Key Concepts and Formulas

• Distributive Property of Cross Product: $\vec u \times (\vec v + \vec w) = \vec u \times \vec v + \vec u \times \vec w$ and $(\vec v + \vec w) \times \vec u = \vec v \times \vec u + \vec w \times \vec u$.
• Cross Product of a Vector with Itself: $\vec v \times \vec v = \vec 0$.
• Vector Triple Product Identity: $\vec p \times (\vec q \times \vec r) = (\vec p \cdot \vec r) \vec q - (\vec p \cdot \vec q) \vec r$.
• Scalar Multiplication with Cross Product: $k(\vec u \times \vec v) = (k\vec u) \times \vec v = \vec u \times (k\vec v)$.

Step-by-Step Solution

Let the given expression be $E$. $E = \left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$

Step 1: Simplify the innermost cross product $\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b$ We use the distributive property of the cross product. $(\vec a - \vec b) \times \vec b = (\vec a \times \vec b) - (\vec b \times \vec b)$ Since the cross product of a vector with itself is the zero vector, $\vec b \times \vec b = \vec 0$. $(\vec a - \vec b) \times \vec b = \vec a \times \vec b$

Step 2: Simplify the next nested cross product $\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)$ Substituting the result from Step 1: $\vec a \times \left( (\vec a - \vec b) \times \vec b \right) = \vec a \times (\vec a \times \vec b)$ This is a vector triple product. Using the identity $\vec p \times (\vec q \times \vec r) = (\vec p \cdot \vec r) \vec q - (\vec p \cdot \vec q) \vec r$, with $\vec p = \vec a$, $\vec q = \vec a$, and $\vec r = \vec b$: $\vec a \times (\vec a \times \vec b) = (\vec a \cdot \vec b) \vec a - (\vec a \cdot \vec a) \vec b$ We know that $\vec a \cdot \vec a = |\vec a|^2$. $\vec a \times (\vec a \times \vec b) = (\vec a \cdot \vec b) \vec a - |\vec a|^2 \vec b$

Step 3: Simplify the next level of the expression $\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b$ Substituting the result from Step 2: $\left( (\vec a \cdot \vec b) \vec a - |\vec a|^2 \vec b \right) \times \vec b$ Using the distributive property: $= (\vec a \cdot \vec b) (\vec a \times \vec b) - |\vec a|^2 (\vec b \times \vec b)$ Since $\vec b \times \vec b = \vec 0$: $= (\vec a \cdot \vec b) (\vec a \times \vec b)$

Step 4: Simplify the entire expression $E$ Substitute the result from Step 3 back into the original expression: $E = (\vec a + \vec b) \times \left( (\vec a \cdot \vec b) (\vec a \times \vec b) \right)$ Using the property of scalar multiplication with cross products, we can move the scalar $(\vec a \cdot \vec b)$ outside: $E = (\vec a \cdot \vec b) \left[ (\vec a + \vec b) \times (\vec a \times \vec b) \right]$

Step 5: Calculate the necessary vector components and dot products Given $\vec a = \widehat i + \widehat j + 2\widehat k$ and $\vec b = - \widehat i + 2\widehat j + 3\widehat k$.

Calculate $\vec a \cdot \vec b$: $\vec a \cdot \vec b = (1)(-1) + (1)(2) + (2)(3) = -1 + 2 + 6 = 7$

Calculate $\vec a + \vec b$: $\vec a + \vec b = (\widehat i + \widehat j + 2\widehat k) + (- \widehat i + 2\widehat j + 3\widehat k) = (1-1)\widehat i + (1+2)\widehat j + (2+3)\widehat k = 0\widehat i + 3\widehat j + 5\widehat k = 3\widehat j + 5\widehat k$

Calculate $\vec a \times \vec b$: $\vec a \times \vec b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 1 & 2 \\ -1 & 2 & 3 \end{vmatrix} = \widehat i(1 \cdot 3 - 2 \cdot 2) - \widehat j(1 \cdot 3 - 2 \cdot (-1)) + \widehat k(1 \cdot 2 - 1 \cdot (-1))$ $= \widehat i(3 - 4) - \widehat j(3 + 2) + \widehat k(2 + 1) = -\widehat i - 5\widehat j + 3\widehat k$

Step 6: Calculate the final cross product $(\vec a + \vec b) \times (\vec a \times \vec b)$ Let $\vec u = \vec a + \vec b = 3\widehat j + 5\widehat k$ and $\vec v = \vec a \times \vec b = -\widehat i - 5\widehat j + 3\widehat k$. $(\vec a + \vec b) \times (\vec a \times \vec b) = \vec u \times \vec v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 0 & 3 & 5 \\ -1 & -5 & 3 \end{vmatrix}$ $= \widehat i(3 \cdot 3 - 5 \cdot (-5)) - \widehat j(0 \cdot 3 - 5 \cdot (-1)) + \widehat k(0 \cdot (-5) - 3 \cdot (-1))$ $= \widehat i(9 + 25) - \widehat j(0 + 5) + \widehat k(0 + 3)$ $= 34\widehat i - 5\widehat j + 3\widehat k$

Step 7: Substitute the calculated values back into the simplified expression for $E$ From Step 4, $E = (\vec a \cdot \vec b) \left[ (\vec a + \vec b) \times (\vec a \times \vec b) \right]$. $E = (7) (34\widehat i - 5\widehat j + 3\widehat k)$ $E = 7(34\widehat i - 5\widehat j + 3\widehat k)$

This result matches option (B).

Common Mistakes & Tips

• Vector Triple Product Expansion: Ensure you correctly recall and apply the vector triple product identity. Mistakes in the signs or order of dot products are common.
• Order of Cross Products: Remember that the cross product is not commutative ($\vec u \times \vec v \neq \vec v \times \vec u$). Pay close attention to the order of vectors in each cross product.
• Simplification First: Resist the urge to substitute the vector components too early. Simplifying the expression algebraically using vector identities significantly reduces the chance of arithmetic errors.

Summary

The problem involves simplifying a complex vector expression with nested cross products. By systematically applying the distributive property of the cross product and the vector triple product identity, the expression was reduced to a scalar multiple of a cross product. The necessary dot and cross products were then computed using the given vector components, leading to the final simplified form.

The final answer is \boxed{\text{7(34\widehat i - 5\widehat j + 3\widehat k)}}.