A community for students.
Here's the question you clicked on:
 0 viewing
caozeyuan
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1@terenzreignz , hey the 99 guy there! do you have any idea?

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

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

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

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

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

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

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

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

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

terenzreignz
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Well, you get the idea :P

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

terenzreignz
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Yep, I understand that. move on

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

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Right? Their magnitudes are equal to 1.

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

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

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

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

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1My brain explodes when things get too complicated

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Let's get its magnitude... \[\Large \vec u  \vec v \]

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1so we have to show it is indeed\[\sqrt{2}\]

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

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

terenzreignz
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Yes， I understand so far

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Are you okay with the rearrangement? Something you didn't get? :)

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1and the next three stuff becomes 1 as well?

terenzreignz
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1See anything familiar...?

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

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Oh my goodness! why am I so STUPID?!

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

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

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

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1LOL JK have a nice day :3
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.