Problem 5 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the volume of parallelopiped formed by the vectors \(\hat{i}+\lambda \hat{j}+\hat{k}, \hat{j}+\lambda \hat{k}\) and \(\lambda \hat{i}+\hat{k}\) is minimum, then \(\lambda\) is equal to: \(\quad\) [April 12, 2019 (I)] (a) \(-\frac{1}{\sqrt{3}}\) (b) \(\frac{1}{\sqrt{3}}\) (c) \(\sqrt{3}\) (d) \(-\sqrt{3}\)

Step-by-Step Solution

Answer

The value of \( \lambda \) is \( \frac{1}{\sqrt{3}} \).

Step 1: Recognize the Determinant Formula for Volume

The volume of a parallelepiped formed by three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is given by the absolute value of the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \). This is equivalent to the determinant of the matrix with \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) as rows.

Step 2: Define the Vectors

Given vectors are: \( \mathbf{a} = \hat{i} + \lambda \hat{j} + \hat{k} \), \( \mathbf{b} = \hat{j} + \lambda \hat{k} \), \( \mathbf{c} = \lambda \hat{i} + \hat{k} \).

Step 3: Set up the Determinant

Set up the 3x3 matrix with these vectors as rows: \[\begin{vmatrix}1 & \lambda & 1 \0 & 1 & \lambda \\lambda & 0 & 1\end{vmatrix}\]

Step 4: Execute the Determinant Calculation

Calculate the determinant:- Expand along the first row: \[1 \begin{vmatrix}1 & \lambda \0 & 1\end{vmatrix} - \lambda \begin{vmatrix}0 & \lambda \\lambda & 1\end{vmatrix} + 1 \begin{vmatrix}0 & 1 \\lambda & 0\end{vmatrix}\]- Calculate sub-determinants: - \(\begin{vmatrix}1 & \lambda \0 & 1\end{vmatrix} = 1\) - \(\begin{vmatrix}0 & \lambda \ \lambda & 1\end{vmatrix} = -\lambda^2\) - \(\begin{vmatrix}0 & 1 \ \lambda & 0\end{vmatrix} = -\lambda\)- Substituting these, we have: \[1(1) - \lambda(-\lambda^2) + 1(-\lambda) = 1 + \lambda^3 - \lambda\]- Which simplifies to: \[ \lambda^3 + 1 - \lambda \]

Step 5: Minimize the Volume

The volume is minimized when \( \lambda^3 + 1 - \lambda \) is minimized. The critical points of this function occur when its derivative is zero. Calculate the derivative: \[ f(\lambda) = \lambda^3 - \lambda + 1 \]\[ f'(\lambda) = 3\lambda^2 - 1 \]Set \( f'(\lambda) = 0 \) to find \( \lambda \): \[ 3\lambda^2 - 1 = 0 \]\[ 3\lambda^2 = 1 \]\[ \lambda^2 = \frac{1}{3} \]\[ \lambda = \pm \frac{1}{\sqrt{3}} \]

Step 6: Choose the Correct Option

The critical values for \( \lambda \) are \( \pm \frac{1}{\sqrt{3}} \). Among the given options (a), (b), (c), and (d), the correct value that matches one of these solutions is (b) \( \frac{1}{\sqrt{3}} \).

Key Concepts

Determinant Calculation

The determinant is a special number that we can calculate from a square matrix. It's especially useful in working with vectors and matrices in geometry, like when dealing with volumes of shapes such as parallelepipeds. In our case, the determinant helps in calculating the volume of the parallelepiped formed by three given vectors.

If you have three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), the volume of the parallelepiped they form is the absolute value of their scalar triple product, represented as \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \). Mathematically, this is equivalent to the determinant of a 3x3 matrix formed by these vectors as rows.

To compute the determinant in this problem, you'll place each of the vectors as a row in a 3x3 matrix and then use expansion along the rows to find its determinant. You'll calculate minor determinants for each element in a row, multiply them by the respective elements, and sum them up, paying close attention to signs. In this exercise, after calculating the full determinant, we are left with \( \lambda^3 + 1 - \lambda \). This expression stays critical for finding where the volume of the parallelepiped is at its minimum, helping us identify the necessary lambda value.

Parallelepiped Volume

A parallelepiped is a three-dimensional shape with six faces, each of which is a parallelogram. Think of it as a more flexible version of a box-like shape. Calculating its volume helps us understand three-dimensional space and vector mathematics better.

In this exercise, we are interested in how the volume of a parallelepiped can be determined using vectors and their products. The volume is linked to the scalar triple product of the three vectors that form it. Once we calculate this triple product via the determinant, the absolute value gives the parallelepiped's volume.

The beauty (and challenge) here is in understanding that minimization of this product's result corresponds to minimizing the actual volume of the shape. Hence, the smaller the determinant (closer to zero while ensuring positive semi-definiteness), the smaller the volume of our geometric shape.

Critical Points

Critical points are instrumental in calculus and optimization problems, where they help identify local maxima, minima, or saddle points for a given function.

These points occur where the derivative of a function is zero or undefined. In the context of this problem, we use derivatives to identify where the volume function, calculated from the determinant, reaches its minimum value.

Consider the volume function \( f(\lambda) = \lambda^3 + 1 - \lambda \), whose derivative \( f'(\lambda) = 3\lambda^2 - 1 \) is used to find the critical points by setting \( f'(\lambda) = 0 \). Solving this provides us with potential lambda values that might cause the volume of our parallelepiped to be smallest. Eventually, these lambda values guide us to the options capable of achieving the reduced volume, helping us find the correct answer choice of \( \frac{1}{\sqrt{3}} \). Hence, understanding critical points is essential for locating extrema of functions, especially in geometric and physical problems like this.