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caozeyuan
 one year 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
caozeyuan
 one year 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

This Question is Open

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.1I 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)

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

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

terenzreignz
 one year ago
Best 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 .. ;)

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

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

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

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

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

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

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

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

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

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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?

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

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

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

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

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

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

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

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

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

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Okay, 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? :)

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1That 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

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

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

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

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

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1You'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

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

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

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

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