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Question
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Solution
The given equation of the plane is 2x - y + z + 1 = 0
A foot of the perpendicular is given by
`("x"-"x"_1)/"a"=("y"-"y"_1)/"b"=("z"-"z"_1)/"c"=-(("ax"_1+"by"_1+"cz"_1-"d"))/("a"^1+"b"^2+"c"^2)`
Therefore, foot of perpendicular Q from point P(3,2,1) to the given plane is given by
`("x"-3)/2=("y"-2)/-1=("z"-1)/1=-((2xx3+(-1)xx2+1xx1+1))/(2^2+(-1)+1^2)=-1`
⇒ x = 1, y = 3, z = 0
Hence, coordinates of foot of perpendicular Q are (1, 3, 0).
Distance PQ`=|((2xx3+(-1)xx1+1xx1+1))/(sqrt(2^2+(-1) ^2 +1^2))|=sqrt6`
Image of the point P is given by
`("x"-"x"_1)/"a"=("y"-"y"_1)/"b"=("z"-"z"_1)/"c"=-(2("ax"_1+"by"_1+"cz"_1-"d"))/("a"^2+"b"^2+"c"^2)`
Therefore, coordinates of image of the point P are given by
`("x"-3)/2=("y"-2)/-1=("z"-1)/1=-(2[(2xx3+(-1)xx2+1xx1+1)])/(2^2+(-1)^2+1^2)=-2`
⇒ x = - 1, y = 4, z = - 1
Hence, coordinates of image of P are (-1, 4, -1)
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