The two lines `x=ay+b,z=cy+d and x=a'y+b', z=c'y +d'` are pendicular to each other if

Options

A. `aa' + alpha =1`
B. `(a)/( a') + (c )/( c') = -1`
C. `(a)/( a') + ( c )/( c') =1`
D. `aa'+ c c'=-1`

Answer

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

### Step 1: Rewrite the equations in symmetric form
The equations of the lines can be rewritten as:
1. For the first line:
\[ \frac{x - b}{a} = y = \frac{z - d}{c} \]
2. For the second line:
\[ \frac{x - b'}{a'} = y = \frac{z - d'}{c'} \]

### Step 2: Identify direction ratios
From the symmetric forms, we can identify the direction ratios of the two lines:
- For the first line, the direction ratios are \( (a, 1, c) \).
- For the second line, the direction ratios are \( (a', 1, c') \).

### Step 3: Use the condition for perpendicularity
Two lines are perpendicular if the dot product of their direction ratios is zero. Therefore, we need to compute:
\[ a \cdot a' + 1 \cdot 1 + c \cdot c' = 0 \]
This simplifies to:
\[ aa' + 1 + cc' = 0 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ aa' + cc' = -1 \]

### Conclusion
Thus, the two lines are perpendicular to each other if:
\[ aa' + cc' = -1 \]