Problem 69 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The equation of a plane containing the line of intersection of the planes \(2 x-y-4=0\) and \(y+2 z-4=0\) and passing through the point \((1,1,0)\) is: (a) \(x-3 y-2 z=-2\) (b) \(2 x-z=2\) (c) \(x-y-z=0\) (d) \(x+3 y+z=4\)

Step-by-Step Solution

Answer

Option (a) \(x - 3y - 2z = -2\) is correct.

Step 1: Understand the Problem

We have two planes given by the equations \(2x - y - 4 = 0\) and \(y + 2z - 4 = 0\). We need to find a plane that contains the line of intersection of these two planes and also passes through the point \((1, 1, 0)\).

Step 2: Finding Normal Vectors

The normal vector to the first plane \(2x - y - 4 = 0\) is \((2, -1, 0)\), and the normal vector to the second plane \(y + 2z - 4 = 0\) is \((0, 1, 2)\).

Step 3: Line of Intersection's Direction Vector

The direction vector of the line of intersection is given by the cross product of the normal vectors: \((2, -1, 0) \times (0, 1, 2) = (-2, -4, 2)\).

Step 4: Equation of the Plane through the Intersection Line

The equation of any plane passing through the line of intersection of two planes is given by the family \(2x - y - 4 + k(y + 2z - 4) = 0\). Simplifying, we get \((2)x + (-1 + k)y + 2kz = 4(1 + k)\).

Step 5: Substitute Point Into the Plane Equation

We substitute the point \((1, 1, 0)\) into the equation obtained: \(2(1) + (-1+k)(1) + 2k(0) = 4(1 + k)\), resulting in \(2 - 1 + k = 4 + 4k\).

Step 6: Solve for \(k\)

Rewriting and solving the equation from the previous step: \(1 + k = 4 + 4k\). Simplifying gives \(-3 = 3k\), thus \(k = -1\).

Step 7: Final Plane Equation

Substitute \(k = -1\) back into the plane equation \(2x - y - 4 + (-1)(y + 2z - 4) = 0\): which simplifies to \(x - 3y - 2z = -2\).

Step 8: Verify Choice

Compare the equation \(x - 3y - 2z = -2\) with given options: Option (a) matches this equation.

Key Concepts

Intersection of Planes

When two planes intersect, they form a line. Imagine two sheets of paper intersecting each other—the line or edge at which they touch represents the intersection line of these planes. In mathematical terms, the intersection of two planes can be described using their equations.

Given two planes, each described by its own equation: - Plane 1: \(2x - y - 4 = 0\) - Plane 2: \(y + 2z - 4 = 0\)
The goal is to find a new plane that contains this line of intersection. Determining such a plane often involves finding additional parameters, like the normal vector and a specific point wherever possible. If we know another point the plane passes through, like \(1, 1, 0\), this can help us determine the final plane's equation.

Normal Vectors

A normal vector is perpendicular to a plane. Each plane can be described using a normal vector that is derived from the coefficients of the variables in its equation. For a plane given by \(Ax + By + Cz = D\), the normal vector is \((A, B, C)\).

For the planes specified:

• Plane 1: \(2x - y - 4 = 0\), Normal Vector: \((2, -1, 0)\)
• Plane 2: \(y + 2z - 4 = 0\), Normal Vector: \((0, 1, 2)\)

These normal vectors are essential because they help us understand the orientation of the planes in space. They are also typically used in cross product operations to find a direction vector for the line of intersection.

Direction Vector

A direction vector indicates the direction along which a line runs. In the case of finding the line of intersection of two planes, this vector is found by taking the cross product of the normal vectors of the two planes.

The steps are as follows:

• Take the normal vector of Plane 1: \((2, -1, 0)\)
• Take the normal vector of Plane 2: \((0, 1, 2)\)
• Find the cross product: \((2, -1, 0) \times (0, 1, 2) = (-2, -4, 2)\)

This result, \((-2, -4, 2)\), represents the direction vector of the line where the two planes intersect. It's pivotal for developing an equation of a plane passing through this line, as it dictates the line's orientation in space.

Cross Product

The cross product is a mathematical operation that can be performed on two vectors in three-dimensional space. It results in another vector that is perpendicular to the plane containing the initial vectors.

To understand the cross product, consider two vectors:

• Vector A: \((a_1, a_2, a_3)\)
• Vector B: \((b_1, b_2, b_3)\)

The cross product is calculated as:\[\begin{vmatrix}i & j & k \a_1 & a_2 & a_3 \b_1 & b_2 & b_3 \\end{vmatrix} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\]In our context:- Normal Vector 1: \((2, -1, 0)\)- Normal Vector 2: \((0, 1, 2)\)Their cross product yields \((-2, -4, 2)\). This vector is integral to forming the equation of a new plane as it defines the line of intersection between the original planes.