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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
 9 months ago
 9 months 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
 9 months ago
 9 months ago

This Question is Open

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

caozeyuanBest ResponseYou've already chosen the best response.1
@terenzreignz , hey the 99 guy there! do you have any idea?
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Which one are you currently working on?
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
I'm working on b. (a) solved
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
My gut feel says no...
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
counterexample needed then
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
What if u is not perpendicular to w?
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
but the question said that!
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Wait nvm... misread it... recalculating..
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
okay, it's true, then :P
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
What's a convenient way to find if two vectors are perpendicular?
 9 months ago

terenzreignzBest 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
 9 months ago

terenzreignzBest 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
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Well, you get the idea :P
 9 months ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.1
(c) doesn't make sense, I think you didn't copy it correctly :)
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Oh, I saw it just now... derp :3
 9 months ago

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

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

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

caozeyuanBest 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)
 9 months ago

terenzreignzBest 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...
 9 months ago

terenzreignzBest 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? :)
 9 months ago

terenzreignzBest 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 .. ;)
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
Yep, I understand that. move on
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Someone's bossy :/
 9 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.1
Right? Their magnitudes are equal to 1.
 9 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.1
How did you figure that? .
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
I like your initiative, but patience... is a virtue :)
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
And no, your reasoning is faulty ;)
 9 months ago

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

caozeyuanBest ResponseYou've already chosen the best response.1
My brain explodes when things get too complicated
 9 months ago

terenzreignzBest 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?
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Let's get its magnitude... \[\Large \vec u  \vec v \]
 9 months ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.1
Ready? The tricky part begins now...
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
oh NO! Don't torture me！
 9 months ago

terenzreignzBest 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... \[\Large = \sqrt{x_u^22x_ux_v+x_v^2+y_u^22y_uy_v+y_v^2+z_u^22z_uz_v+z_v^2}\]
 9 months ago

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

caozeyuanBest ResponseYou've already chosen the best response.1
Yes， I understand so far
 9 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.1
Are you okay with the rearrangement? Something you didn't get? :)
 9 months ago

terenzreignzBest 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? :)
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
and the next three stuff becomes 1 as well?
 9 months ago

terenzreignzBest 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
 9 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.1
See anything familiar...?
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
no, I don't see any familiar
 9 months ago

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

caozeyuanBest ResponseYou've already chosen the best response.1
Oh my goodness! why am I so STUPID?!
 9 months ago

terenzreignzBest 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
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Now say it... Say Terence is awesome :3
 9 months ago

caozeyuanBest ResponseYou've already chosen the best response.1
NO! you are arrogant! LOL!
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Arrogant is not the opposite of awesome. Now say it >:)
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
LOL JK have a nice day :3
 9 months ago
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