Question

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

Answer 1

Then the scalars c1, c2, and c3 are given by the formula: \(\Large c_i=\LARGE \frac{(v*v_i)}{||V_i||^2}\)

Answer 2

v=c1v1+c1v2+c3v3

What happens if you take the scalar product with v1 on both sides?

Answer 3

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

Answer 4

oo this drawing tool is so cool!!

Answer 5

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

Answer 6

yes

Answer 7

So v1*v2=0?

Answer 8

yes. So take the scalar product of

v=c1v1+c1v2+c3v3 and simplify.

Answer 9

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

Answer 10

You mean c1?

Answer 11

Zale are you there?

Answer 12

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

Answer 13

yeah that got a bit confusing @beginnersmind

Answer 14

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?

Answer 15

Yep!

Answer 16

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

Answer 17

Makes sense now, i got confused when you said "take the scale product" and i couldn't know which vectors to deal with.

Answer 18

so then the dot product would be v*v1?

Answer 19

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

Answer 20

v*v1 on the left hand side.
right hand side is

(c1v1 + c2v2 +c3v3)*v1 = c1(v1*v1) + c2(v2*v1) + c3(v3*v1)

Answer 21

Okay. I'm following along.

Answer 22

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

Answer 23

i'm sorry for asking too many questions lol

Answer 24

It means that

\[\Large c_i=\LARGE \frac{(v*v_i)}{||v_i||^2}\]
is actually 3 equations.

\[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]
\[\Large c_2=\LARGE \frac{(v*v_2)}{||v_2||^2}\]
\[\Large c_3=\LARGE \frac{(v*v_3)}{||v_3||^2}\]

Answer 25

OH! That makes sense!! Thank you

Answer 26

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

Answer 27

Now the question makes a lot sense now.

Answer 28

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.

Answer 29

Thanks @beginnersmind :)

Answer 30

No problem. Just to clarify though, do you understand how we get

\[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]

from
\[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1\]

Answer 31

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

Answer 32

when it only had v and v1*

Answer 33

I started with
\[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1 \]

and got

\[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]

Answer 34

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

Answer 35

Oh okay.

Answer 36

Yeah it makes sense now.

Answer 37

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

Answer 38

Yes

Answer 39

lol

Answer 40

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 ?

Answer 41

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

\[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]

Answer 42

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.

Answer 43

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.

Answer 44

No problem, thanks for sticking with it :)

Answer 45

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

\[\Huge c_1=\LARGE \frac{v*v_1}{||v_1||^2}\]