Simplifying Dot Product Equations with Vectors - Need Help!

[jeanf]

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


1. suppose that $$a \bullet b = c \bullet b$$ for all vectors $$\overrightarrow{b}$$. show that $$\overrightarrow{a} = \overrightarrow{c}$$.

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) $$\bullet$$(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.

 

[jeanf]

cross product

my second question is:

2. suppose that $$\overrightarrow{a} X \overrightarrow{b} = \overrightarrow{c} X \overrightarrow{b}$$ for all vectors $$\overrightarrow{b}.$$ how are $$\overrightarrow{a}$$ and $$\overrightarrow{c}$$ 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.

 

[whozum]

For the first, did you know that $\vec{a} \bullet \vec{b} = |a||b|cos(\theta)$? That should be all you need.

There is a similar identity for the cross product.

 

[jeanf]

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.

 

[Curious3141]

For the dot product, here's a hint. Consider the cases where $$\vec{b}$$ is a unit vector first in the direction of $$\vec{a}$$ then in the direction of $$\vec{c}$$. Draw up some equations and see what you can conclude about the angle between vectors $$\vec{a}$$ and $$\vec{c}$$, then what that implies about the relationship between their magnitudes.

 

[Curious3141]

Here's the solution for the dot product :

Let the angle between vectors $$\vec{a}$$ and $$\vec{c}$$ be $$\theta$$.

Let $$\vec{b}$$ be a unit vector in the same direction as $$\vec{a}$$

Then $$\vec{a} \bullet \vec{b} = |\vec{a}|$$

$$\vec{c} \bullet \vec{b} = |\vec{c}|\cos\theta$$

These two are equal, so we have :

$$|\vec{a}| = |\vec{c}|\cos\theta$$ --eqn(1)

Now let $$\vec{b}$$ be a unit vector in the direction of $$\vec{c}$$

Working through it in the same way gives :

$$|\vec{c}| = |\vec{a}|\cos\theta$$ --eqn(2)

Solving those we get :

$$|\vec{c}|\cos^2{\theta} = |\vec{c}|$$

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

or $$\cos^2{\theta} = 1$$

giving $$\cos{\theta} = 1$$ (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 $$\theta = 0$$, and putting that back into either equation 1 or 2, we get $$|\vec{a}| = |\vec{c}|$$.

So the two vectors $$\vec{a}$$ and $$\vec{c}$$ are of the same magnitude and oriented in the same direction. By definition, therefore, $$\vec{a} = \vec{c}$$ (QED)

 

[Galileo]

Write the equation in a different form:

$$(\vec a -\vec c)\times \vec b=\vec 0$$
for any vector $\vec b$.

(You can do the same thing with the dot product also).
What can you infer about $\vec a-\vec c$ from this?

 

[dextercioby]

The first exercise is incorrect.

$$\vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b}$$

Using the distributivity,one gets

$$\left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0$$

Which has the infinite set of solutions:

$$\vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\}$$.

The "V" is a linear space and the $\vec{d}$ and $\vec{b}$ are arbitrary vectors.

So,please,think deeper before giving erroneous advice.

Daniel.

 

[Galileo]

The exercise is to show that $\vec a \cdot \vec b=\vec c\cdot \vec b\forall \vec b \Rightarrow \vec a = \vec c$.
This is clearly true, you just have to pick a clever value for b, say b=a-c

 

[dextercioby]

That's not true.That difference is not zero,but an arbitrary vector.Zero vector is not arbitrary.

Daniel.

 

[Galileo]

Are you sure you have the same understanding of the problem as I do?
$\vec b$ can be any vector.

 

[dextercioby]

Yes.

HINT:$$\vec{c}\cdot\vec{b}=\vec{c}\cdot\left(\vec{b}+\vec{c}\times \vec{d}\right)$$

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

Daniel.

 

[Crosson]

here is how I would handle the two dimensional 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.

 

[arildno]

You are wrong, Daniel.
$$\vec{a},\vec{c}$$ are fixed vectors, and the $$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}$$ is supposed to hold for ALL $$\vec{b}$$
This is equivalent to:
$$(\vec{a}-\vec{c})\cdot\vec{b}=0$$
In particular, this must hold for the choice $$\vec{b}=\vec{a}-\vec{c}$$ from which follows $$\vec{a}=\vec{c}$$

 

[dextercioby]

That's wrong

$$\vec{a}\cdot\vec{b}$$ is a scalar

and hence

$$a_{x}b_{x}+a_{y}b_{y}=c_{x}b_{x}+c_{y}b_{y} \nRightarrow \left\{\begin{array}{c}a_{x}b_{x}=c_{x}b_{x}\\a_{y}b_{y}=c_{y}b_{y} \end{array}\right$$

Okay??

Daniel.

 

[dextercioby]

You are wrong, Daniel.
$$\vec{a},\vec{c}$$ are fixed vectors, and the $$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}$$ is supposed to hold for ALL $$\vec{b}$$
This is equivalent to:
$$(\vec{a}-\vec{c})\cdot\vec{b}=0$$
In particular, this must hold for the particular choice $$\vec{b}=\vec{a}-\vec{c}$$ from which follows $$\vec{a}=\vec{c}$$

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.

 

[arildno]

The first exercise is incorrect.

$$\vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b}$$

Using the distributivity,one gets

$$\left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0$$

Which has the infinite set of solutions:

$$\vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\}$$.

The "V" is a linear space and the $\vec{d}$ and $\vec{b}$ are arbitrary vectors.

So,please,think deeper before giving erroneous advice.

Daniel.

Whatever are you talking about?
$$\vec{a},\vec{c}$$ are FIXED vectors; their difference do not change if $$\vec{b}$$ changes.

 

[dextercioby]

Who said "a" and "c" are fixed?

Daniel.

 

[arildno]

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


1. suppose that $$a \bullet b = c \bullet b$$ for all vectors $$\overrightarrow{b}$$. show that $$\overrightarrow{a} = \overrightarrow{c}$$.

It says for all b here.

 

[dextercioby]

Yes,and my solution exploits the equality of the 2 scalar products.

Daniel.

 

[arildno]

No, it doesn't do any such thing.
For the particular choice $$\vec{b}=\vec{a}-\vec{c}$$
we get
$$(\vec{a}-\vec{c})^{2}=0$$
which holds if and only if $$\vec{a}=\vec{c}$$

 

[HackaB]

The first exercise is incorrect.

$$\vec{a}\cdot\vec{b}=\vec{c}\cdot\vec{b}$$

Using the distributivity,one gets

$$\left(\vec{a}-\vec{c}\right)\cdot\vec{b}=0$$

Which has the infinite set of solutions:

$$\vec{a}-\vec{c}=\left\{\vec{d}\times\vec{b}|,\vec{d}\in V,\vec{b}\in V \right\}$$.

The "V" is a linear space and the $\vec{d}$ and $\vec{b}$ 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.

 

[dextercioby]

Let's look at it more abstractly and u'll see that my solution includes yours.

My solution asserts that

If $\vec{b}\in \mathcal{P}_{1}\subseteq \mathcal{P}$ with $\bar{\mathcal{P}_{1}}=\mathcal{P}$,then

$$\left(\vec{a}-\vec{c}\right) \in \mathcal{P}_{1}^{\perp}$$

and $$\mathcal{P}_{1}\oplus\mathcal{P}_{1}^{\perp}=\mathcal{P}$$

You assumed that

$$\vec{b}\in \mathcal{P}$$.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

 

[dextercioby]

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... :tongue2:
My solution was more general.It had their solution as a particular case.

Daniel.

 

[HackaB]

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.

 

[dextercioby]

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.

I wasn't confusing any issue.I was interpreting a vague issue MY way.It is a mathematics problem and the way to do it correctly is less relevant.
As for not helping,i'm not helping you,maybe.Don't speak for someone else,unless you're paid to.

Jackass.

P.S.I think that MTV show really sukt. :yuck:

 

[mathwonk]

jeanf: here is one way to think about problem 1, (although I know you understand it by now).

if two vectors have the same dot product with evrything, then in particular they are perpendicular to all the same vectors, so they must be parallel.

so one is a scalar times the other. to see what that scalar is just dot both of them with one of them.

i.e. if A = tB, then A.B = tB.B, so t = (A.B)/(B.B). but by hypothesis, A.B = B.B.

so t = 1.

so A = B.

 

[arildno]

You are simply misunderstanding a vaguely stated exercise, Daniel.
If it had been specifically stated that a and c were fixed vectors (rather than tacitly assumed) we wouldn't have this conversation.
Your smartass comments on Hilbert spaces are irrelevant in this context.

 

[dextercioby]

I'm not misuderstanding anything.It was vague.It left room for interpretations.Mine was simply broader.

Daniel.

 

[arildno]

When your interpretation is in conflict with the intended meaning then it is you who have misunderstood, even though it is the author's fault that he didn't state what he meant clearly and without ambiguity.

 

[whozum]

You guys have confused the living **** out of me, so I don't even know what the OP is going through. Besides set a good example. Let's be civilized.

 

[dextercioby]

And what was the "intended meaning"...?I think it's a matter of subjectivity in this case,which has nothing to do with mathematics.

I'll drop it.

Daniel.