Vectors: How to prove the BAC-CAB identity w/o components?

In summary: I have ended up with $$ \textbf{a} \times (\textbf{b} \times \textbf{c}) = \frac{p}{(\textbf{a} \bullet \textbf{c})} [\textbf{b}(\textbf{a}\bullet\textbf{c}) - \textbf{c}(\textbf{a}\bullet\textbf{b})]$$Since you already know that the product of the vectors is equal to the sum of the vectors, the two equations will cancel each other out.
  • #1
Master1022
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Homework Statement


Prove that $$\bf{ a \times ( b \times c ) = \phi [ b(a \bullet c) - c(a \bullet b) ]} $$

for some constant phi

Homework Equations

The Attempt at a Solution


So I have used the unit vectors i, j, and k and found out that phi = 1.

With the main part of the proof, we are not allowed to expand it using components. Also, I am not at the level where we know the tensors or more advanced methods that I have seen elsewhere on the internet- just the basic rules of dot and cross product.

So I know that $$ a \times ( b \times c ) = p c + q b + k(b \times c)$$ for some scalars p,q, and k. I also know that the a x ( b x c) will lie in the plane of bc, so thus k = 0.
$$ \textbf{a} \times ( \textbf{b} \times \textbf{c} ) = p \textbf{c} + q \textbf{b} $$

Our professor then gave the hint of dotting both sides by a and (b x c) separately and going from there.

So when I dotted both sides by a, I got:
$$ 0 = p (\bf{a \bullet c}) + q (\bf{a \bullet b}) $$

Then when I dotted everything by (b x c):
$$ 0 = p (\bf{c \bullet (b \times c)}) + q (\bf{b \bullet (b \times c)}) $$

This is where I don't know what to do, because the second expression goes to 0 = 0 and I cannot really see how to use it.

Any help is greatly appreciated.
 
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  • #2
Master1022 said:
Then when I dotted everything by (b x c)
You dot from the right on the right hand side of the equation. Perhaps try to do that on the lefthand side too ... o:)

[edit] I'll get some coffee first, sorry
 
  • #3
BvU said:
You dot from the right on the right hand side of the equation. Perhaps try to do that on the lefthand side too ... :rolleyes:
The inner product is commutative ...

Master1022 said:
This is where I don't know what to do, because the second expression goes to 0 = 0 and I cannot really see how to use it.
This is what allows you to conclude that there is no ##\vec b \times \vec c## component (i.e., ##k = 0##). Since you have already assumed that, it will come out as 0 = 0.

However, your inner product with ##\vec a## is sufficient for the conclusion that ##\vec a \times (\vec b \times \vec c) = \phi[\vec b(\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)]##.
 
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  • #4
Formula ##\boldsymbol a\times( \boldsymbol b\times\boldsymbol c)=\boldsymbol b(\boldsymbol a,\boldsymbol c)-\boldsymbol c(\boldsymbol a,\boldsymbol b)## follows by direct calculation in a Cartesian frame. This calculation may be little bit simplified if you choose the frame such that ##\boldsymbol a=(a,0,0),\quad \boldsymbol b=(b_1,b_2,0)##
 
  • #5
wrobel said:
Formula ##\boldsymbol a\times( \boldsymbol b\times\boldsymbol c)=\boldsymbol b(\boldsymbol a,\boldsymbol c)-\boldsymbol c(\boldsymbol a,\boldsymbol b)## follows by direct calculation in a Cartesian frame. This calculation may be little bit simplified if you choose the frame such that ##\boldsymbol a=(a,0,0)##
I believe this fails the requirement not to use components as specified in the OP.
Master1022 said:
With the main part of the proof, we are not allowed to expand it using components.
 
  • #6
Orodruin said:
believe this fails the requirement not to use components as specified in the OP.
oh yes, I missed that. Pain without any gain, I see
 
  • #7
Orodruin said:
The inner product is commutative ...
However, your inner product with ##\vec a## is sufficient for the conclusion that ##\vec a \times (\vec b \times \vec c) = \phi[\vec b(\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)]##.

Thank you for your response! Could you explain what you mean by the above statement- I cannot really understand what you mean.

Thanks.
 
  • #8
Master1022 said:
Thank you for your response! Could you explain what you mean by the above statement- I cannot really understand what you mean.

Thanks.
Which of the two statements you quoted?
 
  • #9
Orodruin said:
Which of the two statements you quoted?
The second one
 
  • #10
You have an equation that relates ##p## and ##q##. Use this to express one in terms of the other. This is sufficient to come to the conclusion that you are asked to derive.
 
  • #11
To be more specific, you have
Master1022 said:
So when I dotted both sides by a, I got:
$$ 0 = p (\bf{a \bullet c}) + q (\bf{a \bullet b}) $$
You can solve for either ##p## or ##q## in terms of the other here and insert it into your expression
Master1022 said:
$$ \textbf{a} \times ( \textbf{b} \times \textbf{c} ) = p \textbf{c} + q \textbf{b} $$

wrobel said:
oh yes, I missed that. Pain without any gain, I see
My gym partner assures me that pain is just weakness leaving the body ... :rolleyes:
 
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  • #12
Orodruin said:
To be more specific, you have
You can solve for either ##p## or ##q## in terms of the other here and insert it into your expression

I have ended up with $$ \textbf{a} \times (\textbf{b} \times \textbf{c}) = \frac{p}{(\textbf{a} \bullet \textbf{c})} [\textbf{b}(\textbf{a}\bullet\textbf{c}) - \textbf{c}(\textbf{a}\bullet\textbf{b})]$$

do we just let the $$\frac{p}{(a \bullet c)} = \phi = 1$$

Thanks.
 
  • #13
Master1022 said:
I have ended up with $$ \textbf{a} \times (\textbf{b} \times \textbf{c}) = \frac{p}{(\textbf{a} \bullet \textbf{c})} [\textbf{b}(\textbf{a}\bullet\textbf{c}) - \textbf{c}(\textbf{a}\bullet\textbf{b})]$$

do we just let the $$\frac{p}{(a \bullet c)} = \phi = 1$$
The first part, yes (##\phi = p/(\vec a \cdot \vec c)##). That ##\phi = 1## is something you must deduce by using some reference vectors (such as the basis vectors).
 
  • #14
Orodruin said:
The first part, yes (##\phi = p/(\vec a \cdot \vec c)##). That ##\phi = 1## is something you must deduce by using some reference vectors (such as the basis vectors).
wouldn't that mean for non-basis vectors, phi wouldn't be 1?
 
  • #15
Master1022 said:
wouldn't that mean for non-basis vectors, phi wouldn't be 1?
No. The product is linear in all arguments.
 
  • #16
Orodruin said:
No. The product is linear in all arguments.
That is true... not sure I completely understand why that is, but I have tried it out by setting vectors a, b, and c to random things and they seem to always yield the result that p = (a . c) for the LHS to equal the RHS...

Thanks for all the help!
 
  • #17
Master1022 said:
not sure I completely understand why that is
Why what is? Why the product is linear in all arguments? That follows directly from the cross product being linear in both arguments, i.e., ##\vec a \times (\vec b_1 + \vec b_2) = \vec a \times \vec b_1 + \vec a \times \vec b_2## etc.
 
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FAQ: Vectors: How to prove the BAC-CAB identity w/o components?

1. What is the BAC-CAB identity?

The BAC-CAB identity is a mathematical identity that states that the cross product of three vectors (A, B, and C) can be rewritten as the cross product of two of those vectors (A and B) multiplied by the third vector (C).

2. Why is it important to prove the BAC-CAB identity without using components?

Proving the BAC-CAB identity without using components allows for a more general and elegant proof that can be applied to any vector space. It also provides a deeper understanding of the underlying principles behind the identity.

3. How can the BAC-CAB identity be proven without using components?

The BAC-CAB identity can be proven using the properties of vector triple products, such as the associativity and distributivity properties. It can also be proven using geometric arguments and the properties of vector cross products.

4. Can the BAC-CAB identity be proven using matrices?

Yes, the BAC-CAB identity can also be proven using matrices. This approach involves representing the vectors as column matrices and using matrix multiplication to show that the identity holds.

5. Are there any real-world applications of the BAC-CAB identity?

Yes, the BAC-CAB identity has many applications in physics and engineering, particularly in mechanics and electromagnetism. It is used to solve problems involving rotational motion, magnetic fields, and torque, among others.

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