Why is $(\vec{b} \times \vec{a}) \times \vec{b}$ non zero?

Question

If I use the rule of vector triple product, it becomes:

$$ \vec{b} \times \vec{a} \times \vec{b} = \vec{a} ( |\vec{b}|^2) - \vec{b}(\vec{b} \cdot \vec{a})$$

which is generally non-zero, but suppose I use properties of cross product:

$$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$$

Hence,

$$ \vec{b} \times \vec{a} \times \vec{b} = - \vec{a} \times \vec{b} \times \vec{b} = \vec{a} \times (\vec{b} \times \vec{b})=0$$

What did I do wrong?

The vector cross product is not associative, i.e. $(\vec a \times \vec b) \times \vec c \ne \vec a \times (\vec b \times \vec c)$ — player3236, Feb 16 '21 at 11:56
$b\times a\times b$ is not well defined since the vector product is not associatif. — Surb, Feb 16 '21 at 11:57
Oh, thank you I got the mistake one. If someone writes this as an answer with providing source on where this concept is discussed, I'll accept it. — tryst with freedom, Feb 16 '21 at 11:58
A nice demonstration of the non-associativity: https://math.stackexchange.com/questions/1617751/proof-that-the-cross-product-is-not-associative-without-using-components — player3236, Feb 16 '21 at 11:58
Technically, $b×a×b$ can be rendered uniquely without parentheses as both possibilities are equal. This equality holds iff the end members of the triple cross product are parallel or antiparallel. — Oscar Lanzi, Feb 16 '21 at 12:05

score 1 · Answer 1 · answered Feb 16 '21 at 12:28

1

$\vec S= \vec P \times (\vec Q \times \vec R)= (\vec P. \vec R) \vec Q-(\vec P. \vec Q) \vec R$

So $(\vec b \times a)\times \vec b=-\vec b\times(\vec b \times \vec a)=-[(\vec b. \vec a) \vec b- (\vec b . \vec b) \vec a] \ne 0$

answered Feb 16 '21 at 12:28

Z Ahmed

43,235

score 1 · Accepted Answer · answered Feb 16 '21 at 12:33

Removing the parenthesis makes you perform invalid transformations, because the cross product is not associative.

In fact $$\vec a\times\vec b\times\vec c$$ has no defined meaning.

Unless $\vec a,\vec b$ are parallel, $\vec b\times\vec a$ is orthogonal to $\vec b$, hence not parallel, and $(\vec b\times\vec a)\times\vec b$ is nonzero.

score 1 · Answer 3 · answered Feb 16 '21 at 12:41

If $\mathbf{x}$ and $\mathbf{y}$ are perpendicular, then $\|\mathbf{x} \times \mathbf{y}\| = \|\mathbf{x}\|\;\|\mathbf{y}\|$. So in that case, if the two factors are nonzero, so is the cross product. But $\mathbf{b} \times \mathbf{a}$ is perpendicular to $\mathbf{b}$, so we can conclude: if $\mathbf{b} \times \mathbf{a}$ is nonzero, then so is $(\mathbf{b}\times\mathbf{a})\times \mathbf{b}$.

Note, however, that if $\mathbf{a}$ and $\mathbf{b}$ are parallel, then $\mathbf{b} \times \mathbf{a} = \mathbf{0}$ and your conclusion fails.

Why is $(\vec{b} \times \vec{a}) \times \vec{b}$ non zero?

3 Answers3