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caozeyuan
Group Title
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
 one year ago
 one year ago
caozeyuan Group Title
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
 one year ago
 one year ago

This Question is Open

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
(c) if u and v are perpendicular unit vectors then \[\left uv \right\]=\[\sqrt{2}\]
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
@terenzreignz , hey the 99 guy there! do you have any idea?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Which one are you currently working on?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
I'm working on b. (a) solved
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
My gut feel says no...
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
counterexample needed then
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
What if u is not perpendicular to w?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
but the question said that!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Wait nvm... misread it... recalculating..
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
okay, it's true, then :P
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
What's a convenient way to find if two vectors are perpendicular?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
u.v=0! ofcuz!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Here's the <ehem> convenient truth... if u is perpendicular to v, then... \[\Large \vec u \cdot \vec v = 0\] Yup
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Same goes for w. What then bodes for... \[\Large \vec u \cdot(\vec v + 2\vec w)\] :P
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Well, you get the idea :P
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
If you don't (you probably do, but anyway...) remember that dot product distributes over addition...
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
so how about (c) . I think it should be True, but I cannot give a reason
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
(c) doesn't make sense, I think you didn't copy it correctly :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Oh, I saw it just now... derp :3
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Tough one :D
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Okay, got it :P
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Sorry it took so long, I didn't feel like getting a piece of paper XD
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Let's let \[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_u \vec k \]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
crud, I meant \[\Large \vec v= x_v\vec i + y_v\vec j + z_{\color{red}v}\vec k\]
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Yes, but that's a specific case... you're supposed to prove it for *any* unit vector...
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Things are not that simple, but they are... rather elegant... Let's work out the magic, shall we? :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
\[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k\] Just tell me when you're ready .. ;)
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
Yep, I understand that. move on
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Someone's bossy :/
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Okay, we know u and v are unit vectors, which can only mean... \[\Large x_u^2 + y_u^2 + z_u^2 = 1 = x_v^2 + y_v^2+z_v^2\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Right? Their magnitudes are equal to 1.
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
so that gives sqrt(1^2+1^2)=sqrt(2), right?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
How did you figure that? .
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
I like your initiative, but patience... is a virtue :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
And no, your reasoning is faulty ;)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
So, why don't we acquire u  v... \[\Large \vec u  \vec v =(x_u  x_v)\vec i + (y_uy_v)\vec j + (z_u  z_v )\vec k\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Catch me so far?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
My brain explodes when things get too complicated
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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 \[\Large \vec u  \vec v =(x_u  x_v)\vec i + (y_uy_v)\vec j + (z_u  z_v )\vec k\]so far... do you?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
Yes, I'm following
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Let's get its magnitude... \[\Large \vec u  \vec v \]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
By definition, this is \[\Large =\sqrt{(x_ux_v)^2+(y_uy_v)^2+(z_uz_v)^2}\]
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
so we have to show it is indeed\[\sqrt{2}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Yup :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Ready? The tricky part begins now...
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
oh NO! Don't torture me！
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
This is for your own good <force feeds you lima beans> Now, let's simplify them squares... \[\Large = \sqrt{x_u^22x_ux_v+x_v^2+y_u^22y_uy_v+y_v^2+z_u^22z_uz_v+z_v^2}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Everything good so far? I only simplified the squares... you know... \[\large (ab)^2 = a^22ab+b^2\]
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
Yes， I understand so far
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
I think, I shall rearrange the terms... \[\Large = \sqrt{x_u^2+y_u^2+z_u^2+x_v^2+y_v^2+z_v^22x_ux_v2y_uy_v2z_uz_v}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Are you okay with the rearrangement? Something you didn't get? :)
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
Get it w/o prob.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Okay, by hypothesis, since u is a unit vector, this part \[\Large = \sqrt{\color{red}{x_u^2+y_u^2+z_u^2}+x_v^2+y_v^2+z_v^22x_ux_v2y_uy_v2z_uz_v}\] Is just \[\Large = \sqrt{\color{red}1+x_v^2+y_v^2+z_v^22x_ux_v2y_uy_v2z_uz_v}\] Are you following me so far? :)
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
and the next three stuff becomes 1 as well?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
That is correct, since v is also a unit vector... \[\Large = \sqrt{\color{red}1+\color{green}12x_ux_v2y_uy_v2z_uz_v}\] \[\Large = \sqrt{\color{blue}22x_ux_v2y_uy_v2z_uz_v}\] Are we done here? :P
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Nah, of course not :) Allow me to factor out the 2 here... \[\Large = \sqrt{22\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
See anything familiar...?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
no, I don't see any familiar
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
You're lacking creativity :P \[\Large = \sqrt{22\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\] \[\Large = \sqrt{22\color{blue}{(\vec u \cdot \vec v)}}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
"aha!" moment?
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
Oh my goodness! why am I so STUPID?!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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 \[\Large = \sqrt{22\color{blue}{(0)}}= \sqrt2\] tadaa
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
ok, QED.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Now say it... Say Terence is awesome :3
 one year ago

caozeyuan Group TitleBest ResponseYou've already chosen the best response.1
NO! you are arrogant! LOL!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Arrogant is not the opposite of awesome. Now say it >:)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
LOL JK have a nice day :3
 one year ago
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