Key Concepts and Formulas

• Condition for Coplanarity of Two Lines: Two lines in 3D space are coplanar if and only if the vector connecting a point on the first line to a point on the second line ($\vec{P_1P_2}$) is coplanar with the direction vectors of the two lines ($\vec{d_1}$ and $\vec{d_2}$). Mathematically, their scalar triple product must be zero: $\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2}) = 0$. If the lines are given in symmetric form: Line 1: $\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$ Line 2: $\frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2}$ The coplanarity condition is expressed as the determinant: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
• Converting Non-Standard Line Equations: A line given in the form $x = Ay+B = Cz+D$ represents the intersection of two planes.
  • To find a point on the line, we can set one variable (e.g., $x=0$) and solve for the other two.
  • To find the direction ratios of the line, we recognize that the line is perpendicular to the normal vectors of both intersecting planes. Thus, its direction vector is proportional to the cross product of the normal vectors of these two planes. If the planes are $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$, their normal vectors are $\vec{n_1}=(A_1, B_1, C_1)$ and $\vec{n_2}=(A_2, B_2, C_2)$. The direction vector of the line is $\vec{d} = \vec{n_1} \times \vec{n_2}$.

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Step-by-Step Solution

The problem asks for the condition under which two given lines are coplanar. The lines are not in the standard symmetric (Cartesian) form, so our first task is to convert them to identify a point on each line and their respective direction ratios.

Step 1: Convert Line 1 to Standard Form and Extract Information

The first line is given as $x = ay - 1 = z - 2$. This can be broken down into two equations, representing two planes whose intersection forms the line:

1. $x = ay - 1 \implies x - ay + 1 = 0$
2. $x = z - 2 \implies x - z + 2 = 0$

• Finding a Point on Line 1 ($P_1$): To find a specific point on the line, we can set one coordinate to zero and solve for the others. Let's set $x=0$: From $x = ay - 1 \implies 0 = ay - 1 \implies ay = 1$. Since $ab \ne 0$, we know $a \ne 0$, so $y = \frac{1}{a}$. From $x = z - 2 \implies 0 = z - 2 \implies z = 2$. Thus, a point on Line 1 is $P_1(x_1, y_1, z_1) = (0, \frac{1}{a}, 2)$.

• Finding the Direction Ratios of Line 1 ($\vec{d_1}$): The direction vector of the line is perpendicular to the normal vectors of both planes that define it. The normal vector of the first plane ($x - ay + 1 = 0$) is $\vec{n_A} = (1, -a, 0)$. The normal vector of the second plane ($x - z + 2 = 0$) is $\vec{n_B} = (1, 0, -1)$. The direction vector $\vec{d_1}$ is proportional to the cross product of $\vec{n_A}$ and $\vec{n_B}$: $\vec{d_1} = \vec{n_A} \times \vec{n_B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -a & 0 \\ 1 & 0 & -1 \end{vmatrix}$ Expanding the determinant: $\vec{d_1} = \mathbf{i}((-a)(-1) - 0(0)) - \mathbf{j}(1(-1) - 0(1)) + \mathbf{k}(1(0) - (-a)(1))$ $\vec{d_1} = a\mathbf{i} + \mathbf{j} + a\mathbf{k}$ So, the direction ratios for Line 1 are $(l_1, m_1, n_1) = (a, 1, a)$.

Step 2: Convert Line 2 to Standard Form and Extract Information

The second line is given as $x = 3y - 2 = bz - 2$. This also represents the intersection of two planes:

1. $x = 3y - 2 \implies x - 3y + 2 = 0$
2. $x = bz - 2 \implies x - bz + 2 = 0$

• Finding a Point on Line 2 ($P_2$): Again, we set $x=0$: From $x = 3y - 2 \implies 0 = 3y - 2 \implies 3y = 2 \implies y = \frac{2}{3}$. From $x = bz - 2 \implies 0 = bz - 2 \implies bz = 2$. Since $ab \ne 0$, we know $b \ne 0$, so $z = \frac{2}{b}$. Thus, a point on Line 2 is $P_2(x_2, y_2, z_2) = (0, \frac{2}{3}, \frac{2}{b})$.

• Finding the Direction Ratios of Line 2 ($\vec{d_2}$): The normal vector of the first plane ($x - 3y + 2 = 0$) is $\vec{n_C} = (1, -3, 0)$. The normal vector of the second plane ($x - bz + 2 = 0$) is $\vec{n_D} = (1, 0, -b)$. The direction vector $\vec{d_2}$ is proportional to the cross product of $\vec{n_C}$ and $\vec{n_D}$: $\vec{d_2} = \vec{n_C} \times \vec{n_D} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 0 \\ 1 & 0 & -b \end{vmatrix}$ Expanding the determinant: $\vec{d_2} = \mathbf{i}((-3)(-b) - 0(0)) - \mathbf{j}(1(-b) - 0(1)) + \mathbf{k}(1(0) - (-3)(1))$ $\vec{d_2} = 3b\mathbf{i} + b\mathbf{j} + 3\mathbf{k}$ So, the direction ratios for Line 2 are $(l_2, m_2, n_2) = (3b, b, 3)$.

• Finding the Vector $\vec{P_1P_2}$: Now we calculate the vector connecting $P_1(0, \frac{1}{a}, 2)$ to $P_2(0, \frac{2}{3}, \frac{2}{b})$: $\vec{P_1P_2} = (x_2-x_1, y_2-y_1, z_2-z_1) = \left(0-0, \frac{2}{3} - \frac{1}{a}, \frac{2}{b} - 2\right)$ $\vec{P_1P_2} = \left(0, \frac{2a-3}{3a}, \frac{2-2b}{b}\right)$

Step 3: Apply the Coplanarity Condition

Now we substitute the components of $\vec{P_1P_2}$, $\vec{d_1}$, and $\vec{d_2}$ into the determinant condition for coplanarity: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ $\begin{vmatrix} 0 & \frac{2a-3}{3a} & \frac{2-2b}{b} \\ a & 1 & a \\ 3b & b & 3 \end{vmatrix} = 0$ Expand the determinant along the first row: $0 \cdot (1 \cdot 3 - a \cdot b) - \left(\frac{2a-3}{3a}\right) \cdot (a \cdot 3 - a \cdot 3b) + \left(\frac{2-2b}{b}\right) \cdot (a \cdot b - 1 \cdot 3b) = 0$ Simplify the expression: $- \left(\frac{2a-3}{3a}\right) \cdot (3a - 3ab) + \left(\frac{2-2b}{b}\right) \cdot (ab - 3b) = 0$

Step 4: Solve for the Relationship between $a$ and $b$

Factor out common terms from the parentheses: $- \left(\frac{2a-3}{3a}\right) \cdot 3a(1 - b) + \left(\frac{2-2b}{b}\right) \cdot b(a - 3) = 0$ Since $ab \ne 0$, we know $a \ne 0$ and $b \ne 0$. This allows us to cancel $3a$ from the first term and $b$ from the second term: $- (2a-3)(1 - b) + (2-2b)(a - 3) = 0$ Notice that $2-2b = 2(1-b)$. Substitute this into the equation: $- (2a-3)(1 - b) + 2(1-b)(a - 3) = 0$ Now, $(1-b)$ is a common factor in both terms. Factor it out: $(1-b) [-(2a-3) + 2(a-3)] = 0$ Expand the terms inside the square brackets: $(1-b) [-2a + 3 + 2a - 6] = 0$ $(1-b) [-3] = 0$ $-3(1-b) = 0$ This equation implies that $1-b=0$. Therefore, $b=1$.

The problem states that $ab \ne 0$. Since we found $b=1$, this condition becomes $a(1) \ne 0$, which means $a \ne 0$. So, the lines are coplanar if $b=1$ and $a$ can be any real number except $0$.

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Common Mistakes & Tips

• Incorrectly Identifying Points and Direction Ratios: The most common error is misinterpreting the non-standard form of the line equations. Always take the time to convert them into a form where a point and direction ratios are clearly visible.
• Algebraic Errors in Determinant Expansion: Be meticulous with signs and multiplication when expanding the $3 \times 3$ determinant. Factoring common terms early can prevent larger, more complex expressions.
• Forgetting Constraints: The condition $ab \ne 0$ is important. It ensures that $a$ and $b$ are non-zero, validating divisions by $a$ and $b$ and affecting the final range of values for $a$.

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Summary

To determine the condition for coplanarity, we first converted the given lines from their non-standard form to identify a point and direction vector for each line. This involved setting one coordinate to zero to find a point and using the cross product of the normal vectors of the defining planes to find the direction vector. Once these components were extracted, we applied the coplanarity condition, which states that the determinant formed by the vector connecting the two points and the two direction vectors must be zero. Solving the resulting algebraic equation led to the condition $b=1$. Combined with the given constraint $ab \ne 0$, we concluded that $a$ can be any non-zero real number.

The final answer is $\boxed{b = 1, a \in R - \{0\}}$, which corresponds to option (A).