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

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

Option (A): \(-2 \mathbf{i} + 7 \mathbf{j} + 13 \mathbf{k}\).

Step 1: Identify the Normal Vectors

The given planes can be represented by their normal vectors. For the first plane \( \mathbf{r} \cdot(3 \mathbf{i}-\mathbf{j}+\mathbf{k})= 1 \), the normal vector is \( \mathbf{n_1} = 3 \mathbf{i}-\mathbf{j}+\mathbf{k} \). For the second plane \( \mathbf{r} \cdot(\mathbf{i}+4 \mathbf{j}-2 \mathbf{k})=2 \), the normal vector is \( \mathbf{n_2} = \mathbf{i}+4 \mathbf{j}-2 \mathbf{k} \).

Step 2: Calculate the Direction Vector

The line of intersection of two planes is parallel to the vector obtained by the cross product of the planes' normal vectors. Find \( \mathbf{n_1} \times \mathbf{n_2} \):\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix} \].

Step 3: Simplify the Determinant

Calculate each component of the vector \[= \mathbf{i}((-1)(-2) - 1(4)) - \mathbf{j}(3(-2) - 1(1)) + \mathbf{k}(3(4) - (-1)(1))\]\[= \mathbf{i}(2 - 4) - \mathbf{j}(-6 - 1) + \mathbf{k}(12 + 1)\]\[= -2 \mathbf{i} + 7 \mathbf{j} + 13 \mathbf{k}.\]

Step 4: Match the Result with Options

Refer back to the multiple-choice options:- Option (A): \(-2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\)- Option (B): \(2 \mathbf{i}+7 \mathbf{j}-13 \mathbf{k}\)- Option (C): \(-2 \mathbf{i}-7 \mathbf{j}+13 \mathbf{k}\)- Option (D): \(2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\)The direction vector \(-2 \mathbf{i}+7 \mathbf{j}+13 \mathbf{k}\) matches Option (A).

Key Concepts

Cross Product

The cross product is an operation that can be performed on two vectors in three-dimensional space and results in another vector that is perpendicular to both of the original vectors. This concept is crucial in physics and engineering because it allows us to find vectors that are orthogonal, which helps in applications like finding rotational effects or determining a plane's orientation.

To compute the cross product between two vectors, say \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), we can use the determinant of a matrix constructed with these vectors along with the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). The formula for the cross product \( \mathbf{a} \times \mathbf{b} \) is:

• \( \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2) \mathbf{i} - (a_1b_3 - a_3b_1) \mathbf{j} + (a_1b_2 - a_2b_1) \mathbf{k} \)

This resulting vector gives not only direction but also helps quantify the magnitude of the perpendicular vector. The direction of the cross product is determined by the right-hand rule: if you position your right hand so that your fingers point in the direction of \( \mathbf{a} \) and curl towards \( \mathbf{b} \), your thumb will point in the direction of \( \mathbf{a} \times \mathbf{b} \).

In our example problem, the cross product operation was used to find a direction vector for the line of intersection of two planes, illustrating a practical use of this mathematical operation.

Intersection of Planes

The intersection of two planes is a fundamental concept in vector algebra and geometry. When two planes intersect in three dimensions, they often do so along a line. By understanding the line of intersection, it becomes easier to visualize and solve geometrical problems involving planes.

In our particular problem, two planes are defined by their vector equations involving normal vectors. The normal vector is a vector that is perpendicular to a plane, and it helps characterize the plane's orientation in space. The intersection line can be described by a direction vector, which is parallel to the line.

To find this line, we need the cross product of the two planes' normal vectors, as explained in the preceding section. The vector obtained from this operation is the direction vector of the intersection line.

• Plane 1: \( \mathbf{n_1} = 3 \mathbf{i} - \mathbf{j} + \mathbf{k} \)
• Plane 2: \( \mathbf{n_2} = \mathbf{i} + 4 \mathbf{j} - 2 \mathbf{k} \)

The cross product \( \mathbf{n_1} \times \mathbf{n_2} \) gives the direction vector of the line where these planes meet. Thus, this intersection line inherits properties from both planes, lying in both and not being perpendicular to either.

Understanding plane intersections can assist in numerous applications, such as computer graphics, navigation, and architecture, where spatial relationships are critical.

Normal Vector Calculation

Calculating normal vectors is essential when working with planes because these vectors provide insight into a plane's orientation in three-dimensional space. A normal vector is orthogonal (perpendicular) to the plane, and every point on the plane satisfies a linear equation determined by this vector.

For a plane with an equation \( \mathbf{r} \cdot \mathbf{n} = d \), where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), \( \mathbf{n} \) is the normal vector, and \( d \) is a constant. The normal vector \( \mathbf{n} \) can be extracted directly from the coefficients of \( x, y, \) and \( z \) in the equation.

In our exercise:

• The normal vector of the first plane is given by the equation's coefficients: \( 3 \mathbf{i} - \mathbf{j} + \mathbf{k} \)
• The second plane's normal vector is \( \mathbf{i} + 4 \mathbf{j} - 2 \mathbf{k} \)

Identifying these vectors correctly is crucial because any incorrect calculations may lead to wrong conclusions about the intersection, parallelism, or perpendicularity between planes.

By understanding the role of normal vectors, you can gain comprehensive knowledge about plane positioning and relationships between them in space, which proves highly beneficial in diverse mathematical and physical contexts.