Key Concepts and Formulas

• Co-planarity of Vectors: Three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ are co-planar if their scalar triple product is zero, i.e., $[\vec{A} \ \vec{B} \ \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C}) = 0$.
• Scalar Triple Product using Determinants: If $\vec{A} = a_1\widehat i + a_2\widehat j + a_3\widehat k$, $\vec{B} = b_1\widehat i + b_2\widehat j + b_3\widehat k$, and $\vec{C} = c_1\widehat i + c_2\widehat j + c_3\widehat k$, then their scalar triple product is given by the determinant: $[\vec{A} \ \vec{B} \ \vec{C}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$
• Elementary Row/Column Operations: To simplify determinants, we can use operations like $R_i \to R_i + kR_j$ or $C_i \to C_i + kC_j$. These operations do not change the value of the determinant.

Step-by-Step Solution

Step 1: Identify the given vectors and the condition for co-planarity. Let the three given vectors be $\vec{A}$, $\vec{B}$, and $\vec{C}$: $\vec{A} = (2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k$ $\vec{B} = (1 + b)\widehat i + 2b\widehat j - b\widehat k$ $\vec{C} = (2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k$ For these vectors to be co-planar, their scalar triple product must be zero. This can be represented by the determinant of the matrix formed by their components: $\begin{vmatrix} (2 + a + b) & (a + 2b + c) & -(b + c) \\ (1 + b) & 2b & -b \\ (2 + b) & 2b & (1 - b) \end{vmatrix} = 0$

Step 2: Simplify the determinant using elementary row operations. To make the determinant calculation easier, we apply row operations. We aim to create zeros in the determinant. First, let's perform $R_1 \to R_1 - R_2$: The new first row elements will be: $(2 + a + b) - (1 + b) = a + 1$ $(a + 2b + c) - 2b = a + c$ $-(b + c) - (-b) = -b - c + b = -c$

Next, let's perform $R_3 \to R_3 - R_2$: The new third row elements will be: $(2 + b) - (1 + b) = 1$ $2b - 2b = 0$ $(1 - b) - (-b) = 1 - b + b = 1$

The determinant now becomes: $\begin{vmatrix} (a + 1) & (a + c) & -c \\ (b + 1) & 2b & -b \\ 1 & 0 & 1 \end{vmatrix} = 0$

Step 3: Expand the simplified determinant. We expand the determinant along the third row ($R_3$) because it contains a zero, which simplifies the expansion. The expansion is given by: $1 \cdot \begin{vmatrix} (a + c) & -c \\ 2b & -b \end{vmatrix} - 0 \cdot \begin{vmatrix} (a + 1) & -c \\ (b + 1) & -b \end{vmatrix} + 1 \cdot \begin{vmatrix} (a + 1) & (a + c) \\ (b + 1) & 2b \end{vmatrix} = 0$

Step 4: Calculate the $2 \times 2$ determinants and simplify the equation. The equation simplifies to: $\begin{vmatrix} (a + c) & -c \\ 2b & -b \end{vmatrix} + \begin{vmatrix} (a + 1) & (a + c) \\ (b + 1) & 2b \end{vmatrix} = 0$ Let's evaluate each determinant: The first determinant: $(a + c)(-b) - (-c)(2b) = -ab - cb + 2bc = -ab + bc$ The second determinant: $(a + 1)(2b) - (a + c)(b + 1) = (2ab + 2b) - (ab + a + cb + c)$ $= 2ab + 2b - ab - a - cb - c$ $= ab + 2b - a - cb - c$

Now, substitute these back into the equation: $(-ab + bc) + (ab + 2b - a - cb - c) = 0$

Step 5: Combine like terms and derive the final relation. Combine the terms: $-ab + ab + bc - cb + 2b - a - c = 0$ $0 + 0 + 2b - a - c = 0$ $2b - a - c = 0$ Rearranging this equation, we get: $2b = a + c$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating determinants and performing algebraic manipulations. A single sign error can lead to an incorrect result.
• Determinant Expansion: Expanding along a row or column with the most zeros significantly reduces the complexity and potential for errors.
• Strategic Row Operations: Before calculating, look for ways to create zeros using row/column operations. This is a critical time-saving technique in competitive exams.

Summary

The problem requires applying the condition for co-planarity of three vectors, which is that their scalar triple product must be zero. This condition is mathematically expressed as a determinant of the vectors' components being equal to zero. By strategically using elementary row operations to simplify the determinant and then expanding it, we arrived at a linear relationship between $a$, $b$, and $c$. The simplification process was key to avoiding complex algebraic calculations.

The final answer is $\boxed{\text{2b = a + c}}$.