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

- Zale101

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- 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

v=c1v1+c1v2+c3v3
What happens if you take the scalar product with v1 on both sides?

- phi

i.e. what is the dot product of two orthogonal vectors?

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## More answers

- Jhannybean

|dw:1441576756278:dw| oo this drawing tool is so cool!!

- Zale101

Hi! Sorry for the late reply! The dot product of the two orthogonal vectors is zero?

- Zarkon

yes

- Zale101

So v1*v2=0?

- beginnersmind

yes. So take the scalar product of
v=c1v1+c1v2+c3v3 and simplify.

- beginnersmind

So take the scalar product of v=c1v1+c1v2+c3v3 *with v1* and simplify.

- beginnersmind

You mean c1?

- beginnersmind

Zale are you there?

- Zale101

Yes, i'm sorry i dont understand what you meant :(

- Jhannybean

yeah that got a bit confusing @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

Yep!

- beginnersmind

If not, can you tell me exactly which point is confusing?

- 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

so then the dot product would be v*v1?

- Zale101

Also, we can choose to simplify with either v2, v3 or v1?

- 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

Okay. I'm following along.

- Zale101

When it say i=1,2,3 what does that mean?

- Zale101

i'm sorry for asking too many questions lol

- 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

OH! That makes sense!! Thank you

- beginnersmind

Nah, it's cool. I'd rather you ask if you don't understand something :)

- Zale101

Now the question makes a lot sense now.

- 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

Thanks @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

No, because i dont understand how you got v2, v3? when it only had v and vc in c1=(v*v1)/(||v||)^2

- Zale101

when it only had v and v1*

- 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

c1||v||^2=v*v1

- Zale101

Oh okay.

- Zale101

Yeah it makes sense now.

- beginnersmind

So the question shouldn't be how I got v2,v3, but where did they go? :)

- Zale101

Yes

- Zale101

lol

- 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

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

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

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

No problem, thanks for sticking with it :)

- 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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