## itzmashy 3 years ago 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

1. colorful

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

|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

|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

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

5. itzmashy

|dw:1337148472550:dw|

6. colorful

yeah, that's a good graph :)

7. colorful

all that stuff I said still works though

8. itzmashy

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

9. colorful

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

10. itzmashy

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

11. colorful

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

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

13. colorful

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

14. itzmashy

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

15. itzmashy

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

16. itzmashy

and then crossing them?

17. colorful

exactly :)

18. itzmashy

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

19. colorful

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

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

21. itzmashy

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

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

23. colorful

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

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

25. itzmashy

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

26. colorful

sure :)

27. itzmashy

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

28. colorful

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

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

30. itzmashy

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

31. colorful

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

32. colorful

hold on...

33. itzmashy

i crossed it in incorrectly lol hold on

34. colorful

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

35. colorful

sorry <2,0,-20>

36. colorful

but wait that way still be wrong...

37. itzmashy

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

38. itzmashy

whats your pq and pr? =\

39. colorful

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

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

41. colorful

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

42. colorful

so that is$\vec n$

43. itzmashy

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

44. itzmashy

or.. it doesnt??

45. colorful

my bad told you I'm tired

46. itzmashy

no worries!!! i totally understand hahha

47. itzmashy

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

10x +8y +20z =40?

49. colorful

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

looks right though

51. colorful

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

52. colorful

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

53. itzmashy

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

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

55. colorful

the 40 I mean

56. itzmashy

oops yeah hehe :D thats what i wrote down!

57. itzmashy

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

58. colorful

welcome, and g'night :)

59. itzmashy

sleep well!~!!!!