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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
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

This Question is Open

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

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

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

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

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

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

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

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

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

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

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

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

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

terenzreignz
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Yep, I understand that. move on

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

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

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

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

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

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

terenzreignz
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1My brain explodes when things get too complicated

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

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

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

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

terenzreignz
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Yes， I understand so far

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

terenzreignz
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1and the next three stuff becomes 1 as well?

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

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

terenzreignz
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Oh my goodness! why am I so STUPID?!

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

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

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

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