A) \[\frac{x-b}{a}+\frac{y-1}{1}=\frac{z-d}{c}\]
B) \[\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}\]
C) \[\frac{x-a}{b}=\frac{y-0}{1}=\frac{z-c}{d}\]
D) \[\frac{x-b-a}{b}=\frac{y-1}{0}=\frac{z-d-c}{d}\]
Correct Answer: B
Solution :
Given two planes : x - ay - b = 0 and cy - z + d = 0 Let, l, m, n be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore \[l-am=0\]and \[cm-n=0\] \[\therefore \]\[\frac{l}{a-0}=\frac{m}{0+1}=\frac{n}{c-0}\] Hence, d.R of the required line are a, 1, c. Hence, options (c) and (d) are rejected. Now, the point (a + b, 1, c + d) satisfy the equation of the two given planes. Option (b) is correct.
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