Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

itzmashy

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  2. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1337148180943:dw|so here are our three points if we can find\[\vec u=\overrightarrow{pq}=p-q\]and\[\vec v=\overrightarrow{qr}=r-q\]

    • one year ago
  3. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    |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

    • one year ago
  4. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\vec n=\vec u\times\vec v\]

    • one year ago
  5. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1337148472550:dw|

    • one year ago
  6. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, that's a good graph :)

    • one year ago
  7. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    all that stuff I said still works though

    • one year ago
  8. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    wait.. \[pq = p-q?\] but then you have \[qr = r -q\] i'm confused here:(

    • one year ago
  9. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    typo, my bad I sort of invented my own notation there on the left

    • one year ago
  10. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh okay so it's suppsoed to he pq = q-p, right?

    • one year ago
  11. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  12. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    so then \[u x v = (-4 x 0, -5 x 5, 2 x 0) = < 0, -10, 0> ?\]

    • one year ago
  13. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    no that's not how you take a cross product...

    • one year ago
  14. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    oh wait.. so.. nope you lost me again lol i'm sorry..

    • one year ago
  15. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    so you're picking one point as the middle point, and then creating two vectors from it?

    • one year ago
  16. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    and then crossing them?

    • one year ago
  17. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    exactly :)

    • one year ago
  18. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    and if i'm finding a vector PR, then its r -p, right? second point, minus first point? o.O

    • one year ago
  19. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  20. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    so, pr = < -4, 0, 2> pq = <-4, 5, 0> pr x pq = <-16, 0, 0> ?!

    • one year ago
  21. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  22. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    so getting the cross product of two vectors on a plane, gets me a vector perpendicular to the plane?

    • one year ago
  23. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • one year ago
  24. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    ...so if you want a vector that points from p to r you write r-p

    • one year ago
  25. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    okay! perfect :D thank you!!! one more question?

    • one year ago
  26. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    sure :)

    • one year ago
  27. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    find an equation of the plane containing points P Q and R

    • one year ago
  28. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  29. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant

    • one year ago
  30. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?

    • one year ago
  31. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    oh I didn't check, but that seems wrong...

    • one year ago
  32. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    hold on...

    • one year ago
  33. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    i crossed it in incorrectly lol hold on

    • one year ago
  34. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah I got <2,-20> on the first try, but I'm tired so let me try again

    • one year ago
  35. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry <2,0,-20>

    • one year ago
  36. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    but wait that way still be wrong...

    • one year ago
  37. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    really? pellet.. i got <10 8 20> =\

    • one year ago
  38. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    whats your pq and pr? =\

    • one year ago
  39. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    let me try again\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|\]

    • one year ago
  40. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    this is what i got: (10-0)i - (-8-0)j +(0+20)k is that wrong? :(

    • one year ago
  41. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  42. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    so that is\[\vec n\]

    • one year ago
  43. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    but.. but.. i got 10i + 8j + 20k, doesn't the - distribute to the 8? :(

    • one year ago
  44. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    or.. it doesnt??

    • one year ago
  45. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    my bad told you I'm tired

    • one year ago
  46. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    no worries!!! i totally understand hahha

    • one year ago
  47. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • one year ago
  48. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    10x +8y +20z =40?

    • one year ago
  49. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P

    • one year ago
  50. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    looks right though

    • one year ago
  51. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    ...of course you could have picked any point and you would have gotten the same d

    • one year ago
  52. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    and you can simplify it obviously too 5x +4y +10z =20

    • one year ago
  53. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • one year ago
  54. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    isn't it positive? you said so yourself, and I agree

    • one year ago
  55. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    the 40 I mean

    • one year ago
  56. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    oops yeah hehe :D thats what i wrote down!

    • one year ago
  57. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!

    • one year ago
  58. colorful
    Best Response
    You've already chosen the best response.
    Medals 1

    welcome, and g'night :)

    • one year ago
  59. itzmashy
    Best Response
    You've already chosen the best response.
    Medals 0

    sleep well!~!!!!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.