If the lines
x=ay +b, z=cy+d and `x=a'y + b', z=c'y + d'` are perpendicular, then

Options

A. aa'+cc'=1
B. aa' +cc'`=-1`
C. ab+cd =a'b'+c'd'
D. aa'+bb'-cc'+dd'

Answer

To determine the condition under which the lines given by the equations \( x = ay + b, z = cy + d \) and \( x = a'y + b', z = c'y + d' \) are perpendicular, we can follow these steps:

### Step 1: Identify Direction Ratios
The first line can be expressed in terms of direction ratios. From the equations:
- For the first line:
- \( x = ay + b \) can be rewritten as \( x - ay - b = 0 \)
- \( z = cy + d \) can be rewritten as \( z - cy - d = 0 \)

The direction ratios for the first line can be taken as \( (1, -a, 0) \) and \( (0, -c, 1) \).

- For the second line:
- \( x = a'y + b' \) can be rewritten as \( x - a'y - b' = 0 \)
- \( z = c'y + d' \) can be rewritten as \( z - c'y - d' = 0 \)

The direction ratios for the second line can be taken as \( (1, -a', 0) \) and \( (0, -c', 1) \).

### Step 2: Find Normal Vectors
To find the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \) for the two sets of equations, we can use the determinant of the matrix formed by the direction ratios.

For the first line, the normal vector \( \mathbf{n_1} \) is given by:
\[ \mathbf{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 0 & -c & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \mathbf{n_1} = \hat{i} \cdot (-a \cdot 1 - 0) - \hat{j} \cdot (1 \cdot 1 - 0) + \hat{k} \cdot (1 \cdot -c - 0) = (-a) \hat{i} - \hat{j} - c \hat{k} \]
Thus, \( \mathbf{n_1} = (-a, -1, -c) \).

For the second line, the normal vector \( \mathbf{n_2} \) is given by:
\[ \mathbf{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a' & 0 \\ 0 & -c' & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \mathbf{n_2} = \hat{i} \cdot (-a' \cdot 1 - 0) - \hat{j} \cdot (1 \cdot 1 - 0) + \hat{k} \cdot (1 \cdot -c' - 0) = (-a') \hat{i} - \hat{j} - c' \hat{k} \]
Thus, \( \mathbf{n_2} = (-a', -1, -c') \).

### Step 3: Condition for Perpendicularity
The lines are perpendicular if the dot product of their normal vectors is zero:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]
Calculating the dot product:
\[ (-a)(-a') + (-1)(-1) + (-c)(-c') = 0 \]
This simplifies to:
\[ aa' + 1 + cc' = 0 \]

### Step 4: Rearranging the Equation
Rearranging gives us the required condition:
\[ aa' + cc' = -1 \]

### Final Answer
Thus, the condition for the lines to be perpendicular is:
\[ aa' + cc' = -1 \]

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