Given that the vectors A , B , and C are mutually perpendicular,...

Using the vector triple product expansion:

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

Since $\vec{A}$, $\vec{B}$, and $\vec{C}$ are mutually perpendicular, the dot product of any two distinct vectors is zero:

$\vec{C} \cdot \vec{B} = 0$ and $\vec{C} \cdot \vec{A} = 0$

Substituting these values into the expansion gives:

$\vec{C} \times (\vec{A} \times \vec{B}) = 0\vec{A} - 0\vec{B} = 0$

To check the converse, assume $\vec{C} \times (\vec{A} \times \vec{B}) = 0$. This implies:

$(\vec{C} \cdot \vec{B})\vec{A} - (\vec{C} \cdot \vec{A})\vec{B} = 0$

If $\vec{A}$ and $\vec{B}$ are non-collinear, this equation requires $\vec{C} \cdot \vec{B} = 0$ and $\vec{C} \cdot \vec{A} = 0$, which means $\vec{C}$ is perpendicular to both $\vec{A}$ and $\vec{B}$. However, it places no restriction on the angle between $\vec{A}$ and $\vec{B}$, so $\vec{A}$ and $\vec{B}$ need not be perpendicular. Furthermore, if $\vec{A}$ and $\vec{B}$ are collinear, the condition is satisfied for any vector $\vec{C}$.

Therefore, the converse is not necessarily true.