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Question
Find the equations of the planes parallel to the plane x-2y + 2z-4 = 0, which are at a unit distance from the point (1,2, 3).
Solution
The equation of the planes parallel to the plane x −2y + 2z − 4 = 0
are of the form x-2y+2z+k=0
The distance of a plane ax+by+cz+ λ from a point (x1,y1,z1) is given by
`d=|(9ax_1+by_1+cz_1+lambda)/sqrt(a^2+b^2+c^2)|`
It is given the plane x-2y+2z+k=0 at an unit distance from the point (1, 2, 3).
`d=|(1-2(2)+2(3)+k)/(sqrt(1^2+(-2)^2)+(2)^2)|`
`1=|(k+3)/3|`
∴ |k+3|=|3|
∴ k=0 or k=-6
The equation of the planes parallel to the plane x-2y+2z-4=0
are of x-2y+2z=0 and x-2y+2z=6
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