Key Concepts and Formulas

• Equation of a Line in Symmetric Form: A line in three-dimensional space can be represented by its symmetric form: $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ Here, $(x_1, y_1, z_1)$ is a point on the line, and $(l, m, n)$ are its direction ratios. The direction ratios $(l, m, n)$ represent the components of a vector parallel to the line, often called the direction vector.

• Condition for Perpendicularity of Two Lines: If two lines have direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively, they are perpendicular if and only if the dot product of their direction vectors is zero: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$

Step-by-Step Solution

Step 1: Determine the direction ratios for the first line. The first line is given by the equations: $x = ay + b \quad \text{(Equation 1.1)}$ $z = cy + d \quad \text{(Equation 1.2)}$ Our goal is to convert these equations into the symmetric form $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$. This form directly reveals the direction ratios $(l, m, n)$. We can achieve this by expressing $y$ in terms of $x$ and $z$ from the given equations.

From Equation 1.1, isolate $y$: $x - b = ay$ Assuming $a \neq 0$, we can write: $y = \frac{x - b}{a} \quad \text{(Equation 1.3)}$ If $a=0$, then $x=b$, which means the line lies in the plane $x=b$. In this case, the x-component of the direction vector would be 0. We will address this general case using the symmetric form notation.

From Equation 1.2, isolate $y$: $z - d = cy$ Assuming $c \neq 0$, we can write: $y = \frac{z - d}{c} \quad \text{(Equation 1.4)}$ Similarly, if $c=0$, $z=d$, and the z-component of the direction vector would be 0.

Since both expressions are equal to $y$, and $y$ itself can be written as $\frac{y}{1}$, we can equate them to obtain the symmetric form of the first line: $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$ From this symmetric form, the direction ratios of the first line, $(l_1, m_1, n_1)$, are the denominators: $(l_1, m_1, n_1) = (a, 1, c)$ Explanation: By making $y$ the common parameter, we effectively align the components that scale with changes in $x, y, z$ along the line, which are precisely the direction ratios.

Step 2: Determine the direction ratios for the second line. The second line is given by the equations: $x = a'y + b' \quad \text{(Equation 2.1)}$ $z = c'y + d' \quad \text{(Equation 2.2)}$ We follow the exact same procedure as for the first line.

From Equation 2.1, isolate $y$: $x - b' = a'y$ $y = \frac{x - b'}{a'} \quad \text{(Equation 2.3)}$

From Equation 2.2, isolate $y$: $z - d' = c'y$ $y = \frac{z - d'}{c'} \quad \text{(Equation 2.4)}$

Equating these expressions for $y$ (and $\frac{y}{1}$), we get the symmetric form of the second line: $\frac{x - b'}{a'} = \frac{y}{1} = \frac{z - d'}{c'}$ From this symmetric form, the direction ratios of the second line, $(l_2, m_2, n_2)$, are: $(l_2, m_2, n_2) = (a', 1, c')$ Explanation: This step ensures we correctly identify the orientation of the second line in 3D space, independent of its specific position.

Step 3: Apply the condition for perpendicularity. For the two lines to be perpendicular, the dot product of their direction vectors must be zero. Using the direction ratios found in Step 1 and Step 2: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ Substitute the identified direction ratios $(a, 1, c)$ and $(a', 1, c')$: $(a)(a') + (1)(1) + (c)(c') = 0$ $aa' + 1 + cc' = 0$ Rearranging the terms, we get the condition: $aa' + cc' + 1 = 0$ Explanation: This is the direct application of the fundamental geometric condition for two vectors (and thus two lines they represent) to be orthogonal. The constants $b, d, b', d'$ do not influence the direction of the lines, hence they do not appear in the perpendicularity condition.

Common Mistakes & Tips

• Parametric Form Shortcut: A quicker way to find direction ratios for lines given in the form $x=ay+b, z=cy+d$ is to recognize it as a parametric form where $y$ is the parameter. Let $y=t$. Then $x=at+b$, $y=t$, $z=ct+d$. The direction vector components are the coefficients of $t$, which are $(a, 1, c)$. This can save time in competitive exams.
• Zero Direction Ratios: If any of $a, a', c, c'$ are zero, the corresponding direction ratio is zero. For example, if $a=0$, the first line is $x=b, z=cy+d$. This means the line is parallel to the YZ-plane, and its direction ratios are $(0, 1, c)$. The perpendicularity condition $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ still holds true when some direction ratios are zero.
• Role of Constants: Remember that $b, d, b', d'$ define specific points through which the lines pass. They affect the line's position but not its orientation or direction. Therefore, they do not appear in the condition for perpendicularity.

Summary

To find the condition for perpendicularity of the two given lines, we first converted each line from its non-standard form ($x=ay+b, z=cy+d$) into the symmetric form ($\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$). This allowed us to extract their respective direction ratios: $(a, 1, c)$ for the first line and $(a', 1, c')$ for the second line. Finally, we applied the perpendicularity condition, which states that the dot product of their direction vectors must be zero ($l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$), leading to the condition $aa' + 1 + cc' = 0$.

The final answer is $\boxed{aa' + cc' + 1 = 0}$, which corresponds to option (A).