Zale101
  • Zale101
Show that if v1, v2, and v3 are mutually orthogonal nonzero vectors in 3-space, and if a vector v in 3-space is expressed as: v=c1v1+c1v2+c3v3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Zale101
  • Zale101
Then the scalars c1, c2, and c3 are given by the formula: \(\Large c_i=\LARGE \frac{(v*v_i)}{||V_i||^2}\)
beginnersmind
  • beginnersmind
v=c1v1+c1v2+c3v3 What happens if you take the scalar product with v1 on both sides?
phi
  • phi
i.e. what is the dot product of two orthogonal vectors?

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Jhannybean
  • Jhannybean
|dw:1441576756278:dw| oo this drawing tool is so cool!!
Zale101
  • Zale101
Hi! Sorry for the late reply! The dot product of the two orthogonal vectors is zero?
Zarkon
  • Zarkon
yes
Zale101
  • Zale101
So v1*v2=0?
beginnersmind
  • beginnersmind
yes. So take the scalar product of v=c1v1+c1v2+c3v3 and simplify.
beginnersmind
  • beginnersmind
So take the scalar product of v=c1v1+c1v2+c3v3 *with v1* and simplify.
beginnersmind
  • beginnersmind
You mean c1?
beginnersmind
  • beginnersmind
Zale are you there?
Zale101
  • Zale101
Yes, i'm sorry i dont understand what you meant :(
Jhannybean
  • Jhannybean
yeah that got a bit confusing @beginnersmind
beginnersmind
  • beginnersmind
Ok. 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?
Jhannybean
  • Jhannybean
Yep!
beginnersmind
  • beginnersmind
If not, can you tell me exactly which point is confusing?
Zale101
  • Zale101
Makes sense now, i got confused when you said "take the scale product" and i couldn't know which vectors to deal with.
Zale101
  • Zale101
so then the dot product would be v*v1?
Zale101
  • Zale101
Also, we can choose to simplify with either v2, v3 or v1?
beginnersmind
  • beginnersmind
v*v1 on the left hand side. right hand side is (c1v1 + c2v2 +c3v3)*v1 = c1(v1*v1) + c2(v2*v1) + c3(v3*v1)
Zale101
  • Zale101
Okay. I'm following along.
Zale101
  • Zale101
When it say i=1,2,3 what does that mean?
Zale101
  • Zale101
i'm sorry for asking too many questions lol
beginnersmind
  • beginnersmind
It 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
  • Zale101
OH! That makes sense!! Thank you
beginnersmind
  • beginnersmind
Nah, it's cool. I'd rather you ask if you don't understand something :)
Zale101
  • Zale101
Now the question makes a lot sense now.
Zale101
  • Zale101
So 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
  • Zale101
Thanks @beginnersmind :)
beginnersmind
  • beginnersmind
No 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
  • Zale101
No, because i dont understand how you got v2, v3? when it only had v and vc in c1=(v*v1)/(||v||)^2
Zale101
  • Zale101
when it only had v and v1*
beginnersmind
  • beginnersmind
I 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
  • Zale101
c1||v||^2=v*v1
Zale101
  • Zale101
Oh okay.
Zale101
  • Zale101
Yeah it makes sense now.
beginnersmind
  • beginnersmind
So the question shouldn't be how I got v2,v3, but where did they go? :)
Zale101
  • Zale101
Yes
Zale101
  • Zale101
lol
Zale101
  • Zale101
I 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
  • beginnersmind
Ok, 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): \[\Large v*v_1=c_1||v_1||^2\] or \[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]
beginnersmind
  • beginnersmind
For 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
  • Zale101
I 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
  • beginnersmind
No problem, thanks for sticking with it :)
Zarkon
  • Zarkon
you 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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