A) \[41 or - 42\]
B) \[42 or - 43\]
C) \[- 41 or 43\]
D) \[- 42 or 44\]
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. (a) Option (b) is correct. (c) Given planes are\[4x-2y-4z+1=0\] and \[4x-2y-4z+d=0\] They are parallel. Distance between them is \[\pm 7=\frac{d-1}{\sqrt{16+4+16}}\] \[\Rightarrow \]\[\frac{d-1}{6}=\pm 7\] \[\Rightarrow \]\[d=42+1\]or\[-42+1\] i.e. \[d=-41\]or 43.
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