What Conditions Make These Vector Dot Product Equations True?

[TsAmE]

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.

 

[HallsofIvy]

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 $\mathbf{x}\cdot\mathbf{y}= \mathbf{y}\cdot\mathbf{x}$?

 

[TsAmE]

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?

 

[Mark44]

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.

, 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?

 

[HallsofIvy]

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)?

 

[TsAmE]

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?

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?