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

1. colorful Group Title

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. colorful Group Title

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

3. colorful Group Title

|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

4. colorful Group Title

$\vec n=\vec u\times\vec v$

5. itzmashy Group Title

|dw:1337148472550:dw|

6. colorful Group Title

yeah, that's a good graph :)

7. colorful Group Title

all that stuff I said still works though

8. itzmashy Group Title

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

9. colorful Group Title

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

10. itzmashy Group Title

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

11. colorful Group Title

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

12. itzmashy Group Title

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

13. colorful Group Title

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

14. itzmashy Group Title

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

15. itzmashy Group Title

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

16. itzmashy Group Title

and then crossing them?

17. colorful Group Title

exactly :)

18. itzmashy Group Title

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

19. colorful Group Title

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

20. itzmashy Group Title

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

21. itzmashy Group Title

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

22. itzmashy Group Title

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

23. colorful Group Title

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)

24. colorful Group Title

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

25. itzmashy Group Title

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

26. colorful Group Title

sure :)

27. itzmashy Group Title

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

28. colorful Group Title

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

29. colorful Group Title

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

30. itzmashy Group Title

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

31. colorful Group Title

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

32. colorful Group Title

hold on...

33. itzmashy Group Title

i crossed it in incorrectly lol hold on

34. colorful Group Title

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

35. colorful Group Title

sorry <2,0,-20>

36. colorful Group Title

but wait that way still be wrong...

37. itzmashy Group Title

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

38. itzmashy Group Title

whats your pq and pr? =\

39. colorful Group Title

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

40. itzmashy Group Title

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

41. colorful Group Title

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

42. colorful Group Title

so that is$\vec n$

43. itzmashy Group Title

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

44. itzmashy Group Title

or.. it doesnt??

45. colorful Group Title

my bad told you I'm tired

46. itzmashy Group Title

no worries!!! i totally understand hahha

47. itzmashy Group Title

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?

48. itzmashy Group Title

10x +8y +20z =40?

49. colorful Group Title

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

50. colorful Group Title

looks right though

51. colorful Group Title

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

52. colorful Group Title

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

53. itzmashy Group Title

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?

54. colorful Group Title

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

55. colorful Group Title

the 40 I mean

56. itzmashy Group Title

oops yeah hehe :D thats what i wrote down!

57. itzmashy Group Title

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

58. colorful Group Title

welcome, and g'night :)

59. itzmashy Group Title

sleep well!~!!!!