What Conditions Make These Vector Dot Product Equations True?

AI Thread Summary
The discussion centers on the conditions under which two vector dot product equations hold true. For the first equation, |\mathbf{a}\cdot \mathbf{b}| =|\mathbf{a}||\mathbf{b}|, it is established that this equality is valid when the vectors \mathbf{a} and \mathbf{b} are parallel, corresponding to an angle \theta of 0 degrees. The second equation, (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2, is confirmed to be true for all vectors, as it simplifies correctly regardless of their orientation. The discussion also touches on the properties of the dot product, including its behavior with perpendicular vectors. Overall, the key takeaway is that the first equation requires parallel vectors, while the second holds universally.
TsAmE
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Homework Statement



Under what conditions are the following true?

a) |\mathbf{a}\cdot \mathbf{b}| =|\mathbf{a}||\mathbf{b}|

b) (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2

Homework Equations



None.

The Attempt at a Solution



a) I don't understand why the answer is \mathbf{a} = \lambda \mathbf{b}. Surely it would be equivalent all the time?

b) Also looks equivalent all the time to me.
 
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Are you aware that the dot product of any two perpendicular vectors is 0? It isn't likely that |\mathbf{u}||\mathbf{v}|= 0 in that case is it?

a) One definition of dot product is \mathbf{a}\cdot\mathbf{b}= |\mathbf{a}||\mathbf{b}| cos(\theta) where \theta is the angle between the two vectors. For what \theta is cos(\theta)= 1?

b) (\mathbf{a}+ \mathbf{b})\cdot(\mathbf{a}- \mathbf{b})= \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{a}- \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{b}

Now, is \mathbf{x}\cdot\mathbf{x}= |\mathbf{x}|^2? Is \mathbf{x}\cdot\mathbf{y}= \mathbf{y}\cdot\mathbf{x}?
 
HallsofIvy said:
Are you aware that the dot product of any two perpendicular vectors is 0? It isn't likely that |\mathbf{u}||\mathbf{v}|= 0 in that case is it?

a) One definition of dot product is \mathbf{a}\cdot\mathbf{b}= |\mathbf{a}||\mathbf{b}| cos(\theta) where \theta is the angle between the two vectors. For what \theta is cos(\theta)= 1?

b) (\mathbf{a}+ \mathbf{b})\cdot(\mathbf{a}- \mathbf{b})= \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{a}- \mathbf{a}\cdot\mathbf{a}+ \mathbf{b}\cdot\mathbf{b}

Now, is \mathbf{x}\cdot\mathbf{x}= |\mathbf{x}|^2? Is \mathbf{x}\cdot\mathbf{y}= \mathbf{y}\cdot\mathbf{x}?

Yeah I am aware. For a) I would say it could only be equivalent if the dot product doesn't equal 0, but I don't get why the answer is \mathbf{a} = \lambda \mathbf{b}

b) I agree as x^2 gives you the absolute value and x * y is the same as y * x, but how does this apply to b?
 
TsAmE said:
Yeah I am aware. For a) I would say it could only be equivalent if the dot product doesn't equal 0
No. What if the dot product were 1/2? Would that make the two sides of the equation equal? The two sides of that equation are equal only if the dot product is 1. That's different from what you said.

Look at what HallsOfIvy wrote about the coordinate-free definition of the dot product.
TsAmE said:
, but I don't get why the answer is \mathbf{a} = \lambda \mathbf{b}

b) I agree as x^2 gives you the absolute value and x * y is the same as y * x, but how does this apply to b?
 
TsAmE said:
Yeah I am aware. For a) I would say it could only be equivalent if the dot product doesn't equal 0, but I don't get why the answer is \mathbf{a} = \lambda \mathbf{b}
The dot product not equaling 0 is not relevant. Answer my question: For what \theta is cos(\theta)= 1?

b) I agree as x^2 gives you the absolute value and x * y is the same as y * x, but how does this apply to b?
Then you agree that (a+ b)\cdot(a- b)= a\cdot a+ a\cdot b- b\cdot a- b\cdot b (I miswrote that as "+ b\cdot b before)
= |a|^2+ a\cdot b- a\cdot b- |b|^2. What does that reduce to (for all a and b)?
 
HallsofIvy said:
The dot product not equaling 0 is not relevant. Answer my question: For what \theta is cos(\theta)= 1? When theta = 0, so a) is only equivalent when vector a and b are parallel?
HallsofIvy said:
Then you agree that (a+ b)\cdot(a- b)= a\cdot a+ a\cdot b- b\cdot a- b\cdot b (I miswrote that as "+ b\cdot b before)
= |a|^2+ a\cdot b- a\cdot b- |b|^2. What does that reduce to (for all a and b)?
It reduces to = |a|^2- |b|^2 so the LHS = RHS, but for what condition?
 
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