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

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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.
 

Answers and Replies

  • #2
BvU
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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
Orodruin
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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 ...

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)]##.
 
  • #4
wrobel
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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
Orodruin
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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.
With the main part of the proof, we are not allowed to expand it using components.
 
  • #6
wrobel
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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
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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
Orodruin
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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?
 
  • #10
Orodruin
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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
Orodruin
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To be more specific, you have
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
$$ \textbf{a} \times ( \textbf{b} \times \textbf{c} ) = p \textbf{c} + q \textbf{b} $$

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:
 
  • #12
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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
Orodruin
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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
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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
Orodruin
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wouldn't that mean for non-basis vectors, phi wouldn't be 1?
No. The product is linear in all arguments.
 
  • #16
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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
Orodruin
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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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