Problem 53 - A Complete Resource Book in Mathematics for JEE Main

Question

If the angle \(Q\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+\sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\) the value of \(\lambda\) is \(\quad\) (A) \(\frac{5}{3}\) (B) \(\frac{-3}{5}\) (C) \(\frac{3}{4}\) (D) \(\frac{-4}{3}\)

Step-by-Step Solution

Answer

The value of \( \lambda \) is \( \frac{5}{3} \) (Option A).

Step 1: Identify the direction vector of the line

The given line equation is \( \frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2} \). This can be rewritten in parametric form as: \( x = t - 1 \), \( y = 2t + 1 \), \( z = 2t + 2 \), where \( t \) is a parameter. The direction vector of this line is \( \langle 1, 2, 2 \rangle \).

Step 2: Normal vector of the plane

For the plane \( 2x - y + \sqrt{\lambda}z + 4 = 0 \), the normal vector is \( \langle 2, -1, \sqrt{\lambda} \rangle \).

Step 3: Use the sine formula for the angle between the line and plane

The sine of the angle \( \theta \) between a line and a plane is given by the formula: \( \sin \theta = \frac{|\mathbf{a} \cdot \mathbf{n}|}{\|\mathbf{a}\| \cdot \|\mathbf{n}\|} \), where \( \mathbf{a} \) is the line's direction vector and \( \mathbf{n} \) is the plane's normal vector.

Step 4: Calculate the dot product of direction and normal vectors

Compute the dot product \( \mathbf{a} \cdot \mathbf{n} = \langle 1, 2, 2 \rangle \cdot \langle 2, -1, \sqrt{\lambda} \rangle = 1 \cdot 2 + 2 \cdot (-1) + 2 \cdot \sqrt{\lambda} = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda} \).

Step 5: Calculate the magnitudes of the vectors

The magnitude of \( \mathbf{a} \) is \( \|\mathbf{a}\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \). For the normal vector \( \mathbf{n} \), \( \|\mathbf{n}\| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{4 + 1 + \lambda} = \sqrt{5 + \lambda} \).

Step 6: Substitute into the sine formula and solve for \(\lambda\)

Substitute the values into the sine formula: \( \sin \theta = \frac{|2\sqrt{\lambda}|}{3 \cdot \sqrt{5 + \lambda}} = \frac{1}{3} \). Simplifying gives: \( |2\sqrt{\lambda}| = \sqrt{5 + \lambda} \).

Step 7: Solve the equation \(|2\sqrt{\lambda}| = \sqrt{5 + \lambda}\)

Square both sides: \( 4\lambda = 5 + \lambda \). Rearrange to get \( 3\lambda = 5 \), so \( \lambda = \frac{5}{3} \).

Key Concepts

Direction Vector

When working with problems involving lines in space, understanding the direction vector is fundamental. The direction vector indicates the orientation of the line, guiding its path through space.
For the line represented by the symmetric equation \( \frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2} \), we can express it in a more intuitive form known as the parametric form. Here, \( x = t - 1 \), \( y = 2t + 1 \), and \( z = 2t + 2 \), where \( t \) is a parameter.
From reading these equations, we deduce the direction vector as \( \langle 1, 2, 2 \rangle \) because these values show the increments in \( x \), \( y \), and \( z \) as \( t \) varies. This vector tells us that for each increase in \( t \), \( x \) increases by 1, \( y \) by 2, and \( z \) also by 2. This information helps us in comprehending the spatial characteristics of the line.

Normal Vector

The normal vector is crucial when dealing with planes. It provides the direction that is perpendicular to the plane's surface, defining its orientation in space.
For the plane equation \( 2x - y + \sqrt{\lambda}z + 4 = 0 \), the coefficients of \( x, y, \) and \( z \) are used to derive the normal vector. Hence, the normal vector for this plane is \( \langle 2, -1, \sqrt{\lambda} \rangle \).
This calculated vector is perpendicular to any vector lying on the plane, revealing the spatial nature and orientation of the plane. Understanding normal vectors is essential for calculating angles, distances, and intersections with other geometric objects, like in this problem.

Sine Formula

To find the angle \( \theta \) between the line and the plane, we utilize the sine formula. This formula involves the direction vector of the line and the normal vector of the plane:

• \( \sin \theta = \frac{|\mathbf{a} \cdot \mathbf{n}|}{\|\mathbf{a}\| \cdot \|\mathbf{n}\|} \)

Here, \( \mathbf{a} \) is the line's direction vector \( \langle 1, 2, 2 \rangle \), and \( \mathbf{n} \) is the plane's normal vector \( \langle 2, -1, \sqrt{\lambda} \rangle \). Calculating the dot product gives \( 2\sqrt{\lambda} \).
The magnitudes are \( \|\mathbf{a}\| = \sqrt{9} = 3 \) and \( \|\mathbf{n}\| = \sqrt{5 + \lambda} \).
Applying this to our formula, we find \( \sin \theta = \frac{|2\sqrt{\lambda}|}{3 \cdot \sqrt{5 + \lambda}} \). The given \( \sin \theta = \frac{1}{3} \) helps us to solve the equation, ensuring understanding and verifying the solution for \( \lambda \).