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

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- caozeyuan

(c) if u and v are perpendicular unit vectors then \[\left| u-v \right|\]=\[\sqrt{2}\]

- caozeyuan

@terenzreignz , hey the 99 guy there! do you have any idea?

- terenzreignz

Which one are you currently working on?

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## More answers

- caozeyuan

I'm working on b. (a) solved

- terenzreignz

My gut feel says no...

- caozeyuan

counterexample needed then

- terenzreignz

What if u is not perpendicular to w?

- caozeyuan

but the question said that!

- terenzreignz

Wait nvm... misread it... recalculating..

- terenzreignz

okay, it's true, then :P

- terenzreignz

What's a convenient way to find if two vectors are perpendicular?

- caozeyuan

u.v=0! ofcuz!

- terenzreignz

Here's theconvenient truth... if u is perpendicular to v, then... \[\Large \vec u \cdot \vec v = 0\] Yup

- terenzreignz

Same goes for w. What then bodes for... \[\Large \vec u \cdot(\vec v + 2\vec w)\] :P

- terenzreignz

Well, you get the idea :P

- terenzreignz

If you don't (you probably do, but anyway...) remember that dot product distributes over addition...

- caozeyuan

so how about (c) . I think it should be True, but I cannot give a reason

- terenzreignz

(c) doesn't make sense, I think you didn't copy it correctly :)

- terenzreignz

Oh, I saw it just now... derp :3

- terenzreignz

Tough one :D

- terenzreignz

Okay, got it :P

- terenzreignz

Sorry it took so long, I didn't feel like getting a piece of paper XD

- terenzreignz

Let's let \[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_u \vec k \]

- terenzreignz

crud, I meant \[\Large \vec v= x_v\vec i + y_v\vec j + z_{\color{red}v}\vec k\]

- caozeyuan

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)

- terenzreignz

Yes, but that's a specific case... you're supposed to prove it for *any* unit vector...

- terenzreignz

Things are not that simple, but they are... rather elegant... Let's work out the magic, shall we? :)

- terenzreignz

\[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k\] Just tell me when you're ready .. ;)

- caozeyuan

Yep, I understand that. move on

- terenzreignz

Someone's bossy :/

- terenzreignz

Okay, we know u and v are unit vectors, which can only mean... \[\Large x_u^2 + y_u^2 + z_u^2 = 1 = x_v^2 + y_v^2+z_v^2\]

- terenzreignz

Right? Their magnitudes are equal to 1.

- caozeyuan

so that gives sqrt(1^2+1^2)=sqrt(2), right?

- terenzreignz

How did you figure that? -.-

- terenzreignz

I like your initiative, but patience... is a virtue :)

- terenzreignz

And no, your reasoning is faulty ;)

- terenzreignz

So, why don't we acquire u - v... \[\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k\]

- terenzreignz

Catch me so far?

- caozeyuan

My brain explodes when things get too complicated

- terenzreignz

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 \[\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k\]so far... do you?

- caozeyuan

Yes, I'm following

- terenzreignz

Let's get its magnitude... \[\Large ||\vec u - \vec v ||\]

- terenzreignz

By definition, this is \[\Large =\sqrt{(x_u-x_v)^2+(y_u-y_v)^2+(z_u-z_v)^2}\]

- caozeyuan

so we have to show it is indeed\[\sqrt{2}\]

- terenzreignz

Yup :)

- terenzreignz

Ready? The tricky part begins now...

- caozeyuan

oh NO! Don't torture me！

- terenzreignz

This is for your own goodNow, let's simplify them squares... \[\Large = \sqrt{x_u^2-2x_ux_v+x_v^2+y_u^2-2y_uy_v+y_v^2+z_u^2-2z_uz_v+z_v^2}\]

- terenzreignz

Everything good so far? I only simplified the squares... you know... \[\large (a-b)^2 = a^2-2ab+b^2\]

- caozeyuan

Yes， I understand so far

- terenzreignz

I think, I shall rearrange the terms... \[\Large = \sqrt{x_u^2+y_u^2+z_u^2+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\]

- terenzreignz

Are you okay with the rearrangement? Something you didn't get? :)

- caozeyuan

Get it w/o prob.

- terenzreignz

Okay, by hypothesis, since u is a unit vector, this part \[\Large = \sqrt{\color{red}{x_u^2+y_u^2+z_u^2}+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\] Is just \[\Large = \sqrt{\color{red}1+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\] Are you following me so far? :)

- caozeyuan

and the next three stuff becomes 1 as well?

- terenzreignz

That is correct, since v is also a unit vector... \[\Large = \sqrt{\color{red}1+\color{green}1-2x_ux_v-2y_uy_v-2z_uz_v}\] \[\Large = \sqrt{\color{blue}2-2x_ux_v-2y_uy_v-2z_uz_v}\] Are we done here? :P

- terenzreignz

Nah, of course not :) Allow me to factor out the -2 here... \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\]

- terenzreignz

See anything familiar...?

- caozeyuan

no, I don't see any familiar

- terenzreignz

You're lacking creativity :P \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\] \[\Large = \sqrt{2-2\color{blue}{(\vec u \cdot \vec v)}}\]

- terenzreignz

"aha!" moment?

- caozeyuan

Oh my goodness! why am I so STUPID?!

- terenzreignz

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 \[\Large = \sqrt{2-2\color{blue}{(0)}}= \sqrt2\] tadaa

- caozeyuan

ok, QED.

- terenzreignz

Now say it... Say Terence is awesome :3

- caozeyuan

NO! you are arrogant! LOL!

- terenzreignz

Arrogant is not the opposite of awesome. Now say it >:)

- terenzreignz

LOL JK have a nice day :3

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