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Question
Find the equation of the planes parallel to the plane x + 2y+ 2z + 8 =0 which are at the distance of 2 units from the point (1,1, 2)
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Solution
The equation of the plane parallel to the plane ` x+2y+ 2z + 8 = 0 " is " x + 2 y + 2z +λ = 0 .`
Now the distance of this plane from the point (1, 1, 2)
`=|(1(1)+2(1)+2(2)+lambda)/(sqrt(1^2+2^2+3^2))|`
`=|(1+2+4+lambda)/sqrt9|`
`=|(7+lambda)/sqrt9|`
Given that
`|(7+lambda)/3|=2`
`(7+lambda)/3=+-2`
λ=±6−7
`lambda=+-6-7`
`lambda=-1 or lambda=-13`
Hence eq. of plane x + 2y + 2z -1 = 0 or x + 2y + 2z - 13 = 0
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