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

## Homework Statement

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

for some constant phi

## 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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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 ...

 I'll get some coffee first, sorry

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 ...
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 $\vec a \times (\vec b \times \vec c) = \phi[\vec b(\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)]$.

BvU
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 $\boldsymbol a=(a,0,0),\quad \boldsymbol b=(b_1,b_2,0)$

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.

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

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.

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?

Which of the two statements you quoted?
The second one

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.

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 ...

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.

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).

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?

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.

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!

Orodruin
Staff Emeritus
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.