. KatClaire Group Title
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    well either way, I think I understand it more now! I was confused as to why the cross product was involved and what not. Thanks!!

    • one year ago
  24. amistre64 Group Title
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    crossing vectors created a new vector perp to both crossing 2 plane normals will get you a direction vector that is parallel to the line of insection

    • one year ago
  25. KatClaire Group Title
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    Yes! thank you

    • one year ago
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