How can we prove the bac-cab rule using a geometrical derivation?

[ady]

I've looked up on the solutions to bac-cab rule on the internet but they where all the same Computational, idea free ways.
I wonder if anyone knows any better way.
I'd be very thankful for any ideas.

 

[Mark44]

I've looked up on the solutions to bac-cab rule on the internet but they where all the same Computational, idea free ways.
I wonder if anyone knows any better way.
I'd be very thankful for any ideas.

What is the "bac-cab" rule?

 

[ady]

bac-cab rule:
Ax(BxC)=B(A.C)-C(A.B)
and usual proof is available here: https://www.physicsforums.com/showthread.php?t=65457/
but I'm looking for a new way.

 

[PaulDirac]

Hi,

Have you ever tried to come to a proof using the Levi-civita function??

 

[ady]

good point

Hi,

Have you ever tried to come to a proof using the Levi-civita function??

I have but I can't understand it completely. There are some parts that i haven't seen during my studies.
the parts I mentioned previously are: are Einstein summation convention and Levi-Civita and ... which means I know nothing of that proof's tools.

 

[BruceW]

wikipedia shows two different methods, which are not just the standard computation: use of geometric algebra and use of tensor calculus, which has been mentioned already. http://en.wikipedia.org/wiki/Triple_product#Vector_triple_product
Use of geometric algebra is probably the most elegant, but most abstract. The tensor calculus method is still pretty elegant, compared to the standard computation method.

edit: p.s. you can ignore the stuff about the scalar triple product on that page.

 

[jamaklath]

I've looked up on the solutions to bac-cab rule on the internet but they where all the same Computational, idea free ways.
I wonder if anyone knows any better way.
I'd be very thankful for any ideas.

This video shows a really unique way

 

[TSny]

Here's a fairly geometrical derivation that has some similarities to the video in the previous post.

Want to show $\vec A \times \left( \vec B \times \vec C \right) = \vec B \left(\vec A \cdot \vec C \right) - \vec C \left(\vec A \cdot \vec B\right)$.

Since this identity is clearly zero if anyone of the three vectors is zero, we will assume in the following that none of the three vectors is zero.

It is not hard to see that the component of $\vec A$ that is perpendicular to the plane containing $\vec B$ and $\vec C$ does not contribute to either side of the above identity. That is, we can forget about this component of $\vec A$ when evaluating either the left or right sides. This means that we don't lose any generality of the proof if we assume that $\vec A$ lies in the plane containing $\vec B$ and $\vec C$.

So, in the following we assume $\vec A$, $\vec B$, and $\vec C$ are coplanar.Introduce unit vectors $\hat A$, $\hat B$, and $\hat C$ in the directions of $\vec A$, $\vec B$, and $\vec C$, respectively. It is easy to see that each side of the identity that we want to prove is proportional to $ABC$, where $A$, $B$, and $C$ are the magnitudes of the corresponding vectors. Thus the identity will be true if and only if $$\hat A \times \left( \hat B \times \hat C \right) = \hat B \left(\hat A \cdot \hat C \right) - \hat C \left(\hat A \cdot \hat B \right.)$$
Let $\theta$ be the angle between $\hat A$ and $\hat B$. Let $\phi$ be the angle between $\hat A$ and $\hat C$. For convenience, introduce perpendicular $x$ and $y$ axes lying in the plane of our three vectors with the x-axis along $\hat A$.

Consider $\hat A \times \left( \hat B \times \hat C \right)$. The magnitude of $\hat B \times \hat C$ is $|\sin (\phi - \theta)|$ and the direction of $\hat B \times \hat C$ is perpendicular to the plane. Therefore, the cross product of $\hat A$ with $\hat B \times \hat C$ has magnitude $|\sin (\phi- \theta)|$ and has direction $\pm \hat y$, where $\hat y$ is a unit vector in the positive y direction. The sign depends on whether $\phi > \theta$ or $\theta > \phi$. However, in either case you can check that $$\hat A \times \left( \hat B \times \hat C \right) = -\sin (\phi - \theta) \hat y. $$

Next, we have $$ \hat B \left(\hat A \cdot \hat C \right) = \hat B \cos \phi = \left(\cos \theta \, \hat x + \sin \theta \, \hat y \right)\cos \phi$$
And $$ \hat C \left(\hat A \cdot \hat B \right) = \hat C \cos \theta = \left(\cos \phi \, \hat x + \sin \phi \, \hat y \right)\cos \theta$$
Thus, $$ \hat B \left(\hat A \cdot \hat C \right) - \hat C \left(\hat A \cdot \hat B \right) = \left( \sin \theta \cos \phi - \sin \phi \cos \theta \right) \hat y = -\sin ( \phi - \theta) \hat y = \hat A \times \left( \hat B \times \hat C \right). $$

(Please point out mistakes. I don't claim any originality and I don't suggest that this derivation should be preferred in any way. I show it just for fun.)