Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar if their scalar triple product is zero: $\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \overrightarrow a \cdot (\overrightarrow b \times \overrightarrow c) = 0$. This can be calculated as the determinant of the matrix formed by their components.
• Scalar Triple Product: For vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$, $\overrightarrow b = b_1\widehat i + b_2\widehat j + b_3\widehat k$, and $\overrightarrow c = c_1\widehat i + c_2\widehat j + c_3\widehat k$, the scalar triple product is: $\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \left| {\begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix}} \right|$
• Cross Product: The cross product of two vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$ and $\overrightarrow c = c_1\widehat i + c_2\widehat j + c_3\widehat k$ is given by: $\overrightarrow a \times \overrightarrow c = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{matrix}} \right|$
• Condition for Parallel Vectors: If $\overrightarrow a \times \overrightarrow c = \overrightarrow 0$, then $\overrightarrow a$ and $\overrightarrow c$ are parallel (or one of them is a zero vector).

Step-by-Step Solution

Step 1: Apply the coplanarity condition. The problem states that vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar. This means their scalar triple product must be zero. We will use this condition to find the possible values of $\lambda$. Given vectors are: $\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k$ $\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$ $\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$ The scalar triple product is: $\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \left| {\begin{matrix} 1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & {{\lambda ^2} - 1} \end{matrix}} \right| = 0$

Step 2: Expand the determinant and solve for $\lambda$. We expand the determinant along the first row: $1 \cdot \left| {\begin{matrix} \lambda & 4 \\ 4 & {{\lambda ^2} - 1} \end{matrix}} \right| - 2 \cdot \left| {\begin{matrix} 1 & 4 \\ 2 & {{\lambda ^2} - 1} \end{matrix}} \right| + 4 \cdot \left| {\begin{matrix} 1 & \lambda \\ 2 & 4 \end{matrix}} \right| = 0$ Calculate the $2 \times 2$ determinants: $1 \cdot \left( {\lambda ({{\lambda ^2} - 1}) - 4 \cdot 4} \right) - 2 \cdot \left( {1 \cdot ({{\lambda ^2} - 1}) - 4 \cdot 2} \right) + 4 \cdot \left( {1 \cdot 4 - \lambda \cdot 2} \right) = 0$ $\left( {{\lambda ^3} - \lambda - 16} \right) - 2 \left( {{\lambda ^2} - 1 - 8} \right) + 4 \left( {4 - 2\lambda} \right) = 0$ ${\lambda ^3} - \lambda - 16 - 2 \left( {{\lambda ^2} - 9} \right) + 16 - 8\lambda = 0$ ${\lambda ^3} - \lambda - 16 - 2{\lambda ^2} + 18 + 16 - 8\lambda = 0$ Combine like terms to form a cubic equation: ${\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$ Factor the cubic equation by grouping: ${\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$ $\left( {{\lambda ^2} - 9} \right)\left( {\lambda - 2} \right) = 0$ Factor the difference of squares: $\left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$ The possible values for $\lambda$ are $\lambda = 2$, $\lambda = 3$, and $\lambda = -3$.

Step 3: Use the "non-zero vector" condition to select the correct value of $\lambda$. The problem requires the vector $\overrightarrow a \times \overrightarrow c$ to be non-zero. This implies that $\overrightarrow a$ and $\overrightarrow c$ must not be parallel. We need to check for which values of $\lambda$ this condition is violated. Two vectors are parallel if one is a scalar multiple of the other. Let's assume $\overrightarrow a = k \overrightarrow c$ for some scalar $k$. $\widehat i + 2\widehat j + 4\widehat k = k \left( 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k \right)$ Comparing the coefficients of $\widehat i$: $1 = 2k \implies k = \frac{1}{2}$. Comparing the coefficients of $\widehat j$: $2 = 4k \implies k = \frac{1}{2}$. Now, comparing the coefficients of $\widehat k$: $4 = k \left( {{\lambda ^2} - 1} \right)$. Substitute $k = \frac{1}{2}$: $4 = \frac{1}{2} \left( {{\lambda ^2} - 1} \right)$ $8 = {\lambda ^2} - 1$ ${\lambda ^2} = 9$ $\lambda = \pm 3$ This means that if $\lambda = 3$ or $\lambda = -3$, vectors $\overrightarrow a$ and $\overrightarrow c$ are parallel, and their cross product $\overrightarrow a \times \overrightarrow c$ would be the zero vector. Since the problem states that $\overrightarrow a \times \overrightarrow c$ is a non-zero vector, we must exclude $\lambda = 3$ and $\lambda = -3$. Therefore, the only valid value for $\lambda$ is $\lambda = 2$.

Step 4: Calculate the cross product $\overrightarrow a \times \overrightarrow c$ with the selected value of $\lambda$. Substitute $\lambda = 2$ into the vector $\overrightarrow c$: $\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{2^2} - 1} \right)\widehat k = 2\widehat i + 4\widehat j + \left( {4 - 1} \right)\widehat k = 2\widehat i + 4\widehat j + 3\widehat k$ Now, calculate the cross product $\overrightarrow a \times \overrightarrow c$: $\overrightarrow a \times \overrightarrow c = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{matrix}} \right|$ Expand the determinant: $\overrightarrow a \times \overrightarrow c = \widehat i \left( {2 \cdot 3 - 4 \cdot 4} \right) - \widehat j \left( {1 \cdot 3 - 4 \cdot 2} \right) + \widehat k \left( {1 \cdot 4 - 2 \cdot 2} \right)$ $\overrightarrow a \times \overrightarrow c = \widehat i \left( {6 - 16} \right) - \widehat j \left( {3 - 8} \right) + \widehat k \left( {4 - 4} \right)$ $\overrightarrow a \times \overrightarrow c = -10\widehat i - (-5)\widehat j + 0\widehat k$ $\overrightarrow a \times \overrightarrow c = -10\widehat i + 5\widehat j$ This is a non-zero vector, as required by the problem statement.

Common Mistakes & Tips

• Forgetting the "non-zero" condition: This is a critical condition that eliminates some potential values of $\lambda$. Always check if the calculated cross product is indeed non-zero.
• Algebraic errors in determinant expansion: Careless mistakes in expanding determinants can lead to incorrect values of $\lambda$. Double-check your calculations.
• Factoring cubic equations: If factoring by grouping is not obvious, remember to test integer divisors of the constant term to find roots, then use polynomial division.

Summary

The problem involves three coplanar vectors, which implies their scalar triple product is zero. By setting up the determinant of the components and solving the resulting cubic equation, we found three possible values for $\lambda$. The condition that the cross product $\overrightarrow a \times \overrightarrow c$ must be non-zero allowed us to eliminate two of these values, leaving $\lambda = 2$ as the only valid solution. Finally, we computed the cross product $\overrightarrow a \times \overrightarrow c$ using $\lambda = 2$, yielding the result $-10\widehat i + 5\widehat j$.

The final answer is \boxed{-10\widehat i + 5\widehat j}, which corresponds to option (B).