find the equation of the planes parallel to the plane x-2y+2z-3 = 0 which are at a unit distance from - Maths - Three Dimensional Geometry

Question

find the equation of the planes parallel to the plane x-2y+2z-3 = 0
which are at a unit distance from the point (1,2,3)

Ishwarmani · Accepted Answer

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below the solution to the asked query:

As, the equation of a plane is x-2y+2z-3=0So, the normal vector, N→=i^-2j^+2k^Let the equation of the required plane be r→
N→=DSince, the required plane is parallel to the given planeSo, its normal vector will be N→=i^-2j^+2k^⇒Its equation will be r→
i^-2j^+2k^=Dor x-2y+2z=D            
iAs, the distance of point 1,2,3 from the required plane=1 unit⇒1-22+23-D12+-22+22=1⇒1-4+6-D1+4+4=1⇒3-D9=1⇒3-D3=1⇒3-D=3⇒3-D=±3⇒3-D=3 or 3-D=-3⇒D=3-3 or D=3+3⇒D=0 or D=6By putting the value od D=0 and D=6 in equation i, we getThe equation of the required plane isx-2y+2z=0 or x-2y+2z=6

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find the equation of the planes parallel to the plane x-2y+2z-3 = 0 which are at a unit distance from the point (1,2,3)