# Given vectors v and w, prove |v*w| =< |v|*|w|

1. Feb 5, 2009

### JoeTrumpet

1. The problem statement, all variables and given/known data
Given vectors v and w, show that |v*w| =< |v|*|w|

2. Relevant equations
I know that |v| = sqrt((x1)^2+(x2)^2+...+(xn)^2)
and that |w| = sqrt((y1)^2+(y2)^2+...+(yn)^2)

also |v*w| = x1y1 + x2y2 + ... + xnyn

We were also told
Hint: use |v+c*w|^2
and (v+c*w)(v+c*w)

3. The attempt at a solution
I am not at all sure what to do. I distributed the second expression in the hint to get
w^2 + 2vwc + c^2*v^2, but that doesn't tell me much.

Thanks!

2. Feb 5, 2009

### Dick

(v+cw).(v+cw)>=0, right? Now pick a special value of c=(-v.w)/(w.w). It's basically a trick. It's the value of c that extremizes (v+cw).(v+cw) as a function of c.

3. Feb 5, 2009

### JoeTrumpet

How come (v+cw).(v+cw) >=0? And after finding that value of c, I'm still not sure how that relates back to the original question. Sorry if I'm missing something obvious!

4. Feb 5, 2009

### Dick

The dot product of any vector with itself is >=0. Look at your formula, it's a sum of squares. Put the special value of c in and expand it out.

5. Feb 5, 2009

### JoeTrumpet

Hm, I think I get it. (Vectors are fairly new to me so forgive me if I misunderstand and misuse these operations.)

knowing c = (v.w)/(w.w)

v.v - 2*(v.w)*c + (w.w)*c^2 >= 0

a vector times itself must be the magnitude of itself squared, right? So knowing that and plugging in c gets me

|v|^2 - 2(v.w)(v.w)/(w.w) + (|w|^2)*(v.w)(v.w)/((w.w)(w.w)) >= 0

multiplying everything by |w|^2

|w|^2*|v|^2 - 2(v.w)(v.w) + (v.w)(v.w) >= 0
and
|w|^2*|v|^2 >= (v.w)(v.w)

and I suppose I can take the square root of both sides to get

|w|*|v| >= (v.w)

Is this correct?

6. Feb 5, 2009

### Dick

Yes, but what you actually get after the square root is |w||v|>=|v.w|. When you take square roots of squares you get absolute values. Good job.

7. Feb 5, 2009

### JoeTrumpet

Ah, that makes sense. Thank you very much!

8. Feb 5, 2009

### Dick

No problem. The inequality is called Cauchy-Schwarz, if you didn't already know that.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Given vectors v and w, prove |v*w| =< |v|*|w|

1. Feb 5, 2009

### JoeTrumpet

1. The problem statement, all variables and given/known data
Given vectors v and w, show that |v*w| =< |v|*|w|

2. Relevant equations
I know that |v| = sqrt((x1)^2+(x2)^2+...+(xn)^2)
and that |w| = sqrt((y1)^2+(y2)^2+...+(yn)^2)

also |v*w| = x1y1 + x2y2 + ... + xnyn

We were also told
Hint: use |v+c*w|^2
and (v+c*w)(v+c*w)

3. The attempt at a solution
I am not at all sure what to do. I distributed the second expression in the hint to get
w^2 + 2vwc + c^2*v^2, but that doesn't tell me much.

Thanks!

2. Feb 5, 2009

### Dick

(v+cw).(v+cw)>=0, right? Now pick a special value of c=(-v.w)/(w.w). It's basically a trick. It's the value of c that extremizes (v+cw).(v+cw) as a function of c.

3. Feb 5, 2009

### JoeTrumpet

How come (v+cw).(v+cw) >=0? And after finding that value of c, I'm still not sure how that relates back to the original question. Sorry if I'm missing something obvious!

4. Feb 5, 2009

### Dick

The dot product of any vector with itself is >=0. Look at your formula, it's a sum of squares. Put the special value of c in and expand it out.

5. Feb 5, 2009

### JoeTrumpet

Hm, I think I get it. (Vectors are fairly new to me so forgive me if I misunderstand and misuse these operations.)

knowing c = (v.w)/(w.w)

v.v - 2*(v.w)*c + (w.w)*c^2 >= 0

a vector times itself must be the magnitude of itself squared, right? So knowing that and plugging in c gets me

|v|^2 - 2(v.w)(v.w)/(w.w) + (|w|^2)*(v.w)(v.w)/((w.w)(w.w)) >= 0

multiplying everything by |w|^2

|w|^2*|v|^2 - 2(v.w)(v.w) + (v.w)(v.w) >= 0
and
|w|^2*|v|^2 >= (v.w)(v.w)

and I suppose I can take the square root of both sides to get

|w|*|v| >= (v.w)

Is this correct?

6. Feb 5, 2009

### Dick

Yes, but what you actually get after the square root is |w||v|>=|v.w|. When you take square roots of squares you get absolute values. Good job.

7. Feb 5, 2009

### JoeTrumpet

Ah, that makes sense. Thank you very much!

8. Feb 5, 2009

### Dick

No problem. The inequality is called Cauchy-Schwarz, if you didn't already know that.
user
The text is a discussion thread from a forum where a student is asking for help with a proof involving vectors and the Cauchy-Schwarz inequality. The student provides the problem statement, relevant equations, and an attempt at a solution, and receives guidance from another user.

TYPE: NO_QUESTION

However, it's important to note that while the text does not contain a direct mathematical question in the form of a query or problem to solve, it revolves around a mathematical proof. The structure of the content is more of a dialogue or discussion rather than a question-answer format. If we consider the proof itself as a question, then the type would be:

TYPE: SINGLE_QUESTION_WITH_ANSWER

This is because the problem statement is provided, and through the dialogue, the solution is derived and confirmed.