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itzmashy Group Title

a plane contains points p(4,0,0) q (0,5,0) and r(0,0,2) a - sketch the plane b - find a vector perpendicular to the plane

  • 2 years ago
  • 2 years ago

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  1. colorful Group Title
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    I don't know about sketching it, but if you want a vector perpendicular to it you need to find two vectors, \(\vec u\) and \(\vec v\) that lie in the plane you can do that by finding the vectors between the various points with a common vertex

    • 2 years ago
  2. colorful Group Title
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    |dw:1337148180943:dw|so here are our three points if we can find\[\vec u=\overrightarrow{pq}=p-q\]and\[\vec v=\overrightarrow{qr}=r-q\]

    • 2 years ago
  3. colorful Group Title
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    |dw:1337148370560:dw|\[\vec u=p-q=\langle4,-5,0\rangle\]\[\vec v=r-q=\langle0,-5,2\rangle\]the cross product of these vectors will be normal to the plane

    • 2 years ago
  4. colorful Group Title
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    \[\vec n=\vec u\times\vec v\]

    • 2 years ago
  5. itzmashy Group Title
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    |dw:1337148472550:dw|

    • 2 years ago
  6. colorful Group Title
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    yeah, that's a good graph :)

    • 2 years ago
  7. colorful Group Title
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    all that stuff I said still works though

    • 2 years ago
  8. itzmashy Group Title
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    wait.. \[pq = p-q?\] but then you have \[qr = r -q\] i'm confused here:(

    • 2 years ago
  9. colorful Group Title
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    typo, my bad I sort of invented my own notation there on the left

    • 2 years ago
  10. itzmashy Group Title
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    ohh okay so it's suppsoed to he pq = q-p, right?

    • 2 years ago
  11. colorful Group Title
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    you can do it many ways, but they need a common vertex the vector p-q points toward p from q the vector r-q points toward r from q so this should just be\[pq=p-q\]\[rq=r-q\]as in the distance from p and r to q, respectively

    • 2 years ago
  12. itzmashy Group Title
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    so then \[u x v = (-4 x 0, -5 x 5, 2 x 0) = < 0, -10, 0> ?\]

    • 2 years ago
  13. colorful Group Title
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    no that's not how you take a cross product...

    • 2 years ago
  14. itzmashy Group Title
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    oh wait.. so.. nope you lost me again lol i'm sorry..

    • 2 years ago
  15. itzmashy Group Title
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    so you're picking one point as the middle point, and then creating two vectors from it?

    • 2 years ago
  16. itzmashy Group Title
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    and then crossing them?

    • 2 years ago
  17. colorful Group Title
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    exactly :)

    • 2 years ago
  18. itzmashy Group Title
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    and if i'm finding a vector PR, then its r -p, right? second point, minus first point? o.O

    • 2 years ago
  19. colorful Group Title
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    I sort of am making up the notation... whether it is supposed to be written PR or RP I really don't know all that matters is that you're consistent in your choice of vertex

    • 2 years ago
  20. itzmashy Group Title
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    so, pr = < -4, 0, 2> pq = <-4, 5, 0> pr x pq = <-16, 0, 0> ?!

    • 2 years ago
  21. itzmashy Group Title
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    but when i'm finding any vector, like for example PR, then how would i find it? r - p, or p-r? just in general, not related to this problem

    • 2 years ago
  22. itzmashy Group Title
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    so getting the cross product of two vectors on a plane, gets me a vector perpendicular to the plane?

    • 2 years ago
  23. colorful Group Title
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    let me check you on that cross product... as far as what minus what is concerned, in a-b the vector points toward a so it is vector=(to point)-(from point)

    • 2 years ago
  24. colorful Group Title
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    ...so if you want a vector that points from p to r you write r-p

    • 2 years ago
  25. itzmashy Group Title
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    okay! perfect :D thank you!!! one more question?

    • 2 years ago
  26. colorful Group Title
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    sure :)

    • 2 years ago
  27. itzmashy Group Title
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    find an equation of the plane containing points P Q and R

    • 2 years ago
  28. colorful Group Title
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    thankfully once you have the normal vector to the plane this is a piece of cake just put the numbers in the components of the normal vector as the coefficients

    • 2 years ago
  29. colorful Group Title
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    the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant

    • 2 years ago
  30. itzmashy Group Title
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    and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?

    • 2 years ago
  31. colorful Group Title
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    oh I didn't check, but that seems wrong...

    • 2 years ago
  32. colorful Group Title
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    hold on...

    • 2 years ago
  33. itzmashy Group Title
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    i crossed it in incorrectly lol hold on

    • 2 years ago
  34. colorful Group Title
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    yeah I got <2,-20> on the first try, but I'm tired so let me try again

    • 2 years ago
  35. colorful Group Title
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    sorry <2,0,-20>

    • 2 years ago
  36. colorful Group Title
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    but wait that way still be wrong...

    • 2 years ago
  37. itzmashy Group Title
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    really? pellet.. i got <10 8 20> =\

    • 2 years ago
  38. itzmashy Group Title
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    whats your pq and pr? =\

    • 2 years ago
  39. colorful Group Title
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    let me try again\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|\]

    • 2 years ago
  40. itzmashy Group Title
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    this is what i got: (10-0)i - (-8-0)j +(0+20)k is that wrong? :(

    • 2 years ago
  41. colorful Group Title
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    \[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|=10\hat i+20\hat k-8\hat j\]ok now we agree :)

    • 2 years ago
  42. colorful Group Title
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    so that is\[\vec n\]

    • 2 years ago
  43. itzmashy Group Title
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    but.. but.. i got 10i + 8j + 20k, doesn't the - distribute to the 8? :(

    • 2 years ago
  44. itzmashy Group Title
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    or.. it doesnt??

    • 2 years ago
  45. colorful Group Title
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    my bad told you I'm tired

    • 2 years ago
  46. itzmashy Group Title
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    no worries!!! i totally understand hahha

    • 2 years ago
  47. itzmashy Group Title
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    so then i think i'm supposed to plug in the variables and a point so.. 10(x-4)+8(y-0)+20(z-0)=0?

    • 2 years ago
  48. itzmashy Group Title
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    10x +8y +20z =40?

    • 2 years ago
  49. colorful Group Title
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    yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P

    • 2 years ago
  50. colorful Group Title
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    looks right though

    • 2 years ago
  51. colorful Group Title
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    ...of course you could have picked any point and you would have gotten the same d

    • 2 years ago
  52. colorful Group Title
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    and you can simplify it obviously too 5x +4y +10z =20

    • 2 years ago
  53. itzmashy Group Title
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    so the vector perpendicular to the plane is < 10, 8, 20>, and the equation of the plane containing points p q and r is 10x+8y+20z = -40?

    • 2 years ago
  54. colorful Group Title
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    isn't it positive? you said so yourself, and I agree

    • 2 years ago
  55. colorful Group Title
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    the 40 I mean

    • 2 years ago
  56. itzmashy Group Title
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    oops yeah hehe :D thats what i wrote down!

    • 2 years ago
  57. itzmashy Group Title
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    yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!

    • 2 years ago
  58. colorful Group Title
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    welcome, and g'night :)

    • 2 years ago
  59. itzmashy Group Title
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    sleep well!~!!!!

    • 2 years ago
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