Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}\) and \((2 \lambda-1) \bar{c}\) are non-coplanar for (A) all values of \(\lambda\) (B) all except one value of \(\lambda\) (C) all except two values of \(\lambda\) (D) no value or \(\lambda\)

Step-by-Step Solution

Answer

(B) all except one value of \(\lambda\)

Step 1: Understanding Non-Coplanar Vectors

Vectors are non-coplanar if they do not lie in the same plane. For vectors \(\bar{a}, \bar{b}, \bar{c}\) to be non-coplanar with another set, the new set should not be expressible as a linear combination with certain constraints.

Step 2: Set Up Vector Equation

We have three vectors: \(\bar{x} = \bar{a}+2 \bar{b}+3 \bar{c}\), \(\bar{y} = \lambda \bar{b}+4 \bar{c}\), and \(\bar{z} = (2 \lambda-1) \bar{c}\). The question is whether they are non-coplanar for all \(\lambda\).

Step 3: Condition for Coplanarity

Vectors \(\bar{x}\), \(\bar{y}\), and \(\bar{z}\) are coplanar if there exist scalars \(k_1, k_2, k_3\) (not all zero) such that \(k_1 \bar{x} + k_2 \bar{y} + k_3 \bar{z} = \bar{0}\). We expand this equation to see the condition.

Step 4: Represent Vectors in an Equation

Express this condition explicitly: \[k_1(\bar{a} + 2\bar{b} + 3\bar{c}) + k_2(\lambda \bar{b} + 4\bar{c}) + k_3((2\lambda-1)\bar{c}) = \bar{0}\]Rearrange by grouping coefficients of \(\bar{a}, \bar{b}, \bar{c}\).

Step 5: Solve for Coefficients

This results in three scalar conditions: \[k_1 = 0\](for \(\bar{a}\), as \(\bar{a}\) is not part of other vectors),\[2k_1 + k_2\lambda = 0\](for \(\bar{b}\)), and\[3k_1 + 4k_2 + k_3(2\lambda-1) = 0\](for \(\bar{c}\)).

Step 6: Analyze the Equations

Since \(k_1 = 0\), we simplify our remaining conditions to:\[k_2\lambda = 0\]\[4k_2 + k_3(2\lambda-1) = 0\] The first implies either \(\lambda = 0\) or \(k_2 = 0\). Analyze each case.

Step 7: Evaluate Cases for \(\lambda\)

Case 1: If \(k_2 eq 0\), then \(\lambda = 0\).Case 2: If \(k_2 = 0\), utilize \(k_3 = 0\) since the remaining equation cannot be zero otherwise. Thus, all vectors become linearly independent in this scenario.Conclusion: Non-coplanarity fails only when \(\lambda = 0\).

Key Concepts

Non-coplanarity

To understand non-coplanarity, it is essential to grasp what coplanarity means first. Vectors are considered coplanar if they lie within the same geometric plane. Conversely, vectors are non-coplanar when no single plane can contain all of them. This property of vectors being non-coplanar often ensures that they provide a full span of three-dimensional space.

In mathematical terms, if vectors \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar, there are no scalars \( k_1, k_2, k_3 \) — except all being zero — for which the linear combination \( k_1 \vec{a} + k_2 \vec{b} + k_3 \vec{c} = \vec{0} \) holds. Non-coplanarity is crucial in applications where three dimensions play a significant role, such as in computer graphics, physics simulations, and various engineering tasks. For the problem at hand, understanding non-coplanarity helps us determine which values of \( \lambda \) might cause vectors to lie on the same plane, affecting their linear independence.

Linear independence

Linear independence is a critical concept in vector analysis. Vectors are considered linearly independent if no vector in the set can be represented as a linear combination of the others. This means, for vectors \( \vec{u}, \vec{v}, \vec{w} \), they are linearly independent if the equation \( c_1\vec{u} + c_2\vec{v} + c_3\vec{w} = \vec{0} \) holds only when each scalar \( c_1, c_2, c_3 \) is zero.

In our exercise, the vectors \( \vec{x}, \vec{y}, \vec{z} \) are constructed from \( \vec{a}, \vec{b}, \vec{c} \), and their linear independence depends on the values of \( \lambda \). We explored cases where \( k_1 = 0 \), simplifying to two conditions based on \( k_2 \) and \( k_3 \). The vector set becomes linearly dependent when a specific \( \lambda \) enables a non-zero solution to the linear combination provided. This was specifically noted when \( \lambda = 0 \) in the context of the provided problem, indicating a loss of non-coplanarity.

Scalar multiples

Scalar multiplication involves taking a vector and multiplying it by a scalar (a real number). This operation scales the vector by enlarging or reducing it, depending on the magnitude of the scalar. If the scalar is positive, the direction remains unchanged; if negative, the vector flips.

Scalar multiples play a significant role in determining linear dependence or independence among vectors. If one vector in a set can be expressed as a scalar multiple of another, the vectors are linearly dependent. Regarding our problem, vectors \( \vec{x}, \vec{y}, \vec{z} \) used scalars to generate combinations that potentially reveal dependence or independence. By examining the conditions under which coefficients align, we were able to determine the specific \( \lambda \) affecting coplanarity and linear independence.

Understanding scalar multiples helps in solving vector-related problems that involve transformations, comparing magnitudes, and scaling in both theoretical and practical applications.