Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The vector equation of the line of intersection of the planes \(\mathbf{r} \cdot(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})=0\) and \(\mathbf{r} \cdot(3 \mathbf{i}+2 \mathbf{j}+\mathbf{k})=0\) is \((\) A) \(\mathbf{r}=\lambda(\mathbf{i}+2 \mathbf{i}+\mathbf{k})\) (B) \(\mathbf{r}=\lambda(\mathbf{i}-2 \mathbf{i}+\mathbf{k})\) (C) \(\mathbf{r}=\lambda(\mathbf{i}+2 \mathbf{i}-3 \mathbf{k})\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (D) none of these.

Step 1: Identify the Normal Vectors of the Planes

The given planes are represented in the normal vector form. The normal vector for the first plane \( \mathbf{r} \cdot (\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})=0 \) is \( \mathbf{n_1} = \langle 1, 2, 3 \rangle \). Similarly, for the second plane \( \mathbf{r} \cdot (3\mathbf{i}+2\mathbf{j}+\mathbf{k})=0 \), the normal vector is \( \mathbf{n_2} = \langle 3, 2, 1 \rangle \).

Step 2: Calculate the Direction Vector of the Intersection Line

The direction vector of the line of intersection of two planes can be found by taking the cross product of their normal vectors. Compute \( \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \).\[ \mathbf{d} = \langle 1, 2, 3 \rangle \times \langle 3, 2, 1 \rangle = \mathbf{i}(2 \times 1 - 3 \times 2) - \mathbf{j}(1 \times 1 - 3 \times 3) + \mathbf{k}(1 \times 2 - 2 \times 3) \]Solving these, we have:\[ \mathbf{d} = \mathbf{i}(-4) - \mathbf{j}(-8) + \mathbf{k}(-4) = \langle -4, 8, -4 \rangle \]

Step 3: Simplify the Direction Vector

The direction vector \( \langle -4, 8, -4 \rangle \) can be simplified by dividing by the greatest common divisor, which is 4. Thus, \( \mathbf{d} = \langle -1, 2, -1 \rangle \).

Step 4: Write the Vector Equation of the Line

Since we have our direction vector, the line's vector equation can be expressed as:\[ \mathbf{r} = \lambda \langle -1, 2, -1 \rangle \]This means the line has the equation \( \mathbf{r} = \lambda (-\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \).

Step 5: Compare with Given Options

None of the provided options directly match the derived line equation. Thus, the correct answer should be: (D) none of these.

Key Concepts

Lines and Planes Intersection

When two planes intersect, the result is often a line. This is because planes are two-dimensional surfaces, and when they cross or meet, they do so along a line. Imagine each plane as a piece of cardboard, meeting at a line of contact. Mathematically, the line of intersection of two planes can be found by determining a direction vector for the line, which is perpendicular to the normal vectors of the planes.

• The given planes are typically described through their normal vectors.
• The direction of the intersection line can be calculated using the cross product of the normals.
• The resulting line can then be described using a vector equation.

The vector equation gives a way to express any point on the line in terms of a parameter, typically denoted by \( \lambda \). In summary, the intersection of planes results in a line described by a unique direction vector.

Normal Vector

A normal vector is a vector that is perpendicular to a surface. In the context of plane geometry, it is perpendicular to the plane itself. The equation of a plane generally involves its normal vector.

• For a plane described as \( \mathbf{r} \cdot \mathbf{n} = 0 \), \( \mathbf{n} \) represents the normal vector.
• Normal vectors provide a straightforward way to describe the orientation of a plane in three-dimensional space.

The normal vectors for two specific planes from the exercise are \( \mathbf{n_1} = \langle 1, 2, 3 \rangle \) and \( \mathbf{n_2} = \langle 3, 2, 1 \rangle \). These vectors are crucial for determining how the planes relate to each other, particularly when calculating the line of intersection.

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two other vectors in three-dimensional space. In the case of two intersecting planes, the cross product of their normal vectors yields the direction vector of their intersection line.

• Given two vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \), the cross product is calculated as \( \mathbf{n_1} \times \mathbf{n_2} \).
• The cross product results in a vector perpendicular to both \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

In the exercise, the cross product \( \langle 1, 2, 3 \rangle \times \langle 3, 2, 1 \rangle \) led to \( \langle -4, 8, -4 \rangle \), which was simplified to \( \langle -1, 2, -1 \rangle \). This vector signifies the direction of the line of intersection.

Direction Vector

A direction vector is essential for determining the orientation and direction of a line. In vector equations of lines, the direction vector indicates the line's path through space. For the line formed by the intersection of two planes, the direction vector shows which way the line extends.

• The direction vector is derived from the cross product of the plane's normal vectors.
• In the exercise, this vector is \( \langle -1, 2, -1 \rangle \).
• It tells us that for every unit the line moves in the x-direction, it moves two units in the y-direction and negative one unit in the z-direction.

The vector equation of the line can then be written as \( \mathbf{r} = \lambda \langle -1, 2, -1 \rangle \), helping us to understand all points through which the line passes. The parameter \( \lambda \) can vary to represent different points along this direction.