BAC-CAB vector rule.

Question

These are examples of BAC-CAB rule in a physics book.$$(\vec A \times \vec B) \cdot (\vec C\times \vec D) = (\vec A \cdot \vec C)(\vec B \cdot \vec D) - (\vec A \cdot \vec D)(\vec B \cdot \vec C)\tag 1$$ $$ \vec A \times (\vec B \times (\vec C \times\vec D)) = \vec B (\vec A \cdot (\vec C \times \vec D)) - (\vec A \cdot \vec B)(\vec C \cdot \vec D)\tag 2$$

I tried to get the RHS from LHS but I am unable to do so.

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For the (1), I don't even know where to apply BAC-CAB rule. There rule states $\vec A \times (\vec B \times \vec C) = \vec B (\vec A \cdot \vec C) - \vec C(\vec A\cdot \vec B)$ but in (1) there are no three consecutive vector products.

For (2),

$$\vec A \times (\vec B \times (\vec C \times\vec D)) = \vec A (\vec C(\vec B \cdot \vec D) - \vec D(\vec B \cdot \vec C))$$

Since vector product is distributive,

$$\vec A \times (\vec B \times (\vec C \times\vec D)) = \vec A \times \vec C(\vec B \cdot \vec D) - \vec A \times \vec D(\vec B \cdot \vec C)$$ But I don't know what to do now.

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• What should I do ? please provide some help.

Answer 1

We have $\vec a \cdot (\vec b \times \vec c)=\vec b \cdot (\vec c \times \vec a)$. So the first equation can be manipulated using the substitution $\vec a=\vec A \times \vec B$,$\vec b=\vec C$ & $\vec c=\vec D$ \begin{eqnarray*} (\vec A \times \vec B) \cdot (\vec C\times \vec D) = \vec C \cdot (\vec D\times (\vec A \times \vec B)) \end{eqnarray*} Now use $\vec D\times (\vec A \times \vec B)=\vec A (\vec B \cdot \vec D) - \vec B(\vec A\cdot \vec D)$ & the first equation is done.

The second formula: Note that the LHS is a vector and the first term does not really know what it is and the last term is a scalar ... so something has gone wrong!

Use $\vec a \times (\vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c(\vec a\cdot \vec b)$ with $\vec a=\vec A$,$\vec b=\vec B$ & $\vec c=\vec C \times \vec D$ and we have \begin{eqnarray*} \vec A \times (\vec B \times (\vec C\times \vec D)) = \vec B (\vec A \cdot (\vec C \times \vec D)) -(\vec A \cdot \vec B)(\vec C \times \vec D ) \end{eqnarray*} This probably what should have been written in the question ?

Answer 2

For (1), consider the scalar triple product $X\cdot \left(Y\times Z\right)$. We can swap any two elements in this product at the cost of a minus sign. So:

$$ \begin{align} (\vec A \times \vec B) \cdot (\vec C\times \vec D) &= - \vec C \cdot ((\vec A\times \vec B)\times \vec D) \\ &= \vec C \cdot (\vec D\times (\vec A\times \vec B)) \end{align} $$

The last expression has a vector triple product of the form $\vec X \times (\vec Y \times \vec Z)$, which can be manipulated with the BAC-CAB rule. Can you take it from there?

(2) is simply repeated application of the BAC-CAB rule. Start with the outermost triple vector product. But the RHS can't be right as the last expression is a scalar! You may have copied it down wrong.