Key Concepts and Formulas

• Vector Triple Product (BAC-CAB Rule): For any three vectors $\overrightarrow{\mathrm{A}}$, $\overrightarrow{\mathrm{B}}$, and $\overrightarrow{\mathrm{C}}$, the vector triple product is given by: $\overrightarrow{\mathrm{A}} \times (\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}) = (\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}})\overrightarrow{\mathrm{B}} - (\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})\overrightarrow{\mathrm{C}}$
• Linear Independence of Vectors: If two vectors $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}$ are non-parallel (and non-zero), they are linearly independent. This means that if $p\overrightarrow{\mathrm{u}} + q\overrightarrow{\mathrm{v}} = \overrightarrow{\mathrm{0}}$, then it must be that $p=0$ and $q=0$.
• Dot Product: For vectors $\overrightarrow{\mathrm{x}} = x_1\hat{i} + x_2\hat{j} + x_3\hat{k}$ and $\overrightarrow{\mathrm{y}} = y_1\hat{i} + y_2\hat{j} + y_3\hat{k}$, the dot product is $\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}} = x_1y_1 + x_2y_2 + x_3y_3$.

Step-by-Step Solution

Step 1: Apply the Vector Triple Product Formula

• Why: The left-hand side (LHS) of the given equation is a vector triple product. Applying the BAC-CAB rule is the standard method to simplify such expressions.
• We are given the equation: $\overrightarrow{\mathrm{a}} \times (\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) = \overrightarrow{\mathrm{b}} + \lambda \overrightarrow{\mathrm{c}}$
• Applying the BAC-CAB rule to the LHS, with $\overrightarrow{\mathrm{A}} = \overrightarrow{\mathrm{a}}$, $\overrightarrow{\mathrm{B}} = \overrightarrow{\mathrm{b}}$, and $\overrightarrow{\mathrm{C}} = \overrightarrow{\mathrm{c}}$: $\overrightarrow{\mathrm{a}} \times (\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) = (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}} - (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}$

Step 2: Equate the Expanded LHS with the RHS

• Why: By equating the simplified LHS with the given RHS, we create an equation that relates the vectors $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$, allowing us to solve for $\lambda$.
• Substituting the result from Step 1 into the given equation: $(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}} - (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{b}} + \lambda \overrightarrow{\mathrm{c}}$

Step 3: Rearrange the Equation and Utilize Linear Independence

• Why: To use the property of linear independence, we need to express the equation in the form $p\overrightarrow{\mathrm{b}} + q\overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{0}}$. The problem states that $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel, which means they are linearly independent. This allows us to equate the coefficients of $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ to zero.
• Rearranging the equation from Step 2: $(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{b}} - (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}} - \lambda \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{0}}$
• Factoring out $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$: $(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} - 1)\overrightarrow{\mathrm{b}} + (-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} - \lambda)\overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{0}}$
• Since $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel, they are linearly independent. Therefore, the coefficients must be zero:
  1. Coefficient of $\overrightarrow{\mathrm{b}}$: $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} - 1 = 0 \implies \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} = 1$
  2. Coefficient of $\overrightarrow{\mathrm{c}}$: $-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} - \lambda = 0 \implies \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = -\lambda$

Step 4: Calculate the Dot Product $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$

• Why: From Step 3, we found a direct relationship between $\lambda$ and the dot product $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$. Calculating this dot product will allow us to find the value of $\lambda$.
• We are given $\overrightarrow{\mathrm{a}} = 3\hat{i} + \hat{j}$ and $\overrightarrow{\mathrm{b}} = \hat{i} + 2\hat{j} + \hat{k}$. We can write $\overrightarrow{\mathrm{a}}$ as $3\hat{i} + \hat{j} + 0\hat{k}$ to include the $\hat{k}$ component for the dot product calculation.
• Calculating the dot product: $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = (3)(1) + (1)(2) + (0)(1)$ $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = 3 + 2 + 0$ $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = 5$

Step 5: Determine the Value of $\lambda$

• Why: We have established the equation $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = -\lambda$ in Step 3 and calculated $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$ in Step 4. Substituting the calculated value will give us the value of $\lambda$.
• From Step 3: $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = -\lambda$
• From Step 4: $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} = 5$
• Equating these: $5 = -\lambda$ $\lambda = -5$

Common Mistakes & Tips

• Vector Triple Product Formula Errors: Ensure you have the BAC-CAB rule memorized correctly, especially the order of dot products and the signs. A common mistake is to swap the roles of $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ or to misplace the dot products.
• Misinterpreting Linear Independence: The condition that $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel is crucial. If this condition were absent, or if they were parallel, the method of equating coefficients to zero would not be applicable in this direct manner. Always verify conditions like this.
• Arithmetic Errors: Double-check all calculations, especially dot products. A simple sign error or miscalculation can lead to an incorrect final answer.

Summary

This problem is solved by first simplifying the vector triple product on the LHS using the BAC-CAB rule. Subsequently, the equation is rearranged to group terms involving $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$. Leveraging the given condition that $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel, we establish that they are linearly independent, allowing us to equate the coefficients of $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ to zero. This leads to two equations, one of which directly provides the value of $\lambda$ after calculating the dot product $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$.

The final answer is $\boxed{-5}$.