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Question

The plane ax+by =0 is rotated through an angle α about its line of intersection with the plane z=0. Then the equation of the plane in the new position is


A

axby±za2+b2 cot α=0

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B

ax+by±za2+b2 tan α=0

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C

axby±za2+b2 cot α=0

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D

ax+by±za2+b2 tan α=0

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Solution

The correct option is D

ax+by±za2+b2 tan α=0


Given planes are ax+by=0...........(i)
and z=0 .................(ii)
Equation of any plane passing through the line of intersection of planes (i) and (ii) may be taken as; ax+by+kz=0 .......(iii)
The direction cosines of a normal to the plane (iii) are:
aa2+b2+k2,ba2+b2+k2,ca2+b2+k2
The direction cosines of a normal to the plane (i) are :
aa2+b2,ba2+b2,0
Since the angle between the planes (i) and (ii) is α,
cos α=a.a+b.b+k.0a2+b2+k2a2+b2=a2+b2a2+b2+k2k2 cos2 α=a2(1cos2α)+b2(1cos2α)k2=(a2+b2)sin2αcos2αk=±a2+b2 tan α,
Hence, we get equation of plane as ax+by±za2+b2 tan α=0


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