<p>A symmetrical form of the line of intersection of the planes \(x = ay + b\) and \(z = cy + d\) is</p>
<p>\(\dfrac{x-b}{a} = \dfrac{y-1}{1} = \dfrac{z-d}{c}\)</p>
<p>\(\dfrac{x-b-a}{a} = \dfrac{y-1}{1} = \dfrac{z-d-c}{c}\)</p>
<p>\(\dfrac{x-a}{b} = \dfrac{y-0}{1} = \dfrac{z-c}{d}\)</p>
<p>\(\dfrac{x-b-a}{b} = \dfrac{y-1}{0} = \dfrac{z-d-c}{d}\)</p>
Step-by-Step Solution
Key Concept: The line of intersection of two planes can be found by expressing both x and z in terms of the parameter y, then writing the symmetric form using y as the independent variable to get (x-x₀)/a = (y-y₀)/1 = (z-z₀)/c.
Step 1: From the given planes x = ay + b and z = cy + d, we can express both x and z in terms of y. Step 2: Treat y as the parameter. When y = 0: x = b and z = d, giving point (b, 0, d). Step 3: The direction ratios are obtained from the coefficients: as y increases by 1, x increases by a and z increases by c. Thus direction ratios are (a, 1, c). Step 4: The symmetric form of the line passing through (b, 0, d) with direction ratios (a, 1, c) is: (x - b)/a = y/1 = (z - d)/c ∴ Answer: A
Correct Answer: A