Question

The equation of the plane passing through the point \((1,1,1)\) and perpendicular to the planes \(2 x+y-2 z=5\) and \(3 x-6 y-2 z=7\), is [A] \(14 x+2 y-15 z=1\) [B] \(14 x-2 y+15 z=27\) [C] \(14 x+2 y+15 z=31\) [D] \(-14 x+2 y+15 z=3\)

Short answer

The equation of the plane is \(14 x+2 y+15 z=31\), which corresponds to option [C].

Step-by-step solution

Find the normal vector of the required plane

The normal vector of the plane we seek is perpendicular to the normal vectors of both given planes. So, we can find it by computing the cross product of the normals of the given planes. The given plane equations are in the form \( Ax + By + Cz = D \), where \( (A, B, C) \) is the normal vector. For the first plane, the normal vector is \( (2, 1, -2) \), and for the second one, it is \( (3, -6, -2) \). Perform the cross product of these two vectors.

Calculate the cross product

Using the determinant of a matrix comprising the unit vectors \( i, j, k \) and the components of the two normal vectors, compute the cross product: \[ N = \begin{vmatrix} i & j & k \ 2 & 1 & -2 \ 3 & -6 & -2 \end{vmatrix} \]. This yields the normal vector \( N = (i(1(-2) - (-2)(-6)) - j(2(-2) - (-2)(3)) + k(2(-6) - 1(3))) \), which simplifies to \( N = (14, 2, 15) \).

Use the point-normal form of the plane

A plane's equation can be expressed as \( (\mathbf{x} - P) \cdot N = 0 \), where \( P \) is a point on the plane (here \( P = (1, 1, 1) \)) and \( N \) is the normal vector. Using the normal vector \( (14, 2, 15) \), the equation becomes \( 14(x-1) + 2(y-1) + 15(z-1) = 0 \).

Simplify the equation

Expand and simplify the equation from the previous step: \( 14x - 14 + 2y - 2 + 15z - 15 = 0 \). Combining like terms, we get \( 14x + 2y + 15z = 31 \).

Identify the correct answer

The simplified equation matches one of the answer choices, which is the equation of the plane passing through \( (1,1,1) \) and perpendicular to the given planes.

Key concepts

Cross Product of Vectors

In vector mathematics, the cross product is an operation that takes two vectors in three-dimensional space and returns a vector that is perpendicular to both. This is crucial in the context of planes, as it can be used to find a normal vector, which is necessary to define a plane's orientation.

When given two vectors, like the normal vectors from our example plane equations \( (2, 1, -2) \) and \( (3, -6, -2) \), their cross product is found using the determinant of a matrix that includes the unit vectors \( i, j, k \) along with the components of these two vectors. The resulting vector points in the direction perpendicular to both original vectors, providing a fundamental building block for defining the equation of a plane.

Normal Vector

A normal vector to a plane is a three-dimensional vector that is perpendicular to the surface of the plane. In our exercise, the normal vector \( (14, 2, 15) \) was determined using the cross product of vectors, which effectively combines the characteristics of two other planes to describe a new one.

The importance of the normal vector cannot be overstated; it not only helps in defining the orientation of a plane but is also an essential component in the plane's point-normal form equation, which is used to uniquely define the plane in three-dimensional space.

Plane's Point-Normal Form

The plane's point-normal form is a way to express the equation of a plane using a point on the plane and a normal vector. It's written as \( (\mathbf{x} - P) \cdot N = 0 \), where \( P \) is a known point on the plane, and \( N \) represents the normal vector. This compact form is incredibly powerful, as it uses vector operations to define a plane.

In our problem, with point \( P = (1, 1, 1) \) and the normal vector \( N = (14, 2, 15) \), the point-normal form guides us to the final equation of the plane by representing the orientation and position in three-dimensional space. Understanding this form is of great significance, as it simplifies the process of deriving the plane's equation.

Vector Determinants

To calculate the cross product of vectors and ultimately find the normal vector in our exercise, we use the concept of vector determinants. Determinants are mathematical objects associated with square matrices, in this case, a 3x3 matrix generated by integrating the unit vectors and the components of the two vectors involved.

The determinant gives us a scalar quantity that describes the volume of a parallelepiped formed by the vectors, and, when used in conjunction with the unit vectors, helps to find the direction and magnitude of the cross product. Grasping the method to calculate these determinants is crucial not only for finding normal vectors but also for understanding deeper properties of vectors in higher mathematics.

JEE Advanced Mathematics

Preparing for competitive exams like the JEE Advanced entails a deep understanding of various mathematical concepts, including the equation of a plane. The skills to find the equation of a plane using the point-normal form, understanding the use of cross products, and calculating vector determinants are typical of the level expected in JEE Advanced Mathematics.

Students aiming to excel in this exam should thus focus on mastering these fundamental concepts through solved examples like our exercise. Analyzing such problems helps not only in honing problem-solving skills but also in gaining a profound conceptual understanding essential for acing JEE Advanced Mathematics.