## anonymous one year ago Linear Algebra.... Need help with #6 part III.... Please see attachment

1. anonymous

2. anonymous

the inner product is bilinear, so $$\langle a+b,c+d\rangle=\langle a+b,c\rangle+\langle a+b,d\rangle=\langle a,c\rangle +\langle b,c\rangle+\langle a,d\rangle+\langle b,d\rangle$$

3. anonymous

I am about to upload what I have done... don;t know where to go from there

4. anonymous

5. anonymous

so here we have $$\langle 2u+3v-w,u-v+w\rangle\\\quad =\langle 2u,u\rangle+\langle2u,-v\rangle+\langle 2u,w\rangle\\\qquad +\langle 3v,u\rangle+\langle3v,-v\rangle+\langle 3v,w\rangle\\\qquad+\langle -w,u\rangle+\langle-w,-v\rangle+\langle -w,w\rangle$$

6. anonymous

now we're told that $$u,v,w$$ are unit vectors, so $$\|u\|=\|v\|=\|w\|=1$$ and we're also told that they're each separated from one another by an angle of $$\pi/4$$ in space

7. anonymous

I think your looking at the wrong problem.

8. anonymous

I have that one finished

9. anonymous

I am working on part III

10. anonymous

remember we have that $$\langle x,x\rangle=\|x\|^2$$ and more generally $$\langle x,y\rangle =\|x\|\|y\|\cos\theta$$ where $$\theta$$ is the angle between $$x,y$$, so now evaluate all the inner products

11. anonymous

oh, oops -- I misread the title and thought it said part II

12. anonymous

Lol No big deal, could you look at my work?

13. anonymous

I also eliminated all the inner products that were not of the same vector because they are orthogonal

14. anonymous

$\sqrt{<v,v>+<u,u>+<w,w>}$

15. anonymous

That is what I am left with but I don't know what to do with that because $\sqrt{4}+\sqrt{2}\neq \sqrt{4+2}$

16. anonymous

for part III, consider we have $$\langle x,x\rangle=\|x\|^2$$so it follows that $$\|u+v+w\|^2=\langle u+v+w,u+v+w\rangle\\\quad=\langle u,u\rangle+\langle v,v\rangle+\langle w,w\rangle+2\langle u,v\rangle+2\langle u,w\rangle+2\langle v,w\rangle$$now since we said $$u,v,w$$ are all mutually orthogonal it means those cross terms $$\langle u,v\rangle=\langle u,w\rangle=\langle v,w\rangle=0$$ so: $$\|u+v+w\|^2=\|u\|^2+\|v\|^2+\|w\|^2\\\implies \|u+v+w\|\implies\sqrt{\|u\|^2+\|v\|^2+\|w\|^2}$$ indeed

17. anonymous

So I can't just say that its the magnitude of one plus the magnitude of the other etc.

18. anonymous

now $$\|u\|=\|v\|=2$$ and $$\|w\|=3$$ gives us $$\|u+v+w\|=\sqrt{2^2+2^2+3^2}=\sqrt{17}$$

19. anonymous

I am confused

20. anonymous

indeed, you can't (in general) say that the magnitude of the sum is the sum of the magnitudes

21. anonymous

I am trying to explain my confusion but I am not as familiar with the equation tool as you are so give me one second and I will try to show you where I am confused

22. anonymous

Real quick how do you create the brackets that go around the inner product?

23. anonymous

\langle and \rangle

24. anonymous

$$\text{\langle}\implies\langle\\\text{\rangle}\implies\rangle$$

25. anonymous

$\sqrt{<v,v>+<u,u>+<w,w>}\neq \sqrt{<v,v>}+\sqrt{<w,w>}+\sqrt{<u,u>}$

26. anonymous

But in your explanation you made $<v,v>=\left| v \right|$

27. anonymous

indeed, that is not true

28. anonymous

I meant for that absolute v to have double bars.....

29. anonymous

yep, by definition $$\|v\|^2=\langle v,v\rangle$$ so $$\langle u,u\rangle=2^2=4$$, $$\langle v,v\rangle =2^2=4$$ and $$\langle w,w\rangle=3^2=9$$

30. anonymous

but the definition in my book shows that $\left| \right|v \left| \right|=\sqrt{<v,v>}$

31. anonymous

Oh I see it now, you took the square of both sides to arrive at what you had. I didn't make that connection. THank you