Problem 28 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The number of distinct real values of \(\lambda\) for which the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^{2}}\) and \(\quad \frac{x-3}{1}=\frac{y-2}{\lambda^{2}}=\frac{z-1}{2} \quad\) are coplanar is : \(\quad\) [Online April 10, 2016] (a) 2 (b) 4 (c) 3 (d) 1

Step-by-Step Solution

Answer

The number of distinct real values of \(\lambda\) is 2.

Step 1: Understand Conditions for Coplanarity

Two lines are coplanar if they intersect or are parallel to each other. Here, we will find the condition for them to intersect.

Step 2: Direction Ratios from Parametric Form

The direction ratios for the line \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^{2}}\) are \((1, 2, \lambda^{2})\), and for the line \(\frac{x-3}{1}=\frac{y-2}{\lambda^{2}}=\frac{z-1}{2}\) are \((1, \lambda^{2}, 2)\).

Step 3: Find the Condition for Intersection

For the lines to intersect, the direction vectors and the vector connecting any points on the two lines must be coplanar. Use the scalar triple product: \[ \begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \end{vmatrix} = 0 \] where \(\vec{a} = (1, 2, \lambda^2)\), \(\vec{b} = (1, \lambda^2, 2)\), and \(\vec{c} = (2, 0, 4)\) (obtained from the difference of two arbitrary points on the lines such as \((1, 2, -3)\) and \((3, 2, 1)\)).

Step 4: Evaluate the Scalar Triple Product

Calculate \[\begin{vmatrix} 1 & 2 & \lambda^2 \ 1 & \lambda^2 & 2 \ 2 & 0 & 4 \end{vmatrix} = 1(\lambda^2 \cdot 4 - 2 \cdot 0) - 2(1 \cdot 4 - 2 \cdot 2) + \lambda^2(1 \cdot 0 - 2 \cdot 1)\]Simplifying, \[= 4\lambda^{2} - 8 - 2\lambda^2 = 0 \]leading to \[ 2\lambda^2 - 8 = 0 \] which simplifies to \[ \lambda^2 = 4 \].

Step 5: Solve for \(\lambda\)

Solving \(\lambda^2 = 4\) gives \(\lambda = \pm 2\). Thus, two distinct real values for \(\lambda\) are possible.

Key Concepts

Direction Ratios

Lines in space can be represented in a parametric form, which involves direction ratios or direction numbers. Consider a line described parametrically. The coefficients in the parametric equations of the line represent its direction ratios.

For the line \( \frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda^2} \), the direction ratios are \((1, 2, \lambda^2)\). Similarly, the line \( \frac{x-3}{1} = \frac{y-2}{\lambda^2} = \frac{z-1}{2} \) has the direction ratios \((1, \lambda^2, 2)\).

These direction ratios help us understand the orientation of each line in three-dimensional space. They are central to determining how two lines relate, such as checking for parallelism or finding the intersection condition. With the direction ratios, we can proceed to explore conditions like coplanarity that involves these lines.

Scalar Triple Product

The scalar triple product is a mathematical concept used to determine whether three vectors are coplanar, meaning they lie on the same plane. Mathematically, it is expressed using the determinant of a matrix formed by three vectors. For vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \), their scalar triple product is given by:

\[ \begin{vmatrix} \vec{a} & \vec{b} & \vec{c} \end{vmatrix} = 0 \]

In the context of our problem, \( \vec{a} = (1, 2, \lambda^2) \), \( \vec{b} = (1, \lambda^2, 2) \), and \( \vec{c} = (2, 0, 4) \), are the direction vectors and vector connecting points from the lines, respectively. Setting the scalar triple product to zero ensures the lines and connecting vector are coplanar, which is the condition needed for them to potentially intersect.

Intersection Condition

For two lines in three-dimensional space to intersect, their direction vectors and vectors connecting any two points on each line must be coplanar. This requirement boils down to ensuring the scalar triple product of these vectors equals zero.

By solving the expression \( \begin{vmatrix} 1 & 2 & \lambda^2 \ 1 & \lambda^2 & 2 \ 2 & 0 & 4 \end{vmatrix} = 0 \), we can derive the intersection condition for the lines. Simplifying this determinant:

- Calculate: \( 4\lambda^2 - 8 - 2\lambda^2 = 0 \)
- This results in: \(2\lambda^2 - 8 = 0\)

Solving \(\lambda^2 = 4\) gives the potential values of \(\lambda = \pm 2\), meaning the lines intersect for these specific values, confirming their coplanarity.