Key Concepts and Formulas

• Vector Triple Product Formula: For any three vectors $\overrightarrow x$, $\overrightarrow y$, and $\overrightarrow z$, the vector triple product is given by: $\overrightarrow x \times (\overrightarrow y \times \overrightarrow z) = (\overrightarrow x \cdot \overrightarrow z)\overrightarrow y - (\overrightarrow x \cdot \overrightarrow y)\overrightarrow z$
• Perpendicular Vectors: If two vectors $\overrightarrow a$ and $\overrightarrow b$ are perpendicular, their dot product is zero: $\overrightarrow a \cdot \overrightarrow b = 0$.
• Dot Product of a Vector with Itself: The dot product of a vector with itself is the square of its magnitude: $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2$.
• Scalar Multiplication with Cross Product: For a scalar $k$ and vectors $\overrightarrow u$, $\overrightarrow v$: $k(\overrightarrow u \times \overrightarrow v) = (k\overrightarrow u) \times \overrightarrow v = \overrightarrow u \times (k\overrightarrow v)$.

Step-by-Step Solution

We are asked to evaluate the expression: $E = \overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$ given that $\overrightarrow a$ and $\overrightarrow b$ are perpendicular, which means $\overrightarrow a \cdot \overrightarrow b = 0$.

Step 1: Evaluate the innermost cross product. Let's first simplify $\overrightarrow a \times \overrightarrow b$. Since $\overrightarrow a$ and $\overrightarrow b$ are perpendicular, this cross product results in a vector perpendicular to both $\overrightarrow a$ and $\overrightarrow b$.

Step 2: Evaluate the next nested cross product. Consider the expression $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b)$. We will use the vector triple product formula $\overrightarrow x \times (\overrightarrow y \times \overrightarrow z) = (\overrightarrow x \cdot \overrightarrow z)\overrightarrow y - (\overrightarrow x \cdot \overrightarrow y)\overrightarrow z$. Here, $\overrightarrow x = \overrightarrow a$, $\overrightarrow y = \overrightarrow a$, and $\overrightarrow z = \overrightarrow b$. $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = (\overrightarrow a \cdot \overrightarrow b)\overrightarrow a - (\overrightarrow a \cdot \overrightarrow a)\overrightarrow b$ Using the given condition $\overrightarrow a \cdot \overrightarrow b = 0$ and $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2$: $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = (0)\overrightarrow a - (|\overrightarrow a|^2)\overrightarrow b$ $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = -|\overrightarrow a|^2\overrightarrow b \quad \text{(Result 1)}$ This result shows that the cross product of $\overrightarrow a$ with $(\overrightarrow a \times \overrightarrow b)$ is a scalar multiple of $\overrightarrow b$.

Step 3: Evaluate the next layer of the expression. Now, we need to evaluate $\overrightarrow a \times (\overrightarrow a \times (\overrightarrow a \times \overrightarrow b))$. We can substitute 'Result 1' into this expression: $\overrightarrow a \times (\text{Result 1}) = \overrightarrow a \times (-|\overrightarrow a|^2\overrightarrow b)$ Using the property of scalar multiplication with cross products, we can pull the scalar $-|\overrightarrow a|^2$ out: $\overrightarrow a \times (-|\overrightarrow a|^2\overrightarrow b) = -|\overrightarrow a|^2 (\overrightarrow a \times \overrightarrow b)$ We already know from Step 1 that $\overrightarrow a \times \overrightarrow b$ is some vector perpendicular to $\overrightarrow a$. Let's denote $\overrightarrow a \times \overrightarrow b = \overrightarrow c$. Then, $\overrightarrow a \times \overrightarrow c = \overrightarrow a \times (\overrightarrow a \times \overrightarrow b)$.

Let's re-evaluate this step using the vector triple product formula on the entire expression from Step 2. We have $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = -|\overrightarrow a|^2\overrightarrow b$. Let's consider the expression $\overrightarrow a \times (\overrightarrow a \times (\text{some vector}))$. The expression we need to evaluate is $E = \overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$. Let $\overrightarrow V_1 = \overrightarrow a \times \overrightarrow b$. Then $E = \overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow V_1 } \right)} \right)$. From Step 2, we found $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = -|\overrightarrow a|^2\overrightarrow b$. So, let $\overrightarrow V_2 = \overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = -|\overrightarrow a|^2\overrightarrow b$. The expression becomes $E = \overrightarrow a \times (\overrightarrow a \times \overrightarrow V_2)$. Now we apply the vector triple product formula to $\overrightarrow a \times (\overrightarrow a \times \overrightarrow V_2)$: $\overrightarrow a \times (\overrightarrow a \times \overrightarrow V_2) = (\overrightarrow a \cdot \overrightarrow V_2)\overrightarrow a - (\overrightarrow a \cdot \overrightarrow a)\overrightarrow V_2$ Substitute $\overrightarrow V_2 = -|\overrightarrow a|^2\overrightarrow b$: $\overrightarrow a \times (\overrightarrow a \times (-|\overrightarrow a|^2\overrightarrow b)) = (\overrightarrow a \cdot (-|\overrightarrow a|^2\overrightarrow b))\overrightarrow a - (\overrightarrow a \cdot \overrightarrow a)(-|\overrightarrow a|^2\overrightarrow b)$ $= (-|\overrightarrow a|^2 (\overrightarrow a \cdot \overrightarrow b))\overrightarrow a - (|\overrightarrow a|^2)(-|\overrightarrow a|^2\overrightarrow b)$ Using $\overrightarrow a \cdot \overrightarrow b = 0$: $= (-|\overrightarrow a|^2 (0))\overrightarrow a - (|\overrightarrow a|^2)(-|\overrightarrow a|^2\overrightarrow b)$ $= (0)\overrightarrow a + |\overrightarrow a|^4\overrightarrow b$ $= |\overrightarrow a|^4\overrightarrow b$

Step 4: Final Simplification. The expression $E$ simplifies to $|\overrightarrow a|^4\overrightarrow b$.

Common Mistakes & Tips

• Incorrect application of the vector triple product formula: Ensure the terms are in the correct order: $(\overrightarrow x \cdot \overrightarrow z)\overrightarrow y - (\overrightarrow x \cdot \overrightarrow y)\overrightarrow z$.
• Forgetting the perpendicularity condition: The condition $\overrightarrow a \cdot \overrightarrow b = 0$ is crucial for simplification and should be used as early as possible.
• Sign errors: Be careful with negative signs when pulling scalars out of cross products or during the application of the triple product formula.

Summary

The problem involves repeated application of the vector triple product formula and utilizes the property of perpendicular vectors. By systematically working from the innermost cross product outwards and applying the given condition $\overrightarrow a \cdot \overrightarrow b = 0$ at each appropriate step, the complex expression simplifies significantly. The key is to correctly apply the vector triple product formula and to recognize that $\overrightarrow a \cdot \overrightarrow b = 0$ makes the dot product terms vanish, leaving only terms involving the magnitude of $\overrightarrow a$ and the vector $\overrightarrow b$.

The final answer is $|\overrightarrow a {|^4}\overrightarrow b$. This corresponds to option (D).

The final answer is $\boxed{|\overrightarrow a {|^4}\overrightarrow b}$.