Problem 129 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Distance between two parallel planes \(2 x+y+2 z=8\) and \(4 x+2 y+4 z+5=0\) is (a) \(\frac{9}{2}\) (b) \(\frac{5}{2}\) (c) \(\frac{7}{2}\) (d) \(\frac{3}{2}\)

Step-by-Step Solution

Answer

The distance between the planes is \(\frac{7}{2}\), option (c).

Step 1: Identify the Normal Vector

The normal vector for a plane in the form \(Ax + By + Cz = D\) is \( \langle A, B, C \rangle \). For the plane \(2x + y + 2z = 8\), the normal vector is \( \langle 2, 1, 2 \rangle \). For the plane \(4x + 2y + 4z + 5 = 0\), we simplify it to \(4x + 2y + 4z = -5\) and see that it shares the directional vector \( \langle 4, 2, 4 \rangle \) which is a scalar multiple of the first plane's normal vector: \(2 \times \langle 2, 1, 2 \rangle \). This confirms that the planes are parallel.

Step 2: Normalize the Plane Equation

To find the distance between two parallel planes, we should use the form \(Ax + By + Cz + D = 0\). The first plane \(2x + y + 2z = 8\) can be rewritten as \(2x + y + 2z - 8 = 0\), which means \(D = -8\). The second plane is given as \(4x + 2y + 4z + 5 = 0\), with \(D = 5\). For easy comparison, divide the second plane's equation by 2, yielding \(2x + y + 2z + \frac{5}{2} = 0\). Thus \(D = \frac{5}{2}\).

Step 3: Calculate Distance between the Planes

The formula for the distance \(d\) between two parallel planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is \(d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}\). For the planes \(2x + y + 2z - 8 = 0\) and \(2x + y + 2z + \frac{5}{2} = 0\), \(A = 2\), \(B = 1\), \(C = 2\). Thus, \(D_1 = -8\) and \(D_2 = \frac{5}{2}\). The distance is \[d = \frac{|\frac{5}{2} - (-8)|}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{|\frac{5}{2} + 8|}{\sqrt{9}} = \frac{|\frac{5}{2} + \frac{16}{2}|}{3} = \frac{\frac{21}{2}}{3} = \frac{21}{6} = \frac{7}{2}.\]

Step 4: Select the Correct Answer

The calculated distance between the two parallel planes is \(\frac{7}{2}\), which corresponds to option (c).

Key Concepts

Normal vector of a plane

In the study of geometry, the normal vector is essential when dealing with planes. A normal vector to a plane is a vector that is perpendicular to every vector lying on the plane. In mathematics, the expression of a plane is typically given as:
\[ Ax + By + Cz = D \]This polynomial equation represents the set of all points (x, y, z) that lie on the plane. The coefficients \(A\), \(B\), and \(C\) serve as crucial components because they make up the normal vector of the plane, expressed as:

• \( \langle A, B, C \rangle \)

For example, for the plane equation \( 2x + y + 2z = 8 \), the normal vector is \( \langle 2, 1, 2 \rangle \). It's incredibly important to understand this because it helps in verifying the parallelism between planes.
Planes that are parallel share normal vectors that are scalar multiples of each other, indicating they run in identical directions across the space.

Equation of a plane

The equation of a plane in three-dimensional space gives us a clear picture of how the plane is situated in space relative to the coordinate axes. The standard form is:

• \( Ax + By + Cz + D = 0 \)

The constants \(A\), \(B\), and \(C\) represent the normal vector as covered earlier. The constant \(D\) shifts the plane either closer to or further from the origin, making the entire equation crucial for understanding the plane's location.
For instance, the plane equation \( 2x + y + 2z = 8 \) can be rearranged to \( 2x + y + 2z - 8 = 0 \). This subtle change helps us convert the given plane to a form suitable for distance calculations.
It’s always good to transform any given plane equation into its standard form for easier calculations, particularly when comparing with other planes or when calculating distances.

Distance formula for parallel planes

When you need to determine the distance between two parallel planes, using the distance formula is straightforward yet highly effective. The planes will have equations of the form:

• \( Ax + By + Cz + D_1 = 0 \)
• \( Ax + By + Cz + D_2 = 0 \)

The formula to calculate the distance \(d\) between these two parallel planes is:
\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]Here, \( |D_2 - D_1| \) measures the difference in location along the normal vector between the two planes. Meanwhile, \( \sqrt{A^2 + B^2 + C^2} \) normalizes this difference by taking the magnitude of the normal vector.
In the case of our previously defined planes, applying this formula yields \( d = \frac{21}{6} = \frac{7}{2} \), a simple calculation that demonstrates the beauty of mathematics in providing precise spatial measurements.