## anonymous 4 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. anonymous

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. anonymous

|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. anonymous

|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. anonymous

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

5. anonymous

|dw:1337148472550:dw|

6. anonymous

yeah, that's a good graph :)

7. anonymous

all that stuff I said still works though

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

and then crossing them?

17. anonymous

exactly :)

18. anonymous

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

19. anonymous

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. anonymous

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

21. anonymous

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. anonymous

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

23. anonymous

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. anonymous

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

25. anonymous

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

26. anonymous

sure :)

27. anonymous

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

28. anonymous

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. anonymous

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

30. anonymous

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

31. anonymous

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

32. anonymous

hold on...

33. anonymous

i crossed it in incorrectly lol hold on

34. anonymous

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

35. anonymous

sorry <2,0,-20>

36. anonymous

but wait that way still be wrong...

37. anonymous

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

38. anonymous

whats your pq and pr? =\

39. anonymous

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. anonymous

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

41. anonymous

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

42. anonymous

so that is$\vec n$

43. anonymous

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

44. anonymous

or.. it doesnt??

45. anonymous

my bad told you I'm tired

46. anonymous

no worries!!! i totally understand hahha

47. anonymous

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. anonymous

10x +8y +20z =40?

49. anonymous

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

50. anonymous

looks right though

51. anonymous

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

52. anonymous

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

53. anonymous

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. anonymous

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

55. anonymous

the 40 I mean

56. anonymous

oops yeah hehe :D thats what i wrote down!

57. anonymous

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

58. anonymous

welcome, and g'night :)

59. anonymous

sleep well!~!!!!