Dot product, cross product

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i need help with the following:
note that the big dots represents the dot product


1. suppose that [tex]a \bullet b = c \bullet b [/tex] for all vectors [tex]\overrightarrow{b} [/tex]. show that [tex]\overrightarrow{a} = \overrightarrow{c} [/tex].

i suppose i can't simply divide out the b, right? anyway, i tried writing out the components of each vector - for example a=(a1, a2, a3), b=(b1, b2, b3), c=(c1, c2, c3). i got as far as:

(b1, b2, b3) [tex]\bullet[/tex](a1-c1, a2-c2, a3-c3) = 0

but i don't know how to simplify this further to eliminate the b. any help is appreciated.
 

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  • #2
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cross product

my second question is:

2. suppose that [tex]\overrightarrow{a} X \overrightarrow{b} = \overrightarrow{c} X \overrightarrow{b} [/tex] for all vectors [tex]\overrightarrow{b}. [/tex] how are [tex]\overrightarrow{a} [/tex] and [tex]\overrightarrow{c} [/tex] related?.

i wrote the vectors in terms of its components, took the cross product, and i got as far as:

(b3(a2-c1) + b2(a3-c3), b1(a3-c3) + b3(a1-c1), b2(a1-c1) + b1(a2-c2)) = 0

again, i'm not sure how to simplify this further or what conclusion to draw about the relationship between a and c. any help is greatly appreciated.
 
  • #3
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For the first, did you know that [itex] \vec{a} \bullet \vec{b} = |a||b|cos(\theta) [/itex]? That should be all you need.

There is a similar identity for the cross product.
 
  • #4
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i have considered that, but the angles between a and b, and c and b, are different (presumably). so i end up with

|a| cos(t) = |c| cos (s)

...which doesn't seem to help very much.
 
  • #5
Curious3141
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For the dot product, here's a hint. Consider the cases where [tex]\vec{b}[/tex] is a unit vector first in the direction of [tex]\vec{a}[/tex] then in the direction of [tex]\vec{c}[/tex]. Draw up some equations and see what you can conclude about the angle between vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex], then what that implies about the relationship between their magnitudes.
 
  • #6
Curious3141
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Here's the solution for the dot product :

Let the angle between vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] be [tex]\theta[/tex].

Let [tex]\vec{b}[/tex] be a unit vector in the same direction as [tex]\vec{a}[/tex]

Then [tex]\vec{a} \bullet \vec{b} = |\vec{a}|[/tex]

[tex]\vec{c} \bullet \vec{b} = |\vec{c}|\cos\theta[/tex]

These two are equal, so we have :

[tex]|\vec{a}| = |\vec{c}|\cos\theta[/tex] --eqn(1)

Now let [tex]\vec{b}[/tex] be a unit vector in the direction of [tex]\vec{c}[/tex]

Working through it in the same way gives :

[tex]|\vec{c}| = |\vec{a}|\cos\theta[/tex] --eqn(2)

Solving those we get :

[tex]|\vec{c}|\cos^2{\theta} = |\vec{c}|[/tex]

Either [tex]|\vec{c}| = |\vec{a}| = 0[/tex], meaning both of them are null vectors (in which case they're equal in any case),

or [tex]\cos^2{\theta} = 1[/tex]

giving [tex]\cos{\theta} = 1[/tex] (the negative root is inadmissible because if you look at equations 1 and 2, the magnitudes are positive, so the cosine term can't be negative).

So we have [tex]\theta = 0[/tex], and putting that back into either equation 1 or 2, we get [tex]|\vec{a}| = |\vec{c}|[/tex].

So the two vectors [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] are of the same magnitude and oriented in the same direction. By definition, therefore, [tex]\vec{a} = \vec{c}[/tex] (QED)
 
  • #7
Galileo
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Write the equation in a different form:

[tex](\vec a -\vec c)\times \vec b=\vec 0[/tex]
for any vector [itex]\vec b[/itex].

(You can do the same thing with the dot product also).
What can you infer about [itex]\vec a-\vec c [/itex] from this?
 
  • #8
dextercioby
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The first exercise is incorrect.

[tex] \vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b} [/tex]

Using the distributivity,one gets

[tex] \left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0 [/tex]

Which has the infinite set of solutions:

[tex] \vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\} [/tex].

The "V" is a linear space and the [itex] \vec{d}[/itex] and [itex] \vec{b} [/itex] are arbitrary vectors.

So,please,think deeper before giving erroneous advice.

Daniel.
 
  • #9
Galileo
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The exercise is to show that [itex]\vec a \cdot \vec b=\vec c\cdot \vec b\forall \vec b \Rightarrow \vec a = \vec c[/itex].
This is clearly true, you just have to pick a clever value for b, say b=a-c
 
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  • #10
dextercioby
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That's not true.That difference is not zero,but an arbitrary vector.Zero vector is not arbitrary.

Daniel.
 
  • #11
Galileo
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Are you sure you have the same understanding of the problem as I do?
[itex]\vec b[/itex] can be any vector.
 
  • #12
dextercioby
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Yes.

HINT:[tex]\vec{c}\cdot\vec{b}=\vec{c}\cdot\left(\vec{b}+\vec{c}\times \vec{d}\right) [/tex]

,where \vec{d} is an arbitrary vector...:wink:

Daniel.
 
  • #13
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here is how I would handle the two dimentional case:

a * b = (ax + ay)(bx + by) = ax*bx + ay*by

c * b = (cx + cy)(bx + by) = cx*bx + cy*by

Two vectors are equal only if their corresponding components are equal. So if a * b = c *b then we have:

ax*bx = cx*bx

ay*by=cy*by

Which shows that a and c are equal.
 
  • #14
arildno
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You are wrong, Daniel.
[tex]\vec{a},\vec{c}[/tex] are fixed vectors, and the [tex]\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}[/tex] is supposed to hold for ALL [tex]\vec{b}[/tex]
This is equivalent to:
[tex](\vec{a}-\vec{c})\cdot\vec{b}=0[/tex]
In particular, this must hold for the choice [tex]\vec{b}=\vec{a}-\vec{c}[/tex] from which follows [tex]\vec{a}=\vec{c}[/tex]
 
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  • #15
dextercioby
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That's wrong

[tex] \vec{a}\cdot\vec{b} [/tex] is a scalar

and hence

[tex] a_{x}b_{x}+a_{y}b_{y}=c_{x}b_{x}+c_{y}b_{y} \nRightarrow \left\{\begin{array}{c}a_{x}b_{x}=c_{x}b_{x}\\a_{y}b_{y}=c_{y}b_{y} \end{array}\right [/tex]

Okay??

Daniel.
 
  • #16
dextercioby
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arildno said:
You are wrong, Daniel.
[tex]\vec{a},\vec{c}[/tex] are fixed vectors, and the [tex]\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}[/tex] is supposed to hold for ALL [tex]\vec{b}[/tex]
This is equivalent to:
[tex](\vec{a}-\vec{c})\cdot\vec{b}=0[/tex]
In particular, this must hold for the particular choice [tex]\vec{b}=\vec{a}-\vec{c}[/tex] from which follows [tex]\vec{a}=\vec{c}[/tex]
I'm not wrong,my solution does not exclude the 0 vector.That "d" is arbitrary,just as "b",so it can be the 0 vector.

Daniel.
 
  • #17
arildno
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dextercioby said:
The first exercise is incorrect.

[tex] \vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b} [/tex]

Using the distributivity,one gets

[tex] \left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0 [/tex]

Which has the infinite set of solutions:

[tex] \vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\} [/tex].

The "V" is a linear space and the [itex] \vec{d}[/itex] and [itex] \vec{b} [/itex] are arbitrary vectors.

So,please,think deeper before giving erroneous advice.

Daniel.
Whatever are you talking about???
[tex]\vec{a},\vec{c}[/tex] are FIXED vectors; their difference do not change if [tex]\vec{b}[/tex] changes.
 
  • #18
dextercioby
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Who said "a" and "c" are fixed?

Daniel.
 
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  • #19
arildno
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jeanf said:
i need help with the following:
note that the big dots represents the dot product


1. suppose that [tex]a \bullet b = c \bullet b [/tex] for all vectors [tex]\overrightarrow{b} [/tex]. show that [tex]\overrightarrow{a} = \overrightarrow{c} [/tex].
It says for all b here.
 
  • #20
dextercioby
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Yes,and my solution exploits the equality of the 2 scalar products.

Daniel.
 
  • #21
arildno
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No, it doesn't do any such thing.
For the particular choice [tex]\vec{b}=\vec{a}-\vec{c}[/tex]
we get
[tex](\vec{a}-\vec{c})^{2}=0[/tex]
which holds if and only if [tex]\vec{a}=\vec{c}[/tex]
 
  • #22
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dextercioby said:
The first exercise is incorrect.

[tex] \vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b} [/tex]

Using the distributivity,one gets

[tex] \left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0 [/tex]

Which has the infinite set of solutions:

[tex] \vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\} [/tex].

The "V" is a linear space and the [itex] \vec{d}[/itex] and [itex] \vec{b} [/itex] are arbitrary vectors.

So,please,think deeper before giving erroneous advice.

Daniel.
To be in a state of complete misunderstanding, and say something like that. And then continue to argue about it. It takes balls. I respect that.

You did not give a "solution". The problem was to give a proof that a = c , and that's what Galileo and arildno did. You just didn't follow it.
 
  • #23
dextercioby
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Let's look at it more abstractly and u'll see that my solution includes yours.

My solution asserts that

If [itex] \vec{b}\in \mathcal{P}_{1}\subseteq \mathcal{P} [/itex] with [itex] \bar{\mathcal{P}_{1}}=\mathcal{P} [/itex],then

[tex] \left(\vec{a}-\vec{c}\right) \in \mathcal{P}_{1}^{\perp} [/tex]

and [tex] \mathcal{P}_{1}\oplus\mathcal{P}_{1}^{\perp}=\mathcal{P} [/tex]

You assumed that

[tex]\vec{b}\in \mathcal{P} [/tex].Since any improper preHilbert subspace is closed and its orthogonal complement is made up from the zero vector,your solution follows from mine.

Daniel.

P.S.Arildno :wink:
 
  • #24
dextercioby
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HackaB said:
To be in a state of complete misunderstanding, and say something like that. And then continue to argue about it. It takes balls. I respect that.

You did not give a "solution". The problem was to give a proof that a = c , and that's what Galileo and arildno did. You just didn't follow it.

Maybe there was a missunderstanding,the text of the problem was vague,specifically that "for all b".I got balls... :surprised :cool: :tongue2:
My solution was more general.It had their solution as a particular case.

Daniel.
 
  • #25
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No, what you posted was an attempt to confuse the issue, and then show off some of your new found functional analysis jargon. You aren't helping anyone. You are just being a jackass.
 

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