Question

An equation of the plane parallel to the plane x – 2y + 2z = 0 and at a unit distance from the origin is:
$\left( a \right)$ x – 2y + 2z – 3 = 0
$\left( b \right)$ x – 2y + 2z + 1 = 0
$\left( c \right)$ x – 2y + 2z - 1 = 0
$\left( d \right)$ x – 2y + 2z + 5 = 0

Answer 1

Hint: In this particular question use the concept that parallel planes has equal coefficients of x, y, and z only difference is in the constant term, and use the concept that the perpendicular distance from any point $\left( {{x_1},{y_1},{z_1}} \right)$ on the plane ax + by + cz = 0 is given as, $d = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Given plane
x – 2y + 2z = 0
Now it is given that another plane is parallel to this plane.
Now as we know that parallel planes have equal coefficients of x, y, and z only difference is in the constant term.
So let the equation of the parallel plane be, x – 2y + 2z + p = 0, where p is some arbitrary constant.
Now it is given that the above plane is at unit distance from the origin.
Therefore, d = 1, and the coordinates of the origin is (0, 0, 0)
Now as we all know that the perpendicular distance from any point $\left( {{x_1},{y_1},{z_1}} \right)$ on the plane ax + by + cz + p = 0 is given as, $d = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + p} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
Therefore, $\left( {{x_1},{y_1},{z_1}} \right)$ = (0, 0, 0)
And a = 1, b = -2, c = 2
Therefore, $d = 1 = \dfrac{{\left| {0 + 0 + 0 + p} \right|}}{{\sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {2^2}} }}$
$ \Rightarrow 1 = \dfrac{{\left| p \right|}}{{\sqrt {1 + 4 + 4} }} = \dfrac{{\left| p \right|}}{{\sqrt 9 }}$
\[ \Rightarrow \left| p \right| = \sqrt 9 = \pm 3\]
Therefore, p = 3, or p = -3
So the equation of parallel plane become,
$ \Rightarrow x - 2y + 2z - 3 = 0$ or $x - 2y - 2z + 3 = 0$
So this is the required answer.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of perpendicular distance from any general point on the plane which is stated above, also recall the conditions of parallel planes, so assume any parallel plane equation corresponding to the given plane as above then apply the perpendicular distance formula and find out the value of constant term, substitute this value in the assumed plane, so the resultant plane is the required answer.