Question

Linear Algebra....
Need help with #6 part III.... Please see attachment

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

I think your looking at the wrong problem.

Answer 6

I have that one finished

Answer 7

I am working on part III

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

\[\sqrt{<v,v>+<u,u>+<w,w>}\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

I am confused

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

\langle and \rangle

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

indeed, that is not true

Answer 26

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

Answer 27

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 \(\langle w,w\rangle=3^2=9\)

Answer 28

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

Answer 29

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