## caozeyuan 2 years ago True or false (give a reason if true or a counterexample if false): (a) If u is perpendicular (in three dimensions) to v and w, those vectors v and w are parallel. " (b) If u is perpendicular to v and w, then u is perpendicular to v + 2 w, (c) If u and v are perpendicular unit vectors then II u - v" = ,.,fi, g

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1. caozeyuan

(c) if u and v are perpendicular unit vectors then $\left| u-v \right|$=$\sqrt{2}$

2. caozeyuan

@terenzreignz , hey the 99 guy there! do you have any idea?

3. terenzreignz

Which one are you currently working on?

4. caozeyuan

I'm working on b. (a) solved

5. terenzreignz

My gut feel says no...

6. caozeyuan

counterexample needed then

7. terenzreignz

What if u is not perpendicular to w?

8. caozeyuan

but the question said that!

9. terenzreignz

10. terenzreignz

okay, it's true, then :P

11. terenzreignz

What's a convenient way to find if two vectors are perpendicular?

12. caozeyuan

u.v=0! ofcuz!

13. terenzreignz

Here's the <ehem> convenient truth... if u is perpendicular to v, then... $\Large \vec u \cdot \vec v = 0$ Yup

14. terenzreignz

Same goes for w. What then bodes for... $\Large \vec u \cdot(\vec v + 2\vec w)$ :P

15. terenzreignz

Well, you get the idea :P

16. terenzreignz

If you don't (you probably do, but anyway...) remember that dot product distributes over addition...

17. caozeyuan

so how about (c) . I think it should be True, but I cannot give a reason

18. terenzreignz

(c) doesn't make sense, I think you didn't copy it correctly :)

19. terenzreignz

Oh, I saw it just now... derp :3

20. terenzreignz

Tough one :D

21. terenzreignz

Okay, got it :P

22. terenzreignz

Sorry it took so long, I didn't feel like getting a piece of paper XD

23. terenzreignz

Let's let $\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k$$\Large \vec v= x_v\vec i + y_v\vec j + z_u \vec k$

24. terenzreignz

crud, I meant $\Large \vec v= x_v\vec i + y_v\vec j + z_{\color{red}v}\vec k$

25. caozeyuan

I got an idea! It is given that u and v are unit vectors so I can say that they are ( 1,0,0) and (0,1,0) respectively. the difference gives the vector(1,-1,0) and the length of that vector is clearly sqrt(2)

26. terenzreignz

Yes, but that's a specific case... you're supposed to prove it for *any* unit vector...

27. terenzreignz

Things are not that simple, but they are... rather elegant... Let's work out the magic, shall we? :)

28. terenzreignz

$\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k$$\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k$ Just tell me when you're ready .. ;)

29. caozeyuan

Yep, I understand that. move on

30. terenzreignz

Someone's bossy :/

31. terenzreignz

Okay, we know u and v are unit vectors, which can only mean... $\Large x_u^2 + y_u^2 + z_u^2 = 1 = x_v^2 + y_v^2+z_v^2$

32. terenzreignz

Right? Their magnitudes are equal to 1.

33. caozeyuan

so that gives sqrt(1^2+1^2)=sqrt(2), right?

34. terenzreignz

How did you figure that? -.-

35. terenzreignz

I like your initiative, but patience... is a virtue :)

36. terenzreignz

And no, your reasoning is faulty ;)

37. terenzreignz

So, why don't we acquire u - v... $\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k$

38. terenzreignz

Catch me so far?

39. caozeyuan

My brain explodes when things get too complicated

40. terenzreignz

Well, there is no avoiding this, so brace your brain :) Don't worry, it's not going to be THAT MUCH complicated... I just need to know that you understand everything up to here $\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k$so far... do you?

41. caozeyuan

Yes, I'm following

42. terenzreignz

Let's get its magnitude... $\Large ||\vec u - \vec v ||$

43. terenzreignz

By definition, this is $\Large =\sqrt{(x_u-x_v)^2+(y_u-y_v)^2+(z_u-z_v)^2}$

44. caozeyuan

so we have to show it is indeed$\sqrt{2}$

45. terenzreignz

Yup :)

46. terenzreignz

Ready? The tricky part begins now...

47. caozeyuan

oh NO! Don't torture me！

48. terenzreignz

This is for your own good <force feeds you lima beans> Now, let's simplify them squares... $\Large = \sqrt{x_u^2-2x_ux_v+x_v^2+y_u^2-2y_uy_v+y_v^2+z_u^2-2z_uz_v+z_v^2}$

49. terenzreignz

Everything good so far? I only simplified the squares... you know... $\large (a-b)^2 = a^2-2ab+b^2$

50. caozeyuan

Yes， I understand so far

51. terenzreignz

I think, I shall rearrange the terms... $\Large = \sqrt{x_u^2+y_u^2+z_u^2+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}$

52. terenzreignz

Are you okay with the rearrangement? Something you didn't get? :)

53. caozeyuan

Get it w/o prob.

54. terenzreignz

Okay, by hypothesis, since u is a unit vector, this part $\Large = \sqrt{\color{red}{x_u^2+y_u^2+z_u^2}+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}$ Is just $\Large = \sqrt{\color{red}1+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}$ Are you following me so far? :)

55. caozeyuan

and the next three stuff becomes 1 as well?

56. terenzreignz

That is correct, since v is also a unit vector... $\Large = \sqrt{\color{red}1+\color{green}1-2x_ux_v-2y_uy_v-2z_uz_v}$ $\Large = \sqrt{\color{blue}2-2x_ux_v-2y_uy_v-2z_uz_v}$ Are we done here? :P

57. terenzreignz

Nah, of course not :) Allow me to factor out the -2 here... $\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}$

58. terenzreignz

See anything familiar...?

59. caozeyuan

no, I don't see any familiar

60. terenzreignz

You're lacking creativity :P $\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}$ $\Large = \sqrt{2-2\color{blue}{(\vec u \cdot \vec v)}}$

61. terenzreignz

"aha!" moment?

62. caozeyuan

Oh my goodness! why am I so STUPID?!

63. terenzreignz

You're not... Just a little exhausted, maybe :) We haven't used the fact that they are perpendicular yet, and since they are, their dot product is zero :P $\Large = \sqrt{2-2\color{blue}{(0)}}= \sqrt2$ tadaa

64. caozeyuan

ok, QED.

65. terenzreignz

Now say it... Say Terence is awesome :3

66. caozeyuan

NO! you are arrogant! LOL!

67. terenzreignz

Arrogant is not the opposite of awesome. Now say it >:)

68. terenzreignz

LOL JK have a nice day :3