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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

Linear Algebra
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(c) if u and v are perpendicular unit vectors then \[\left| u-v \right|\]=\[\sqrt{2}\]
@terenzreignz , hey the 99 guy there! do you have any idea?
Which one are you currently working on?

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Other answers:

I'm working on b. (a) solved
My gut feel says no...
counterexample needed then
What if u is not perpendicular to w?
but the question said that!
Wait nvm... misread it... recalculating..
okay, it's true, then :P
What's a convenient way to find if two vectors are perpendicular?
u.v=0! ofcuz!
Here's the convenient truth... if u is perpendicular to v, then... \[\Large \vec u \cdot \vec v = 0\] Yup
Same goes for w. What then bodes for... \[\Large \vec u \cdot(\vec v + 2\vec w)\] :P
Well, you get the idea :P
If you don't (you probably do, but anyway...) remember that dot product distributes over addition...
so how about (c) . I think it should be True, but I cannot give a reason
(c) doesn't make sense, I think you didn't copy it correctly :)
Oh, I saw it just now... derp :3
Tough one :D
Okay, got it :P
Sorry it took so long, I didn't feel like getting a piece of paper XD
Let's let \[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_u \vec k \]
crud, I meant \[\Large \vec v= x_v\vec i + y_v\vec j + z_{\color{red}v}\vec k\]
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)
Yes, but that's a specific case... you're supposed to prove it for *any* unit vector...
Things are not that simple, but they are... rather elegant... Let's work out the magic, shall we? :)
\[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k\] Just tell me when you're ready .. ;)
Yep, I understand that. move on
Someone's bossy :/
Okay, we know u and v are unit vectors, which can only mean... \[\Large x_u^2 + y_u^2 + z_u^2 = 1 = x_v^2 + y_v^2+z_v^2\]
Right? Their magnitudes are equal to 1.
so that gives sqrt(1^2+1^2)=sqrt(2), right?
How did you figure that? -.-
I like your initiative, but patience... is a virtue :)
And no, your reasoning is faulty ;)
So, why don't we acquire u - v... \[\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k\]
Catch me so far?
My brain explodes when things get too complicated
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?
Yes, I'm following
Let's get its magnitude... \[\Large ||\vec u - \vec v ||\]
By definition, this is \[\Large =\sqrt{(x_u-x_v)^2+(y_u-y_v)^2+(z_u-z_v)^2}\]
so we have to show it is indeed\[\sqrt{2}\]
Yup :)
Ready? The tricky part begins now...
oh NO! Don't torture me!
This is for your own good Now, let's simplify them squares... \[\Large = \sqrt{x_u^2-2x_ux_v+x_v^2+y_u^2-2y_uy_v+y_v^2+z_u^2-2z_uz_v+z_v^2}\]
Everything good so far? I only simplified the squares... you know... \[\large (a-b)^2 = a^2-2ab+b^2\]
Yes, I understand so far
I think, I shall rearrange the terms... \[\Large = \sqrt{x_u^2+y_u^2+z_u^2+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\]
Are you okay with the rearrangement? Something you didn't get? :)
Get it w/o prob.
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? :)
and the next three stuff becomes 1 as well?
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
Nah, of course not :) Allow me to factor out the -2 here... \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\]
See anything familiar...?
no, I don't see any familiar
You're lacking creativity :P \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\] \[\Large = \sqrt{2-2\color{blue}{(\vec u \cdot \vec v)}}\]
"aha!" moment?
Oh my goodness! why am I so STUPID?!
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
ok, QED.
Now say it... Say Terence is awesome :3
NO! you are arrogant! LOL!
Arrogant is not the opposite of awesome. Now say it >:)
LOL JK have a nice day :3

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