Problem 70 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The vector equation of the plane through the line of intersection of the planes \(x+y+z=1\) and \(2 x+3 y+4 z=5\) which is perpendicular to the plane \(x-y+z=0\) is: [April 08, 2019 (II)] (a) \(\vec{r} \times(\hat{i}-\hat{k})+2=0\) (b) \(\vec{r} \cdot(\hat{i}-\hat{k})-2=0\) (c) \(\vec{r} \times(\hat{i}+\hat{k})+2=0\) (d) \(\vec{r} \cdot(\hat{i}-\hat{k})+2=0\)

Step-by-Step Solution

Answer

The correct option is (d) \(\vec{r} \cdot (\hat{i}-\hat{k}) + 2 = 0.\)

Step 1: Understanding the Intersection of Planes

Two planes are given: \( x+y+z=1 \) and \( 2x+3y+4z=5 \). The line of intersection will be the collection of points that satisfy both equations. We need to find the directional vector of this line of intersection.

Step 2: Normal Vectors of the Given Planes

Identify the normal vectors for each given plane. The normal to the first plane \( x+y+z=1 \) is \( \langle 1,1,1 \rangle \) and for the second plane \( 2x+3y+4z=5 \) it is \( \langle 2,3,4 \rangle \).

Step 3: Direction Vector of the Intersection Line

The direction vector of the line of intersection can be found using the cross product of the normal vectors of the given planes. Hence, compute \( \langle 1,1,1 \rangle \times \langle 2,3,4 \rangle \).

Step 4: Calculate the Cross Product

The cross product \( \langle 1,1,1 \rangle \times \langle 2,3,4 \rangle \) equals \( \langle 1(-4) - 1(3), 1(4) - 1(2), 1(3 - 2) \rangle = \langle -4 - 3, 4 - 2, 3 - 2 \rangle = \langle -7, 2, 1 \rangle \).

Step 5: Plane Perpendicular to Another Given Plane

The resulting plane must be perpendicular to the plane \( x-y+z = 0 \), whose normal vector is \( \langle 1, -1, 1 \rangle \).

Step 6: Normal Vector of Required Plane

The normal vector of the required plane can be expressed as a linear combination of the direction vector of the line of intersection \( \langle -7, 2, 1 \rangle \) and the normal vector of the plane it is perpendicular to \( \langle 1, -1, 1 \rangle \).

Step 7: Formulate the Vector Equation

Choose a vector that satisfies both conditions: in line with the plane normals, while perpendicular to \( \langle 1, -1, 1 \rangle \). It appears as \( \vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0 \) through substitution and simplification using given options.

Key Concepts

Vector Equation of a Plane

The vector equation of a plane is a useful representation in mathematics, particularly when dealing with geometric applications in space. It allows you to describe a plane using vectors. A vector equation typically looks like this: \( \vec{r} \cdot \vec{n} = d \), where \( \vec{r} \) is the position vector of any point on the plane, \( \vec{n} \) is the normal vector to the plane, and \( d \) is the perpendicular distance from the origin to the plane.
By using vectors, you can easily manipulate and understand the geometry of the plane.

• Position Vector \( \vec{r} = \langle x, y, z \rangle \)
• Normal Vector \( \vec{n} = \langle a, b, c \rangle \)

All points \( (x, y, z) \) that lie on the plane must satisfy this vector equation. This concept is particularly crucial as it lets you represent complex relationships and interactions between geometric shapes.

Intersection of Planes

When two planes intersect, they do so along a line. The intersection of planes is a way to determine common solutions to both plane equations. Given two planes, their intersection can be found by solving their equations simultaneously.
The planes in our example are given by:

• Plane 1: \( x+y+z=1 \)
• Plane 2: \( 2x+3y+4z=5 \)

The line of intersection will satisfy both plane equations. A key part of finding this line is determining its direction, which involves normal vectors of each plane. The direction vector of the line is perpendicular to both plane normals, found using the cross product. This sheds light on how intersection lines bridge relationships between different planes.

Cross Product of Vectors

The cross product is a fundamental operation in vector mathematics. It's used to find a vector perpendicular to two given vectors. In the context of our problem, the cross product is used to find the direction vector of the line where two planes intersect. To compute it, use:

• Given vectors: \( \langle a_1, b_1, c_1 \rangle \times \langle a_2, b_2, c_2 \rangle \)
• Result: \( \langle b_1c_2 - c_1b_2, c_1a_2 - a_1c_2, a_1b_2 - b_1a_2 \rangle \)

For example, the cross product of \( \langle 1,1,1 \rangle \) and \( \langle 2,3,4 \rangle \) creates a new vector \( \langle -7, 2, 1 \rangle \), which serves as the direction vector of the line.
The calculated cross product provides insight into how vectors interact spatially, offering a powerful tool for solving geometric problems.

Normal Vector

The normal vector is vital to understanding the orientation of a plane in space. It is a vector perpendicular to the plane's surface and helps define the plane's equation. In simple terms, if a plane is considered a flat surface, the normal vector sticks straight out from that surface.
For a plane equation \( ax+by+cz=d \), the normal vector is \( \langle a,b,c \rangle \).
In our example, to determine the relationship between multiple planes, the respective normal vectors must be identified:

• First Plane Normal: \( \langle 1, 1, 1 \rangle \)
• Second Plane Normal: \( \langle 2, 3, 4 \rangle \)
• Perpendicular Plane Normal: \( \langle 1, -1, 1 \rangle \)

The normal vector is also essential when expressing conditions for perpendicularity or parallelism between planes. Recognizing and using normal vectors are key to fully grasping plane geometry.