A plane passes through the point P(4,0,0) and Q(0,0,4) and is parallel to the y-axis. The distance of the plane from the origin is
A
2
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B
4
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C
√2
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D
2√2
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Solution
The correct option is D2√2 General equation of a plane parallel to y-axis is given by, ax+bz+c=0 Since this plane is also passes through (4,0,0) and (0,0,4) ⇒4a+c=0⇒a=−c4 and 4b+c=0⇒b=−c4 Hence, required plane is, x+z−4=0 Therefore distance of origin from this plane is =d=∣∣
∣∣−4√12+12∣∣
∣∣=2√2