## Zale101 one year ago 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

1. Zale101

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

2. beginnersmind

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

3. phi

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

4. Jhannybean

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

5. Zale101

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

6. Zarkon

yes

7. Zale101

So v1*v2=0?

8. beginnersmind

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

9. beginnersmind

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

10. beginnersmind

You mean c1?

11. beginnersmind

Zale are you there?

12. Zale101

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

13. Jhannybean

yeah that got a bit confusing @beginnersmind

14. 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?

15. Jhannybean

Yep!

16. beginnersmind

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

17. Zale101

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

18. Zale101

so then the dot product would be v*v1?

19. Zale101

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

20. 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)

21. Zale101

Okay. I'm following along.

22. Zale101

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

23. Zale101

i'm sorry for asking too many questions lol

24. beginnersmind

It means that $\Large c_i=\LARGE \frac{(v*v_i)}{||v_i||^2}$ is actually 3 equations. $\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}$ $\Large c_2=\LARGE \frac{(v*v_2)}{||v_2||^2}$ $\Large c_3=\LARGE \frac{(v*v_3)}{||v_3||^2}$

25. Zale101

OH! That makes sense!! Thank you

26. beginnersmind

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

27. Zale101

Now the question makes a lot sense now.

28. 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.

29. Zale101

Thanks @beginnersmind :)

30. 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 $v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1$

31. Zale101

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

32. Zale101

when it only had v and v1*

33. beginnersmind

I started with $v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1$ and got $\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}$

34. Zale101

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

35. Zale101

Oh okay.

36. Zale101

Yeah it makes sense now.

37. beginnersmind

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

38. Zale101

Yes

39. Zale101

lol

40. 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 ?

41. 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 $\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}$

42. 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.

43. 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.

44. beginnersmind

No problem, thanks for sticking with it :)

45. 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 $\Huge c_1=\LARGE \frac{v*v_1}{||v_1||^2}$