<p>If \(|\vec{a}| = 2\) and \(|\vec{b}| = 3\) and \(\vec{a} \cdot \vec{b} = 0\), then \(\left(\vec{a} \times \left(\vec{a} \times \left(\vec{a} \times \left(\vec{a} \times \vec{b}\right)\right)\right)\right)\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula repeatedly: $\vec{u} \times (\vec{v} \times \vec{w}) = \vec{v}(\vec{u} \cdot \vec{w}) - \vec{w}(\vec{u} \cdot \vec{v})$. Since $\vec{a} \perp \vec{b}$, each application simplifies significantly by eliminating dot product terms.
Step 1: Apply vector triple product formula to $\vec{a} \times (\vec{a} \times \vec{b})$ $\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a}(\vec{a} \cdot \vec{b}) - \vec{b}(\vec{a} \cdot \vec{a})$ Since $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4$: $\vec{a} \times (\vec{a} \times \vec{b}) = \vec{0} - 4\vec{b} = -4\vec{b}$ Step 2: Compute $\vec{a} \times (-4\vec{b}) = -4(\vec{a} \times \vec{b})$ Step 3: Apply triple product formula to $\vec{a} \times (-4(\vec{a} \times \vec{b}))$ $\vec{a} \times (-4(\vec{a} \times \vec{b})) = -4[\vec{a}(\vec{a} \cdot \vec{b}) - \vec{b}(\vec{a} \cdot \vec{a})]$ $= -4[\vec{0} - 4\vec{b}] = 16\vec{b}$ Step 4: Apply triple product formula to $\vec{a} \times (16\vec{b})$ $\vec{a} \times (16\vec{b}) = 16(\vec{a} \times \vec{b})$, which is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ with magnitude $16 \cdot 2 \cdot 3 = 96$ ∴ The final answer is $\boxed{-48\vec{b}}$ or equivalent form depending on options
Correct Answer: A