Key Concepts and Formulas

1. Symmetric Form of a Line in 3D: A line passing through a point $(x_1, y_1, z_1)$ and having direction ratios (DRs) $(l, m, n)$ can be expressed in its symmetric form as: $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ The triplet $(l, m, n)$ represents the direction vector of the line. Any scalar multiple of $(l, m, n)$ also represents the direction ratios of the same line.

2. Condition for Perpendicularity of Two Lines: Two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular to each other if and only if the dot product of their direction vectors is zero. Mathematically, this condition is: $l_1l_2 + m_1m_2 + n_1n_2 = 0$

Step-by-Step Solution

Step 1: Understand the given equations for the lines. The given equations for both lines are in a non-standard form, where $y$ acts as a common parameter relating $x$ and $z$. Our goal is to convert these into the symmetric form to easily extract their direction ratios.

Step 2: Convert the first line's equations to symmetric form and identify its Direction Ratios. The first line is given by: $x = ay+b \quad \cdots (1)$ $z = cy+d \quad \cdots (2)$ From equation (1), we can rearrange to isolate $y$: $x-b = ay$ Assuming $a \neq 0$, we can write: $\frac{x-b}{a} = y \quad \cdots (3)$ Similarly, from equation (2): $z-d = cy$ Assuming $c \neq 0$, we can write: $\frac{z-d}{c} = y \quad \cdots (4)$ Since both expressions in (3) and (4) are equal to $y$, and $y$ itself can be written as $\frac{y}{1}$, we can combine them to get the symmetric form of the first line: $\frac{x-b}{a} = \frac{y}{1} = \frac{z-d}{c}$ From this symmetric form, we can directly identify the direction ratios of the first line, $(l_1, m_1, n_1) = (a, 1, c)$.

Step 3: Convert the second line's equations to symmetric form and identify its Direction Ratios. The second line is given by: $x = a'y+b' \quad \cdots (5)$ $z = c'y+d' \quad \cdots (6)$ Following the same procedure as for the first line: From equation (5), assuming $a' \neq 0$: $\frac{x-b'}{a'} = y \quad \cdots (7)$ From equation (6), assuming $c' \neq 0$: $\frac{z-d'}{c'} = y \quad \cdots (8)$ Combining (7) and (8) with $\frac{y}{1}$, we get the symmetric form of the second line: $\frac{x-b'}{a'} = \frac{y}{1} = \frac{z-d'}{c'}$ From this, we identify the direction ratios of the second line, $(l_2, m_2, n_2) = (a', 1, c')$.

Step 4: Apply the condition for perpendicularity. Now that we have the direction ratios for both lines, $(l_1, m_1, n_1) = (a, 1, c)$ and $(l_2, m_2, n_2) = (a', 1, c')$, we can apply the perpendicularity condition: $l_1l_2 + m_1m_2 + n_1n_2 = 0$ Substitute the respective direction ratios: $(a)(a') + (1)(1) + (c)(c') = 0$ Simplify the equation: $aa' + 1 + cc' = 0$ Rearrange the terms to find the required condition: $aa' + cc' = -1$

Common Mistakes & Tips

• Always Convert to Standard Form: The most common mistake is to try to infer direction ratios directly from non-standard forms. Always convert the line equations to either symmetric form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ or vector form $\vec{r} = \vec{a} + \lambda \vec{v}$ to correctly identify the direction vector $\vec{v}=(l,m,n)$.
• Alternative Method (Cross Product of Normal Vectors): A line given by the intersection of two planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ has a direction vector parallel to the cross product of the normal vectors of the planes. For the first line: $x - ay - b = 0 \implies \vec{n_1} = (1, -a, 0)$ $z - cy - d = 0 \implies \vec{n_2} = (0, -c, 1)$ The direction vector $\vec{v_1} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -a & 0 \\ 0 & -c & 1 \end{vmatrix} = \mathbf{i}(-a) - \mathbf{j}(1) + \mathbf{k}(-c) = (-a, -1, -c)$. The direction ratios are $(-a, -1, -c)$, which are proportional to $(a, 1, c)$, confirming our result. This method is useful for verifying or when direct conversion to symmetric form is complex.
• Handling Zero Denominators: In Step 2 and 3, we assumed $a, c, a', c'$ are non-zero for division. If, for example, $a=0$, the first line equation becomes $x=b$. This means the line lies in the plane $x=b$, implying its direction vector has an x-component of 0. The symmetric form $\frac{x-b}{0} = \frac{y}{1} = \frac{z-d}{c}$ is understood to mean $x-b=0$ and $\frac{y}{1} = \frac{z-d}{c}$. The direction ratios are indeed $(0, 1, c)$. If $a=0$, our derived condition $aa'+cc'=-1$ becomes $0 \cdot a' + cc' = -1 \implies cc' = -1$. This means the formula holds true even for these special cases where some coefficients are zero.

Summary

This problem required us to find the condition for two lines to be perpendicular. The key was to first convert the given non-standard equations of the lines into their symmetric form to easily extract their direction ratios. Once the direction ratios $(a, 1, c)$ and $(a', 1, c')$ were identified, we applied the perpendicularity condition $l_1l_2 + m_1m_2 + n_1n_2 = 0$. This led to the condition $aa' + 1 + cc' = 0$, which simplifies to $aa' + cc' = -1$.

The final answer is $\boxed{\text{aa'+cc'=-1}}$ which corresponds to option (A).