Problem 66 - A Complete Resource Book in Mathematics for JEE Main

Question

An equation of a plane parallel to the plane \(x-2 y+2 z\) \(=5\) and at a unit distance from the origin is \([\mathbf{2 0 1 2}]\) (A) \(x-2 y+2 z-3=0\) (B) \(x-2 y+2 z+1=0\) (C) \(x-2 y+2 z-1=0\) (D) \(x-2 y+2 z+5=0\)

Step-by-Step Solution

Answer

(A) \(x - 2y + 2z - 3 = 0\)

Step 1: Understanding Parallel Planes

A plane parallel to the plane given by \(x - 2y + 2z = 5\) will have the same normal vector \(\langle 1, -2, 2 \rangle\). This means any plane parallel to the original plane will be of the form \(x - 2y + 2z + d = 0\), where \(d\) is a constant.

Step 2: Determine Equation Coefficient Using Distance

To find the correct plane, it must be at a unit distance from the origin. The distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(ax + by + cz + d = 0\) is given by \(D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}\). For the origin \((0, 0, 0)\), the distance formula becomes \(D = \frac{|d|}{\sqrt{1^2 + (-2)^2 + 2^2}}\).

Step 3: Calculate Simplified Distance Coefficient

Substitute the plane equation parameters into the distance formula: \(D = \frac{|d|}{\sqrt{1 + 4 + 4}} = \frac{|d|}{\sqrt{9}} = \frac{|d|}{3}\). For a unit distance from the origin, set \(\frac{|d|}{3} = 1\). Solve for \(|d|\) to obtain \(|d| = 3\).

Step 4: Identify Possible Values for Coefficient

Based on \(|d| = 3\), the possible values for \(d\) are 3 or -3. This leads to potential plane equations: \(x - 2y + 2z + 3 = 0\) or \(x - 2y + 2z - 3 = 0\).

Step 5: Select Correct Answer From Options

Check the provided options: (A) \(x - 2y + 2z - 3 = 0\) (B) \(x - 2y + 2z + 1 = 0\) (C) \(x - 2y + 2z - 1 = 0\) (D) \(x - 2y + 2z + 5 = 0\). The matching answer is: (A) \(x - 2y + 2z - 3 = 0\)

Key Concepts

Parallel Planes

In coordinate geometry, parallel planes are two or more planes that never intersect. This is because they have the same normal vector but different constant terms in their equations. For example, two planes described by equations of the form

• \(ax + by + cz + d_1 = 0\)
• \(ax + by + cz + d_2 = 0\)

are parallel. The presence of equal coefficients \(a\), \(b\), and \(c\) ensures they do not intersect. Only the value of the constant \(d\) differs.

In the example exercise, the equation of the plane is given as \(x - 2y + 2z = 5\). A parallel plane shares its normal vector \(\langle 1, -2, 2 \rangle\). Thus, the equation of any plane parallel to this one can be expressed in the form \(x - 2y + 2z + d = 0\). The parameter \(d\) here represents how far vertically this plane shifts while remaining parallel.

Distance from a Point to a Plane

The distance from a point to a plane is a key concept in coordinate geometry. It tells us how far a given point is from the plane. For a plane described by the equation \(ax + by + cz + d = 0\) and a point

• \((x_0, y_0, z_0)\),

the shortest distance \(D\) from the point to the plane can be calculated using the formula:

\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

This equation calculates the perpendicular distance between the point and the plane.

In our problem scenario, the point is the origin, \((0, 0, 0)\), which simplifies calculations. The formula reduces to \(D = \frac{|d|}{\sqrt{9}} = \frac{|d|}{3}\). To ensure the plane is exactly one unit away from the origin, \(D\) is set to 1, leading to \(|d| = 3\). Hence, the possible values for \(d\) are \(\pm 3\).

Normal Vector of a Plane

The normal vector of a plane is a vector that is perpendicular to the surface of the plane. It plays a critical role in defining the orientation and position of the plane in space. A plane's equation, \(ax + by + cz + d = 0\), has a normal vector of \(\langle a, b, c \rangle\). This vector is crucial since:

• It dictates the direction perpendicular to the plane.
• For parallel planes, the normal vector remains unchanged.

In the exercise given, the plane \(x - 2y + 2z = 5\) has a normal vector \(\langle 1, -2, 2 \rangle\). This vector ensures that any plane with this same normal vector is parallel to the original plane.

For any computations involving orientation, such as determining parallelism or calculating distances, the normal vector is essential. In this exercise, maintaining the normal vector \(\langle 1, -2, 2 \rangle\) is necessary to explore any parallel planes.