Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(\quad\) [2005](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 \( \frac{5}{3} \).

Step 1: Identify the direction vector of the line

From the given line equation \( \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} \), the direction vector can be derived by taking the coefficients in the denominators. Hence, the direction vector of the line is \( \mathbf{d} = \langle 1, 2, 2 \rangle \).

Step 2: Identify the normal vector of the plane

The given plane equation is \( 2x - y + \sqrt{\lambda} z + 4 = 0 \). The coefficients of \(x\), \(y\), and \(z\) give the normal vector of the plane: \( \mathbf{n} = \langle 2, -1, \sqrt{\lambda} \rangle \).

Step 3: Use the angle formula between a line and a plane

The angle \( \theta \) between a line and a plane is given by \( \sin \theta = \frac{\text{direction vector} \, \cdot \, \text{normal vector}}{\| \text{direction vector} \| \, \| \text{normal vector} \|} \), where \( \cdot \) denotes the dot product. We know \( \sin \theta = \frac{1}{3} \).

Step 4: Calculate the dot product

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

Step 5: Calculate the magnitudes of the vectors

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

Step 6: Set up the equation for \(\sin \theta\) and solve for \(\lambda\)

Set up the equation using \( \sin \theta = \frac{1}{3} \):\[ \frac{|2\sqrt{\lambda}|}{3 \sqrt{5 + \lambda}} = \frac{1}{3} \]Multiply both sides by \(3 \sqrt{5 + \lambda}\):\[ |2\sqrt{\lambda}| = \sqrt{5 + \lambda} \]Square both sides to remove the absolute value and solve:\[ 4\lambda = 5 + \lambda \]\[ 3\lambda = 5 \]\[ \lambda = \frac{5}{3} \].

Step 7: Verify the solution

Check the solution by substituting \( \lambda = \frac{5}{3} \) back into the angle formula to ensure that \( \sin \theta = \frac{1}{3} \). This confirms that \( \lambda = \frac{5}{3} \) is indeed correct.

Key Concepts

Angle between Line and Plane

In mathematics, particularly in geometry, understanding the angle between a line and a plane is essential for various applications. When we talk about the angle between a line and a plane, we are often referring to the complementary angle formed between the line and the perpendicular (normal) to the plane. The angle is typically found using trigonometric relationships particularly involving the sine or cosine of the angle.

To find this angle, we use the sine identity \( \sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{\| \mathbf{d} \| \| \mathbf{n} \|} \). Here, \( \mathbf{d} \) is the direction vector of the line, and \( \mathbf{n} \) is the normal vector to the plane. The formula is derived from the projection of the line on the normal to the plane, effectively giving the inclination relative to the plane.

• \( \theta \) is the angle between the line and the plane.
• The dot product \( \mathbf{d} \cdot \mathbf{n} \) gives a sense of direction regarding the angle.
• The magnitudes \( \| \mathbf{d} \| \) and \( \| \mathbf{n} \| \) help standardize the calculation to avoid directional biases.

Knowing how to calculate this angle is essential, especially in engineering applications and physics, where the direction of forces and components can significantly impact real-world scenarios.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It provides a measure of two vectors' directional relationship by resulting in a scalar quantity. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]This operation provides vital information about the vectors:

• If \( \mathbf{a} \cdot \mathbf{b} = 0 \), then the vectors are perpendicular.
• The dot product gives the cosine of the angle when normalized by the magnitudes of the vectors.
• It is used in projection calculations to find components in specified directions.

In this context, using the dot product of the direction vector of the line and the normal vector from the plane helps establish the line's inclination relative to the plane’s normal. The simplification or manipulation of this scalar output helps deduce relationships among unknowns, such as \( \lambda \) in the original problem.

Direction and Normal Vectors

Vectors describe direction and magnitude. In geometry, the direction vector of a line indicates its alignment. It is derived from the line's parametric or symmetric form. For instance, the direction vector \( \mathbf{d} = \langle 1, 2, 2 \rangle \) in this context represents steps of 1 unit in x, 2 units in y, and 2 units in z.

A normal vector, on the other hand, is perpendicular to a plane's surface. It is conveniently derived from the coefficients of the equation of the plane. For the plane \( 2x - y + \sqrt{\lambda} z + 4 = 0 \), the normal vector is \( \mathbf{n} = \langle 2, -1, \sqrt{\lambda} \rangle \).

• Direction vectors determine the orientation of lines in space and are pivotal in calculating intersections and angles.
• Normal vectors are critical in defining planes and computing angles, projections, and reflections.

Understanding the physical interpretation of these vectors allows for graphical and analytical solutions to complex geometric problems. By combining these vectors with mathematical operations such as dot products, one can solve for unknowns, verify conditions, and better understand spatial relations.