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

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

From the properties of the cross product, $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

Similarly, $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.

This implies that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are mutually orthogonal. Thus, Statement I is correct.

Since the vectors are mutually orthogonal, the angle between any two of them is 90^{}^{\circ}.

Taking the magnitude of the first equation:

$|\vec{a} \times \vec{b}| = |\vec{c}| \Rightarrow |\vec{a}||\vec{b}| = |\vec{c}|$

Taking the magnitude of the second equation:

$|\vec{b} \times \vec{c}| = |\vec{a}| \Rightarrow |\vec{b}||\vec{c}| = |\vec{a}|$

Substituting $|\vec{c}|$ from the first equation into the second:

$|\vec{b}|(|\vec{a}||\vec{b}|) = |\vec{a}| \Rightarrow |\vec{a}||\vec{b}|^2 = |\vec{a}|$

For non-zero vectors, we get $|\vec{b}|^2 = 1 \Rightarrow |\vec{b}| = 1$.

This gives $|\vec{a}| = |\vec{c}|$. However, $|\vec{a}|$ and $|\vec{c}|$ can be any positive real number and are not necessarily equal to $1$.

Thus, Statement II is incorrect.