Problem 4 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The line of intersection of the planes \(\mathbf{r} \cdot(3 \mathbf{i}-\mathbf{j}+\mathbf{k})=\) 1 and \(\mathbf{r} \cdot(\mathbf{i}+4 \mathbf{j}-2 \mathbf{k})=2\) is parallel to the vector (A) \(-2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\) (B) \(2 \mathbf{i}+7 \mathbf{j}-13 \mathbf{k}\) (C) \(-2 \mathbf{i}-7 \mathbf{j}+13 \mathbf{k}\) (D) \(2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\)

Step-by-Step Solution

Answer

The line is parallel to vector (A): \(-2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\).

Step 1: Identify Normal Vectors

The equation of a plane in vector form is given by \( \mathbf{r} \cdot \mathbf{n} = d \), where \( \mathbf{n} \) is the normal vector to the plane. For the planes we have:- Plane 1: \( \mathbf{n}_1 = 3\mathbf{i} - \mathbf{j} + \mathbf{k} \)- Plane 2: \( \mathbf{n}_2 = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \)

Step 2: Find Direction Vector of Line of Intersection

The line of intersection between two planes is parallel to the vector obtained by the cross product of their normal vectors.Compute the cross product \( \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 \).

Step 3: Calculate Cross Product

Calculate the cross product:\(\mathbf{d} = (3\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (\mathbf{i} + 4\mathbf{j} - 2\mathbf{k})\)Using the determinant method:\[\mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \3 & -1 & 1 \1 & 4 & -2\end{vmatrix}\]Expanding the determinant, we find:\(\mathbf{d} = (2 - 4)\mathbf{i} - (-(6 + 1))\mathbf{j} + (12 + 1)\mathbf{k}\)Which simplifies to:\(\mathbf{d} = -2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k}\)

Step 4: Choose the Correct Answer

Compare the computed direction vector \(-2\mathbf{i} + 7\mathbf{j} + 13\mathbf{k}\) with the given options. The vector matches option (A).

Key Concepts

Cross Product

The cross product is a critical operation in vector algebra, especially when working in three-dimensional space. It allows us to find a vector that is perpendicular to two given vectors. This operation is unique to three dimensions, and the resulting vector provides valuable insights into the geometry and orientation of the original vectors.

To compute the cross product of two vectors, we use a determinant involving unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as placeholders for the vector components. For example, if we have vectors \( \mathbf{u} = a_1\mathbf{i} + b_1\mathbf{j} + c_1\mathbf{k} \) and \( \mathbf{v} = a_2\mathbf{i} + b_2\mathbf{j} + c_2\mathbf{k} \), their cross product \( \mathbf{u} \times \mathbf{v} \) is:

• \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} \]
• Calculating the determinant, the result is: \( (b_1c_2 - c_1b_2)\mathbf{i} - (a_1c_2 - c_1a_2)\mathbf{j} + (a_1b_2 - b_1a_2)\mathbf{k} \)

This formula highlights the interaction between the components of the vectors and succinctly expresses their combined perpendicular vector.

Normal Vectors

Normal vectors are essential when dealing with planes in 3D geometry, as they define the orientation of a plane. A normal vector is perpendicular to every line on the plane, which makes it an ideal tool for defining and manipulating the plane mathematically.

In the equation of a plane given by \( \mathbf{r} \cdot \mathbf{n} = d \), \( \mathbf{n} \) represents the normal vector. This vector is unique to each plane and simplifies the process of determining the angle between planes or calculating the line of intersection when two planes meet. For example:

• For the plane equation \( 3\mathbf{i} - \mathbf{j} + \mathbf{k} = 1 \), the normal vector is \( 3\mathbf{i} - \mathbf{j} + \mathbf{k} \).
• Similarly, \( \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \) serves as the normal vector for another plane.

Understanding normal vectors can greatly aid in visualizing geometric configurations and leads to more efficient problem-solving in vector algebra applications.

Line of Intersection

When two planes intersect, they form a line, known as the line of intersection. This line plays a crucial role in 3D geometry, revealing how two planes relate spatially.

To find the line of intersection of two planes, we first determine the direction of this line. This direction is found by calculating the cross product of the normal vectors of the planes involved. The resultant vector from the cross product is parallel to the line of intersection. Once the direction vector is confirmed, determining a specific point on the line will complete the equation of the line in vector form. Here's how:

• Take the normal vectors of each plane.
• Compute the cross product to find the direction vector.

This method not only establishes the line's direction but also affirms its parallelism to the computed vector, offering a systematic approach to handling plane intersections.

Planes in 3D Geometry

Planes in 3D geometry are flat surfaces extending indefinitely, characterized by equations that employ vectors. These planes serve as fundamental elements in fields like computer graphics, physics, and engineering, where spatial analysis is crucial.

A plane can be defined using a point and a normal vector. Its equation takes the form \( \mathbf{r} \cdot \mathbf{n} = d \), where \( \mathbf{r} \) is a position vector on the plane, \( \mathbf{n} \) is the normal vector, and \( d \) is a scalar. This structure is quite flexible, allowing for a diverse range of applications, such as:

• Determining the angle between planes by examining their normal vectors.
• Finding lines of intersection which are crucial in modeling real-world objects in three dimensions.
• Solving optimization problems where space and alignment are factors.

By making sense of plane equations and their properties, we unlock the vast possibilities of manipulating and understanding space in mathematics and beyond.