Lagrange's Identity Homework (Boas 3rd ed Ch 3 Sec 4, Problem 24)

[kq6up]

Homework Statement

From Mary Boas' "Mathematical Methods in the Physical Sciences" 3rd ed. Ch 3 Sec 4 problem 24

Where A and B are vectors. What is the value of (AXB)^2+(A dot B)^2=? Comment: This is a special case of Lagrange's Identity.

Homework Equations

Cross product and dot product definitions.

The Attempt at a Solution

I get a nasty mess that ends up equaling A^2*B^2. That is what is supposed to equal, but I got such a large unwieldy result that I would have no idea that it all could be collapsed down to A^2*B^2 by simply looking at it. So I looked up the answer and saw that my mess is supposed to equal A^2*B^2. I tried it in SAGE and sure enough they are identical. However, I am not sure what the purpose of the exercise is supposed to help me understand other than expanding both sides to see that Lagrange's Identity indeed works.

Am I missing something here? I have noticed with working with Mary Boas' book most solutions are very elegant and don't require a rat's nest of algebra. Usually if a mess results, I am doing it wrong :D

Thanks,
Chris Maness

 

[LCKurtz]

Homework Statement

From Mary Boas' "Mathematical Methods in the Physical Sciences" 3rd ed. Ch 3 Sec 4 problem 24

Where A and B are vectors. What is the value of (AXB)^2+(A dot B)^2=? Comment: This is a special case of Lagrange's Identity.

[

If A X B is a vector, what does it mean to square it? Or do you mean |AXB|^2?

 

[kq6up]

The book presents it that way, but she states if you see a vector^2 it is actually the dot product with itself. I cheated and peaked at the solutions manual and she actually didn't give the answer, but a fat hint.

Use AXB=|A||B|*sin(x), and A dot B=|A||B|*cos(x).

It is indeed very elegant with this hint.

Thanks,
Chris Maness

 

[LCKurtz]

The book presents it that way, but she states if you see a vector^2 it is actually the dot product with itself. I cheated and peaked at the solutions manual and she actually didn't give the answer, but a fat hint.

Use AXB=|A||B|*sin(x), and A dot B=|A||B|*cos(x).

It is indeed very elegant with this hint.

Thanks,
Chris Maness

You need to be more careful about the distinction between vectors and their magnitudes. Your statement above that I have highlighted makes no sense because you have a vector on the left and a scalar on the right. I think you made the same mistake in your original statement of the problem. Don't confuse AXB with |AXB|.

 

[kq6up]

Yes, I see the issue. She does write it just like that in the text with that statement for clarity.

Thanks,
Chris Maness

 

[lurflurf]

If A X B is a vector, what does it mean to square it? Or do you mean |AXB|^2?

It is not confusion, it is operator overloading.

It is very common (almost standard) to define.
$$\mathbf{A}^2=\mathbf{A}\cdot \mathbf{A}=|\mathbf{A}|^2$$
no confusion is likely to result the other self products are written
$$\mathbf{A} \times \mathbf{A}=\mathbf{0} \\
\mathbf{A} \mathbf{A}
$$

The hint form is slightly troubling it could be written unambiguously, for example
$$(\mathbf{A}\times \mathbf{B})^2=\mathbf{A}^2 \, \mathbf{B}^2 \, \sin^2(x)=(|\mathbf{A}|\, |\mathbf{B}| \sin(x))^2 \\
(\mathbf{A}\cdot \mathbf{B})^2=\mathbf{A}^2 \, \mathbf{B}^2 \, \cos^2(x)=(|\mathbf{A}|\, |\mathbf{B}| \cos(x))^2$$

 

[kq6up]

Yes, your first LaTeX line is in her text, and then it goes on to write (|AXB|)^2 as (AXB)^2.

Chris

 

[LCKurtz]

It is not confusion, it is operator overloading.

It is very common (almost standard) to define.
$$\mathbf{A}^2=\mathbf{A}\cdot \mathbf{A}=|\mathbf{A}|^2$$
no confusion is likely to result the other self products are written
$$\mathbf{A} \times \mathbf{A}=\mathbf{0} \\
\mathbf{A} \mathbf{A}
$$

The hint form is slightly troubling it could be written unambiguously, for example
$$(\mathbf{A}\times \mathbf{B})^2=\mathbf{A}^2 \, \mathbf{B}^2 \, \sin^2(x)=(|\mathbf{A}|\, |\mathbf{B}| \sin(x))^2 \\
(\mathbf{A}\cdot \mathbf{B})^2=\mathbf{A}^2 \, \mathbf{B}^2 \, \cos^2(x)=(|\mathbf{A}|\, |\mathbf{B}| \cos(x))^2$$

Yes, I have seen that before too, and I don't care for it. But I have never seen it acceptable to write$$
A\times B = |A||B|\sin\theta$$Students have enough trouble keeping vectors and scalars straight without overloading (abusing) the notation.