Problem 92 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\vec{a}, \vec{b}, \vec{c}\) arenon coplanar vectors and \(\lambda\) is a real number then \(\quad[2005]\) \(\left[\lambda(\vec{a}+\vec{b}) \lambda^{2} \bar{b} \lambda \vec{c}\right]=\left[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{b}\end{array}\right]\) for (a) exactly one value of \(\lambda\) (b) no value of \(\lambda\) (c) exactly three values of \(\lambda\) (d) exactly two values of \(\lambda\)

Step-by-Step Solution

Answer

(b) no value of \(\lambda\).

Step 1: Understanding Mixed Product

We need to equate mixed products of two different sets of vectors. In the equation, \([ \lambda(\vec{a} + \vec{b}), \lambda^2 \vec{b}, \lambda \vec{c} ] = [ \vec{a}, \vec{b} + \vec{c}, \vec{b} ]\), the terms represent the scalar triple product or the volume of the parallelepiped formed by three vectors.

Step 2: Apply Scalar Triple Product Identity

According to the identity, \([\vec{x}, \vec{y}, \vec{z}]\) equals \([\vec{x}, \vec{z}, \vec{y}]\), but when there's a switch between non-adjacent vectors, the sign changes: \([\vec{x}, \vec{y}, \vec{z}] = -[\vec{x}, \vec{z}, \vec{y}]\). Therefore, our equation can be re-evaluated considering permutations and signs.

Step 3: Calculate Determinants of Given Sets

Evaluate the determinants on both sides. For the left side: \([\lambda(\vec{a}+\vec{b}), \lambda^2 \vec{b}, \lambda \vec{c}] = \lambda^4 [\vec{a}+\vec{b}, \vec{b}, \vec{c}]\). For the right side: \([\vec{a}, \vec{b} + \vec{c}, \vec{b}]\).

Step 4: Equating Volumes

Set the determinant equivalence: \(\lambda^4 [\vec{a} + \vec{b}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b} + \vec{c}, \vec{b}]\). Since \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, we can discuss cases depending on \(\lambda\).

Step 5: Solve for \(\lambda\)

The determinant \([\vec{a}+\vec{b}, \vec{b}, \vec{c}]\) expands to zero because \(\vec{b}\) is a common vector in two slots, implying zero volume or coplanarity. However, for real values solving \([\vec{a}, \vec{b} + \vec{c}, \vec{b}]\), the constant terms cannot compare to a zero determinant on left, resulting in no solution for \(\lambda\).

Step 6: Conclude Solution

Since the only logical assessment is that the left-hand side's determinant consistently yields zero due to the linear dependence portrayed, but the right lacks a dependence equivalent scenario, it means no solution exists since volumes equating to a zero assumption do not match. Thus, the answer aligns with the concept of perpetually having zero left volume.

Key Concepts

Scalar Triple Product

The scalar triple product is a fundamental concept in vector algebra. It involves three vectors, often denoted as \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), and is expressed as \([\vec{a}, \vec{b}, \vec{c}]\). This expression represents the volume of the parallelepiped shaped by these vectors.

The calculation of the scalar triple product is commonly done using the determinant of a 3x3 matrix, where the vectors form the rows. Its value can be computed using the formula:

• \([\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})\)

This formula reveals the link between the triple product and the cross and dot products. It essentially measures how much one vector lies in the plane defined by the other two. When the scalar triple product is zero, it indicates that the vectors are coplanar.

The properties of the scalar triple product allow changes in the order of vectors, affecting only the sign of the result. Hence:

• \([\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]\)
• \([\vec{a}, \vec{c}, \vec{b}] = -[\vec{a}, \vec{b}, \vec{c}]\)

Understanding these properties is crucial when manipulating expressions involving scalar triple products in vector problems.

Non-Coplanar Vectors

Vectors are termed non-coplanar when they do not all lie in the same plane. This implies that no linear combination of two vectors can yield the third. In three-dimensional space, non-coplanar vectors are inherently linearly independent.

The significance of non-coplanar vectors becomes evident in the context of the scalar triple product. If three vectors \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, their scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) will not be zero. This aligns with the fact that non-coplanarity signals the enclosure of volume, whereas coplanar vectors would define zero volume due to their confinement to a plane.

In practical vector operations, ensuring vectors are non-coplanar is crucial when attempting to calculate volumes or establish independence. In some exercises, the condition of non-coplanarity is given, as it guarantees certain properties, such as having unique solutions when relating to specific algebraic equations involving scalar triple products.

• Significance: Non-coplanarity assures a volumetric space, indicated by non-zero scalar triple products.
• Applications: Used in solving vector equations, defining spaces, and ensuring three-dimensionality in geometric contexts.

Determinants in Vector Operations

Determinants play a crucial role in vector algebra, serving as the backbone for calculations involving vector products, especially the scalar triple product.

In vector operations, the determinant is used to find the area or volume enclosed by vectors. For three vectors in space, this is determined using a 3x3 matrix where each vector forms a row:

• The determinant of this matrix provides the scalar triple product value, hence the volume of the parallelepiped formed.
• A zero determinant for vectors \(\vec{a}, \vec{b}, \vec{c}\) implies linear dependence, or coplanarity, making the enclosed volume zero.

Determinants reveal critical properties like indicating when vectors are coplanar. For non-coplanar vectors, the determinant is non-zero, underscoring their ability to span three-dimensional space.

Using determinants, we can also solve systems of linear equations represented by vectors. It acts as a test for independence and the existence of solutions, rooting from their ability to ascertain unique solutions when dealing with vector equations.

• Calculation Tip: Swap any pair of rows in the determinant to reverse its sign. Leave swapped rows unmatched, and the result remains consistent for volume or planar assessments.
• Practical Use: Employed in geometry, physic simulations, and solving complex algebraic expressions involving vector products.