Problem 15 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the plane which contains the origin and the line of intersection of the planes \(\mathbf{r} \cdot \mathbf{a}=p\) and \(\mathbf{r} \cdot \mathbf{b}=q\) is (A) \(\mathbf{r} \cdot(p \mathbf{a}-q \mathbf{b})=0\) (B) \(\mathbf{r} \cdot(p \mathbf{a}+q \mathbf{b})=0\) (C) \(\mathbf{r} \cdot(q \mathbf{a}+p \mathbf{b})=0\) (D) \(\mathbf{r} \cdot(q \mathbf{a}-p \mathbf{b})=0\)

Step-by-Step Solution

Answer

The correct answer is (A) \( \mathbf{r} \cdot (p\mathbf{a} - q\mathbf{b}) = 0 \).

Step 1: Identify Given Planes

The problem involves two given planes: \( \mathbf{r} \cdot \mathbf{a} = p \) and \( \mathbf{r} \cdot \mathbf{b} = q \). These planes can be written in the normal form where \( \mathbf{a} \) and \( \mathbf{b} \) are the normal vectors, respectively. The line of intersection of these two planes is what we need to focus on for constructing the new plane.

Step 2: Find the Line of Intersection

The line of intersection of two planes can be expressed as a vector equation. Since the planes intersect, the direction of their line of intersection is given by the cross product of their normals, \( \mathbf{a} \times \mathbf{b} \). This vector will be parallel to the line of intersection.

Step 3: Determine the Plane Containing the Origin

The plane that passes through the origin and the line of intersection of the two planes will have a normal vector perpendicular to the direction vector found in step 2. Since the plane must contain the origin, the equation of such a plane can be written as \( \mathbf{r} \cdot \mathbf{n} = 0 \) where \( \mathbf{n} = p\mathbf{a} - q\mathbf{b} \).

Step 4: Calculate the Normal Vector for New Plane

The new plane containing the origin and the line of intersection of the original planes should be perpendicular to both given planes' normals. Since the line of intersection is given by \( \mathbf{a} \times \mathbf{b} \), the correct normal vector for the new plane is a linear combination \( p\mathbf{a} - q\mathbf{b} \), reflecting the equation \( \mathbf{r} \cdot (p\mathbf{a} - q\mathbf{b}) = 0 \).

Step 5: Conclusion and Selection of Correct Option

Based on the requirement that the plane contains the origin and is defined by the intersecting line, the correct plane equation is \( \mathbf{r} \cdot (p\mathbf{a} - q\mathbf{b}) = 0 \). Thus, the correct choice is (A) \( \mathbf{r} \cdot(p \mathbf{a}-q \mathbf{b})=0 \).

Key Concepts

Line of Intersection of Planes

The line of intersection of two planes is where they meet in space. Imagine two sheets of paper crossing over each other on a desk. The line where they touch is like their line of intersection. This concept is essential in three-dimensional geometry.
To find the line of intersection between two planes, you need the normal vectors of the planes. The normal vectors tell you the direction in which the plane extends. In the problem, the normal vectors are given as \( \mathbf{a} \) and \( \mathbf{b} \) for the two planes \( \mathbf{r} \cdot \mathbf{a} = p \) and \( \mathbf{r} \cdot \mathbf{b} = q \) respectively.
The line of intersection will be parallel to the vector you get from the cross product of these normals. So, you'll calculate \( \mathbf{a} \times \mathbf{b} \) to find the direction of the intersection line. Understanding this concept helps you grasp how different planes interact in three-dimensional spaces.

Vector Cross Product

The vector cross product is a mathematical operation that plays a crucial role in finding the direction of the line of intersection between two planes. When you have two vectors, like the normal vectors of our planes \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) gives you a third vector. This third vector is perpendicular to both of the original vectors.

Why is this important? Well, the planes’ intersection line direction aligns with this cross product vector. The cross product’s result, \( \mathbf{a} \times \mathbf{b} \), indicates the line's direction that you need for forming a new plane equation. It's fascinating how just using three numbers, describing the components of each vector, can help us solve such geometrical mysteries.

Equation of Plane Through Origin

In geometry, a plane equation describes a flat, two-dimensional surface extending infinitely in three-dimensional space. When a plane passes through the origin, its equation simplifies greatly, forming a special relationship. This is critical in problems like ours, where a new plane must be defined through the origin.

To construct the equation of a plane through the origin, you need a normal vector. This normal vector is perpendicular to every line that lies within the plane. Our plane's normal vector, shown as \( p\mathbf{a} - q\mathbf{b} \) from the problem, derives from factors linked to how the original intersecting planes are defined and mixed.

• The relation \( \mathbf{r} \cdot (p\mathbf{a} - q\mathbf{b}) = 0 \) stems from this idea, as no constant term is involved, ensuring it passes through the origin.
• The form \( \mathbf{n} = p\mathbf{a} - q\mathbf{b} \) describes all vectors that are orthogonal to the plane's surface.

This concept helps in creating the equations of planes that sit flatly through specific points, like the origin of a coordinate system.