Passage: Let a × b = c and b × c = a . Question: Consider the...

Given $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$.

Taking the dot product of the first equation with $\vec{c}$:
$(\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{c} \cdot \vec{c} \Rightarrow [\vec{a} \vec{b} \vec{c}] = |\vec{c}|^2$

Taking the dot product of the second equation with $\vec{a}$:
$(\vec{b} \times \vec{c}) \cdot \vec{a} = \vec{a} \cdot \vec{a} \Rightarrow [\vec{b} \vec{c} \vec{a}] = |\vec{a}|^2$

Using the properties of the scalar triple product, $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$.
Thus, $|\vec{a}|^2 = |\vec{c}|^2 = [\vec{a} \vec{b} \vec{c}]$.

Evaluating Statement I:
The left-hand side is $(\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{a} = [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] = 2[\vec{a} \vec{b} \vec{c}]$.
The right-hand side is $(\vec{c} \times \vec{a}) \cdot \vec{b} = [\vec{c} \vec{a} \vec{b}] = [\vec{a} \vec{b} \vec{c}]$.
For the statement to be true, $2[\vec{a} \vec{b} \vec{c}] = [\vec{a} \vec{b} \vec{c}] \Rightarrow [\vec{a} \vec{b} \vec{c}] = 0$, which implies $|\vec{c}|^2 = 0$. This is not generally true for non-zero vectors. Hence, Statement I is incorrect.

Evaluating Statement II:
Substituting the given cross products into the expression:
$\{(\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c})\} \cdot \vec{b} = (\vec{c} \times \vec{a}) \cdot \vec{b} = [\vec{c} \vec{a} \vec{b}] = [\vec{a} \vec{b} \vec{c}] = |\vec{c}|^2$
This expression equals $1$ only if $|\vec{c}| = 1$. Since there is no condition restricting the magnitude of $\vec{c}$ to $1$, Statement II is also incorrect.

Therefore, neither Statement I nor Statement II is correct.

Answer: Neither I nor II