Problem 114 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A equation of a plane parallel to the plane \(x-2 y+2 z-5=0\) and at a unit distance from the origin is : [2012] (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

The correct answer is (a) \(x-2y+2z-3=0\).

Step 1: Understanding Parallel Planes

Planes are parallel if they have the same normal vector. The given plane is \(x - 2y + 2z - 5 = 0\), so any plane parallel to it will have the form \(x - 2y + 2z - D = 0\), where \(D\) is a constant that we need to determine.

Step 2: Using Distance Formula for a Plane

The distance from the origin \((0, 0, 0)\) to a plane \(Ax + By + Cz + D = 0\) is given by \( \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \). For a parallel plane at a unit distance from the origin, we set this distance to 1.

Step 3: Calculate the Normal Vector Magnitude

For our plane, the coefficients \(A, B, C\) are 1, -2, and 2 respectively. Thus, the magnitude of the normal vector is \(\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3\).

Step 4: Substitute Values into Distance Formula

Substituting \(A=1\), \(B=-2\), \(C=2\), and \(\operatorname{distance} = 1\) into the distance formula \( \frac{|D|}{3} = 1 \), we solve for \(|D|\):\( |D| = 3 \).

Step 5: Determine Possible Planes

Since \(|D| = 3\), the possible values of \(D\) are 3 and -3. Thus, the equations of the parallel planes can be \(x - 2y + 2z - 3 = 0\) or \(x - 2y + 2z + 3 = 0\).

Step 6: Match Possible Planes with Options

Comparing these possible parallel planes with the given options, \(x - 2y + 2z - 3 = 0\) corresponds to option (a). Since it is at a unit distance from the origin, option (a) is correct.

Key Concepts

Parallel Planes

Parallel planes share an important feature: they have identical normal vectors. If two planes have the same normal vector, they will never meet, no matter how far you extend their surfaces. For the plane given by the equation \(x - 2y + 2z - 5 = 0\), any plane parallel to it must have the same coefficients of \(x\), \(y\), and \(z\). This means that any parallel plane will be of the form \(x - 2y + 2z - D = 0\), where \(D\) is some constant. The normal vector in this case is \((1, -2, 2)\). This concept helps establish that variations in the plane equation only occur in the constant term \(D\), as altering the coefficients of \(x\), \(y\), or \(z\) would affect the direction of the plane resulting in it no longer being parallel.

Distance Between a Point and a Plane

To find the distance between a point and a plane, a simple formula can be used:

• Consider a plane given by the equation \(Ax + By + Cz + D = 0\).
• To find the distance to a point \((x_1, y_1, z_1)\), use the formula \(\frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\).

This formula essentially measures how far the point is from the plane along the direction perpendicular to the plane itself, relying on the concept of a normal vector.
In the exercise solution, we are tasked with determining a plane at a unit distance from the origin \((0,0,0)\). By substituting into the formula, if we set the distance to 1, then: \(\frac{|D|}{3} = 1\), which leads to the result \(|D|=3\). This means our potential parallel planes have their constant \(D\) as either \(3\) or \(-3\), demonstrating that the planes are located one unit away from the origin in opposite directions.

Normal Vector Calculation

The normal vector of a plane plays a crucial role in both identifying parallel planes and calculating distances to points or other planes. The normal vector of a plane equation \(Ax + By + Cz + D = 0\) is \((A, B, C)\).
In the current exercise, for the plane equation \(x - 2y + 2z - 5 = 0\), the normal vector is \((1, -2, 2)\). Calculating the magnitude of this vector involves using the formula for the magnitude of a vector \((a, b, c)\): \[\sqrt{a^2 + b^2 + c^2}\].Calculating the magnitude helps in understanding the geometry of the plane and comes into play when calculating distances.
For our vector, we compute its magnitude as \[\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\]. This scalar quantity is essential in the distance formula, allowing us to ascertain how far a plane is from a given point by adjusting the constant \(D\) appropriately.