$\vec{a}\times(\vec{a}\times\vec{R})-\vec{b}\times(\vec{b}\times\vec{R})$

Question

I have $\vec{a}\times(\vec{a}\times\vec{R})-\vec{b}\times(\vec{b}\times\vec{R})$, my textbook says that this equals $((\vec{a}\times\vec{a})-(\vec{b}\times\vec{b}))\times\vec{R}=-(a^2-b^2)\vec{R}$. I know that the final answer is correct, but is the first step correct as well? (I don't think so, since the cross product of a vector with itself is zero) What is going wrong here or what don't I understand?

Answer 1

As others have said, if that's what the textbook says then it's wrong because the vector product is not associative . This may have arisen due to some confusion between $\times$ and $\cdot$, though.

Generally, there is a formula for the vector triple product in terms of the scalar product, given by $$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c) \vec b - (\vec a \cdot \vec b) \vec c$$

Applying this formula here gives

$$\vec a \times (\vec a \times \vec R) = (\vec a \cdot \vec R) \vec a - a^2 \vec R$$ where $a=\lVert \vec a \rVert$. Likewise for the other term; so your expression in fact equals $$-(a^2-b^2)\vec R + (\vec a \cdot \vec R) \vec a - (\vec b \cdot \vec R) \vec b$$ If $\vec a$ and $\vec b$ are perpendicular to $\vec R$ (are they?), then $\vec a \cdot \vec R = \vec b \cdot \vec R = 0$, so this equals $$-(a^2-b^2) \vec R$$ as claimed.

Answer 2

You're right. The first step is incorrect.

Answer 3

There must be some mistake in the book since the cross product is not associative see https://proofwiki.org/wiki/Cross_Product_is_not_Associative

Answer 4

There must be some mistake in the book since the cross product is not associative see https://proofwiki.org/wiki/Cross_Product_is_not_Associative

And $a\times a=0$