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Zale101
 one year ago
Show that if v1, v2, and v3 are mutually orthogonal nonzero vectors in 3space, and if a vector v in 3space is expressed as: v=c1v1+c1v2+c3v3
Zale101
 one year ago
Show that if v1, v2, and v3 are mutually orthogonal nonzero vectors in 3space, and if a vector v in 3space is expressed as: v=c1v1+c1v2+c3v3

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Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Then the scalars c1, c2, and c3 are given by the formula: \(\Large c_i=\LARGE \frac{(v*v_i)}{V_i^2}\)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1v=c1v1+c1v2+c3v3 What happens if you take the scalar product with v1 on both sides?

phi
 one year ago
Best ResponseYou've already chosen the best response.0i.e. what is the dot product of two orthogonal vectors?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441576756278:dw oo this drawing tool is so cool!!

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Hi! Sorry for the late reply! The dot product of the two orthogonal vectors is zero?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1yes. So take the scalar product of v=c1v1+c1v2+c3v3 and simplify.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1So take the scalar product of v=c1v1+c1v2+c3v3 *with v1* and simplify.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Zale are you there?

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Yes, i'm sorry i dont understand what you meant :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that got a bit confusing @beginnersmind

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Ok. I'll try to exlain: We were given v = c1v1+c2v2+c3v3. Both sides of the equation represent the same vector (v) written in different ways. So if we take the scalar product of v with v1 and c1v1+c2v2+c3v3 with v1 we get the same number. I.e v*v1 = ( c1v1+c2v2+c3v3)*v1 Clear so far?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1If not, can you tell me exactly which point is confusing?

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Makes sense now, i got confused when you said "take the scale product" and i couldn't know which vectors to deal with.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0so then the dot product would be v*v1?

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Also, we can choose to simplify with either v2, v3 or v1?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1v*v1 on the left hand side. right hand side is (c1v1 + c2v2 +c3v3)*v1 = c1(v1*v1) + c2(v2*v1) + c3(v3*v1)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Okay. I'm following along.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0When it say i=1,2,3 what does that mean?

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry for asking too many questions lol

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1It means that \[\Large c_i=\LARGE \frac{(v*v_i)}{v_i^2}\] is actually 3 equations. \[\Large c_1=\LARGE \frac{(v*v_1)}{v_1^2}\] \[\Large c_2=\LARGE \frac{(v*v_2)}{v_2^2}\] \[\Large c_3=\LARGE \frac{(v*v_3)}{v_3^2}\]

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0OH! That makes sense!! Thank you

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Nah, it's cool. I'd rather you ask if you don't understand something :)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Now the question makes a lot sense now.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0So the simplifications you did above when you used v*v1 can also be done with the rest: v*v2, v*v3 I though i'd have to do something different to get i.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @beginnersmind :)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1No problem. Just to clarify though, do you understand how we get \[\Large c_1=\LARGE \frac{(v*v_1)}{v_1^2}\] from \[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1\]

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0No, because i dont understand how you got v2, v3? when it only had v and vc in c1=(v*v1)/(v)^2

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0when it only had v and v1*

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1I started with \[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1 \] and got \[\Large c_1=\LARGE \frac{(v*v_1)}{v_1^2}\]

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Yeah it makes sense now.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1So the question shouldn't be how I got v2,v3, but where did they go? :)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0I dont know if i can do this because it's different from your approach. Where it says C2=(v*v2)/(v^2) is it possible for me to use v=c1v1+c2v2+c3v3 and for v2, i solve for v2 by using v=c1v1+c2v2+c3v3 ?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Ok, I'll write it out properly then: We start with \(v= (c_1v_1+c_2v_2+c_3v_3)\) (this was given) \(v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1\) \(v*v1= c_1(v_1*v_1)+c_2(v_2*v_1)+c_3(v_3*v_1)\) now we note that \((v_1*v_2) = 0\) and \((v_1*v_3) = 0\) because v1 is orthogonal to v2 and v3. This was the key step. So \(v*v1= c_1(v_1*v_1)\) remembering that a vector's scalar product with itself is the square of its length (v1*v1 = v1^2): \[\Large v*v_1=c_1v_1^2\] or \[\Large c_1=\LARGE \frac{(v*v_1)}{v_1^2}\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1For c2 you do the same thing, except now you take the scalar product with v2. This gets rid of the terms with v1 and v3.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0I had to keep in my that it states that the orthogonal of v1*v2 is zero. Alright, thanks for all the beautiful and detailed explanation.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1No problem, thanks for sticking with it :)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you should use \cdot. it looks nicer and also use \ v_1\ for \(\v_1\\) \[\Huge c_1=\LARGE \frac{v\cdot v_1}{\v_1\^2}\] vs \[\Huge c_1=\LARGE \frac{v*v_1}{v_1^2}\]
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