Question

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

Answer 1

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

Answer 2

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

Answer 3

Which one are you currently working on?

Answer 4

I'm working on b. (a) solved

Answer 5

My gut feel says no...

Answer 6

counterexample needed then

Answer 7

What if u is not perpendicular to w?

Answer 8

but the question said that!

Answer 9

Wait nvm...
misread it...
recalculating..

Answer 10

okay, it's true, then :P

Answer 11

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

Answer 12

u.v=0! ofcuz!

Answer 13

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

Answer 14

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

Answer 15

Well, you get the idea :P

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Tough one :D

Answer 21

Okay, got it :P

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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)

Answer 26

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

Answer 27

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

Answer 28

\[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k\]

Just tell me when you're ready .. ;)

Answer 29

Yep, I understand that. move on

Answer 30

Someone's bossy :/

Answer 31

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

Answer 32

Right?
Their magnitudes are equal to 1.

Answer 33

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

Answer 34

How did you figure that? -.-

Answer 35

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

Answer 36

And no, your reasoning is faulty ;)

Answer 37

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

Answer 38

Catch me so far?

Answer 39

My brain explodes when things get too complicated

Answer 40

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?

Answer 41

Yes, I'm following

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

Yup :)

Answer 46

Ready?
The tricky part begins now...

Answer 47

oh NO! Don't torture me！

Answer 48

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

Answer 49

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

Answer 50

Yes， I understand so far

Answer 51

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

Answer 52

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

Answer 53

Get it w/o prob.

Answer 54

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? :)

Answer 55

and the next three stuff becomes 1 as well?

Answer 56

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

Answer 57

Nah, of course not :)

Allow me to factor out the -2 here...
\[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\]

Answer 58

See anything familiar...?

Answer 59

no, I don't see any familiar

Answer 60

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

Answer 61

"aha!" moment?

Answer 62

Oh my goodness! why am I so STUPID?!

Answer 63

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

Answer 64

ok, QED.

Answer 65

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

Answer 66

NO! you are arrogant! LOL!

Answer 67

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

Answer 68

LOL JK

have a nice day :3