1. Key Concepts and Formulas

• Vector Triple Product Identity: For any three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$, the identity is given by: $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$
• Dot Product Properties: The dot product of a vector with itself is the square of its magnitude: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.
• Cross Product Properties: The cross product of a vector with itself is the zero vector: $\vec{a} \times \vec{a} = \vec{0}$. The basic cross products of unit vectors are: $\widehat{i} \times \widehat{j} = \widehat{k}$, $\widehat{j} \times \widehat{k} = \widehat{i}$, $\widehat{k} \times \widehat{i} = \widehat{j}$, and $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.

2. Step-by-Step Solution

Step 1: Utilize the Vector Triple Product Identity Our goal is to find $\vec{a} - 6\vec{b}$. We are given $\vec{a}$, $\vec{a} \cdot \vec{b}$, and $\vec{a} \times \vec{b}$. A common strategy when encountering $\vec{a}$ and $\vec{a} \times \vec{b}$ is to take the cross product of $\vec{a}$ with the given cross product $\vec{a} \times \vec{b}$. This allows us to apply the vector triple product identity. We are given $\vec{a} \times \vec{b} = \widehat{i} - \widehat{j}$. Taking the cross product of $\vec{a}$ with this equation: $\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times (\widehat{i} - \widehat{j})$ Now, we apply the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$ to the left side, with $\vec{A} = \vec{a}$, $\vec{B} = \vec{a}$, and $\vec{C} = \vec{b}$. $(\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b} = \vec{a} \times (\widehat{i} - \widehat{j})$

Step 2: Calculate the Right-Hand Side (RHS) of the Equation We need to compute $\vec{a} \times (\widehat{i} - \widehat{j})$. Given $\vec{a} = - \widehat{i} - \widehat{j} + \widehat{k}$: $\vec{a} \times (\widehat{i} - \widehat{j}) = (-\widehat{i} - \widehat{j} + \widehat{k}) \times (\widehat{i} - \widehat{j})$ Expanding this cross product: $= (-\widehat{i} \times \widehat{i}) + (-\widehat{i} \times -\widehat{j}) + (-\widehat{j} \times \widehat{i}) + (-\widehat{j} \times -\widehat{j}) + (\widehat{k} \times \widehat{i}) + (\widehat{k} \times -\widehat{j})$ Using the properties of cross products ($\widehat{i} \times \widehat{i} = \vec{0}$, $\widehat{i} \times \widehat{j} = \widehat{k}$, $\widehat{j} \times \widehat{i} = -\widehat{k}$, $\widehat{k} \times \widehat{i} = \widehat{j}$, $\widehat{k} \times \widehat{j} = -\widehat{i}$): $= \vec{0} + \widehat{k} - (-\widehat{k}) + \vec{0} + \widehat{j} - (-\widehat{i})$ $= \widehat{k} + \widehat{k} + \widehat{j} + \widehat{i}$ $= \widehat{i} + \widehat{j} + 2\widehat{k}$

Step 3: Calculate the Scalar Dot Products We need the values of $\vec{a} \cdot \vec{b}$ and $\vec{a} \cdot \vec{a}$. We are given $\vec{a} \cdot \vec{b} = 1$. For $\vec{a} \cdot \vec{a}$: $\vec{a} \cdot \vec{a} = (-1)(-1) + (-1)(-1) + (1)(1) = 1 + 1 + 1 = 3$

Step 4: Formulate and Solve for $\vec{b}$ Substitute the calculated dot products and the RHS into the equation from Step 1: $(1)\vec{a} - (3)\vec{b} = \widehat{i} + \widehat{j} + 2\widehat{k}$ $\vec{a} - 3\vec{b} = \widehat{i} + \widehat{j} + 2\widehat{k}$ We want to find $\vec{a} - 6\vec{b}$. We can rearrange the equation to isolate $3\vec{b}$: $3\vec{b} = \vec{a} - (\widehat{i} + \widehat{j} + 2\widehat{k})$ To find $6\vec{b}$, we multiply both sides by 2: $6\vec{b} = 2\left( \vec{a} - (\widehat{i} + \widehat{j} + 2\widehat{k}) \right)$ $6\vec{b} = 2\vec{a} - 2\widehat{i} - 2\widehat{j} - 4\widehat{k}$

Step 5: Calculate the Target Expression $\vec{a} - 6\vec{b}$ Now, substitute the expression for $6\vec{b}$ into the expression we need to find: $\vec{a} - 6\vec{b} = \vec{a} - (2\vec{a} - 2\widehat{i} - 2\widehat{j} - 4\widehat{k})$ $\vec{a} - 6\vec{b} = \vec{a} - 2\vec{a} + 2\widehat{i} + 2\widehat{j} + 4\widehat{k}$ $\vec{a} - 6\vec{b} = -\vec{a} + 2\widehat{i} + 2\widehat{j} + 4\widehat{k}$ Substitute the given value of $\vec{a} = - \widehat{i} - \widehat{j} + \widehat{k}$: $\vec{a} - 6\vec{b} = -(-\widehat{i} - \widehat{j} + \widehat{k}) + 2\widehat{i} + 2\widehat{j} + 4\widehat{k}$ $\vec{a} - 6\vec{b} = (\widehat{i} + \widehat{j} - \widehat{k}) + 2\widehat{i} + 2\widehat{j} + 4\widehat{k}$ Combine like terms: $\vec{a} - 6\vec{b} = (1+2)\widehat{i} + (1+2)\widehat{j} + (-1+4)\widehat{k}$ $\vec{a} - 6\vec{b} = 3\widehat{i} + 3\widehat{j} + 3\widehat{k}$ Factoring out 3: $\vec{a} - 6\vec{b} = 3(\widehat{i} + \widehat{j} + \widehat{k})$ This result matches option (A).

3. Common Mistakes & Tips

• Vector Triple Product Order: Be extremely careful with the order of vectors in the vector triple product identity. The correct form is $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. Swapping the dot product terms will lead to an incorrect answer.
• Cross Product Arithmetic: When calculating cross products, especially with negative components and multiple terms, carefully apply the rules for unit vectors and the anti-commutative property ($\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$).
• Algebraic Simplification: Ensure accurate manipulation of vector equations. When solving for $6\vec{b}$, double-check that the signs are handled correctly during subtraction and multiplication.

4. Summary

The problem was solved by strategically applying the vector triple product identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$. By taking the cross product of $\vec{a}$ with the given $\vec{a} \times \vec{b}$, we formed an equation relating $\vec{a}$ and $\vec{b}$. After calculating the necessary dot products and the cross product on the right side, we rearranged the equation to find an expression for $6\vec{b}$. Substituting this into the target expression $\vec{a} - 6\vec{b}$ and simplifying yielded the final answer.

The final answer is $\boxed{3\left( {\widehat i + \widehat j + \widehat k} \right)}$.