Question

what can we say about 2 non-zero vector u and v , that satisfy ||u +v|| = ||u|| + ||v|| .

Answer 1

||u +v|| = ||u|| + ||v|| + 2 ||u|| ||v|| cos theta, where theta = angle between the vector u and v.
If u and v are both non-zero vectors, that means cos theta is 0. At what angle will cos (theta) be zero?!

Answer 2

sorry , I'm not quite sure if this is what you need

Answer 3

please leave those numbers alone, we don't need them

Answer 4

Hmm.. Why is it so complicated :S

Answer 5

@Callisto , I try to figure out your formula in text book, but I cannot see. would you please tell me the name of it? or theorem or some thing about that

Answer 6

My bad!!! I mixed up something!! Sorry!! Ignore my comments!!

Answer 7

were you writing : u dot v=|u| x|v|x cos theta?

Answer 8

@ranxu6j3 I see your pic, thanks for reply but the logic is not stable.how can you get U1 +V1 become U1,U2? or I misunderstand something from yours? if so, please, correct me

Answer 9

thanks for letting me know your confusion,
I was taking the 2 vectors apart into verticality and parallel
then I can proof the 2 sides of = are the same

Answer 10

@ranxu6j3, we don't know whether
\[u,v\in\mathbb{R}^2\]

Answer 11

@ranxu6j3 you mean \(\vec u\) parallel to \(\vec v\)?

Answer 12

@ranxu6j3 even though we have they are parallel, we must base on some thing to prove, right? some formula or concept or something else...right?

Answer 13

do we? sorry , I thought with para and vert we can proof that?

Answer 14

sorry I didn't see your last reply, I didn't mean v and u are para

Answer 15

@SithsAndGiggles , can you explain to me what does R^2 means? tks

Answer 16

@ranxu6j3 the proof must be strong and logic, not intuitive or "trivially"

Answer 17

Triangle inequality? We know that, in general,
\[||u+v||\le ||u||+||v||.\]
I think some manipulation of this will help.

@ranxu6j3,
\[\mathbb{R}^2=\mathbb{R}\times\mathbb{R}, \text{ defined by }\\
\mathbb{R}^2:=\left\{(x,y):x,y\in\mathbb{R}\right\}\]
(the x stands for Cartesian product)

Answer 18

sorry I am not a native speaker so that I might misunderstand you guys when Math in English. I apologize for making you confused, but i'll try.
could you tell me that should we try to proof the "="(since I am not sure what exactly you want)

Answer 19

thanks @SithsAndGiggles , i got your meaning

Answer 20

My point is that, although your logic/intuition may work for R², it may not necessarily work for higher dimensions or other vector spaces.

Answer 21

@ranxu6j3 means = (equal)
@SithsAndGiggles that is inequality formula, but it doesn't help because it has no condition to get extremely = or extremely <

Answer 22

yeah, i got it , @SithsAndGiggles , so do we have any limitation or condition in this case, @Hoa ?

Answer 23

nope

Answer 24

does the case says we are in 2d ?

Answer 25

I post exactly what the book print

Answer 26

nope

Answer 27

oh that's why my pic doesn't work

Answer 28

Okay, I'm pretty sure I know what the answer is, but I'm not sure how to formally obtain it through a proof...

Answer 29

@SithsAndGiggles write it out, others can contribute

Answer 30

If u=v, then the equality holds:\[||u+v||=||u+u||=2||u||=||u||+||u||\]

Answer 31

I'll see if I can write something up...

Answer 32

I think it's just a fact that comes from the triangle inequality; the equality holds when the given vectors are equal, and doesn't when they're not.

Answer 33

everybody, what do you think?

Answer 34

my reply was met a problem , if , for example : u=(1,2) v=(1,-1) , then those i thought would be broken , and i am not quite sure how do we proof it with triangle inequality

Answer 35

The fact does come from the triangle inequality. However, the vectors do not need to be equal. They just need to be linearly dependent (or collinear). I think I have a proof for it, give me a minute to check it though.

Answer 36

i have an idea, everybody check whether it is right or not\[||\vec u +\vec v|| = \sqrt{(a+d)^2 +(b+e)^2 +(c+f)^2}where (\vec u = (a,b,c) and \vec v =(d,e,f)\]

Answer 37

@Hoa, if you are right, you'd only be making the case for R³, and not for a general vector space.

Answer 38

||\(\vec u\)||+||\(\vec v\)|| = \[\sqrt{a^2 + b^2 +c^2 } + \sqrt{d^2 +e^2 +f^2}\]

Answer 39

i think this is the same as what i wrote on my pic, but this time in 3D

Answer 40

make them equal and find out the condition to get that become true

Answer 41

what do you think?

Answer 42

I understand what you wrote, but it must be mathematically proved, right?

Answer 43

yeah so it's good that you make it into math
tks, and I think what you wrote can be used

Answer 44

i can finally get back here, but i can't see those you post
still probs, sorry for not continuing, i'll try later
@hoa

Answer 45

||u +v|| = ||u|| + ||v||
Square both sides
||u +v||^2 = (||u|| + ||v|| )^2
||u +v||^2 = ||u||^2 + ||v||^2 + 2 ||u|| ||v||
Consider the left side
||u +v||^2 = \((u+v)\cdot (u+v)\)
= \(||u||^2 +||v||^2 + 2u\cdot v\)
= ||u||^2 + ||v||^2 + 2 ||u|| ||v|| cos theta
So, we get
\(||u||^2 + ||v||^2 + 2 ||u||\ ||v|| cos \theta = ||u||^2 + ||v||^2 + 2 ||u|| \ ||v||\)
Cancel the common terms:
\(2 ||u||\ ||v|| cos \theta =2 ||u|| \ ||v||\)
Cancel the common factors:
\[cos \theta =1\]
Wow!! This tells you something, doesn't it?

Answer 46

by using geometry , I have 2 results, \(\vec u\)parallel \(\vec v\) either same direction and opposite , I still have the same result since they are ||...|| , is it right?

Answer 47

I draw 3 circles with the same radius and let them cut each other, like

Answer 48

If the direction doesn't matter, why not just write "u and v are parallel to each other"??

Answer 49

but if they don't have the same initial point, how the result comes true? I mean if they // but overlap a while

Answer 50

oh oh, my bad, you are right

Answer 51

the length. thanks a lot

Answer 52

Welcome :)
Sorry for my mistakes earlier!!

Answer 53

@Callisto thanks for your idea.