Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the line \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\) intersects the plane \(2 x+3 y-z+13=0\) at a point \(P\) and the plane \(3 x+y+4 z=\) 16 at a point \(Q\), then \(P Q\) is equal to: [April 12, 2019 (I)] (a) 14 (b) \(\sqrt{14}\) (c) \(2 \sqrt{7}\) (d) \(2 \sqrt{14}\)

Step-by-Step Solution

Answer

The distance PQ is \(2\sqrt{14}\), option (d).

Step 1: Parametrize the Line

The given line equation is \( \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1} = t \), where \( t \) is a parameter. From this, we can express \( x, y, \) and \( z \) as:\[ x = 3t + 2, \quad y = 2t - 1, \quad z = -t + 1 \]

Step 2: Find Intersection with Plane 1 for Point P

Substitute \( x = 3t + 2, y = 2t - 1, z = -t + 1 \) into the plane equation \( 2x + 3y - z + 13 = 0 \):\[ 2(3t + 2) + 3(2t - 1) - (-t + 1) + 13 = 0 \] Simplify and solve for \( t \):\[ 6t + 4 + 6t - 3 + t - 1 + 13 = 0 \]\[ 13t + 13 = 0 \]\[ t = -1 \]

Step 3: Calculate Point P

Using \( t = -1 \), substitute back into \( x = 3t + 2, y = 2t - 1, z = -t + 1 \):\[ x = 3(-1) + 2 = -1 \]\[ y = 2(-1) - 1 = -3 \]\[ z = -(-1) + 1 = 2 \]Thus, point \( P \) is \((-1, -3, 2)\).

Step 4: Find Intersection with Plane 2 for Point Q

Substitute \( x = 3t + 2, y = 2t - 1, z = -t + 1 \) into the plane equation \( 3x + y + 4z = 16 \):\[ 3(3t + 2) + (2t - 1) + 4(-t + 1) = 16 \] Simplify and solve for \( t \):\[ 9t + 6 + 2t - 1 - 4t + 4 = 16 \]\[ 7t + 9 = 16 \]\[ 7t = 7 \]\[ t = 1 \]

Step 5: Calculate Point Q

Using \( t = 1 \), substitute back into \( x = 3t + 2, y = 2t - 1, z = -t + 1 \):\[ x = 3(1) + 2 = 5 \]\[ y = 2(1) - 1 = 1 \]\[ z = -(1) + 1 = 0 \]Thus, point \( Q \) is \((5, 1, 0)\).

Step 6: Calculate the Distance Between Points P and Q

The distance \( PQ \) can be found using the distance formula:\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]Substitute \( P(-1, -3, 2) \) and \( Q(5, 1, 0) \):\[ PQ = \sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2} \]\[ PQ = \sqrt{(6)^2 + (4)^2 + (-2)^2} \]\[ PQ = \sqrt{36 + 16 + 4} \]\[ PQ = \sqrt{56} \]\[ PQ = 2\sqrt{14} \]

Step 7: Final Answer

Since the calculated distance between points \( P \) and \( Q \) is \( 2\sqrt{14} \), option (d) is correct.

Key Concepts

Intersection of Line and Plane

The intersection of a line and a plane is a fundamental concept in coordinate geometry. This occurs when a line passes through a plane at exactly one point. To find this point, we use the parametric equation of the line and substitute these expressions into the plane's equation.
For the line \( \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1} \), the parametric form is \( x = 3t + 2 \), \( y = 2t - 1 \), and \( z = -t + 1 \).
By substituting into the plane \( 2x + 3y - z + 13 = 0 \), we find the specific value of \( t \) where the line intersects this plane. Similar steps apply to find intersections on different planes. Understanding these steps is crucial for visualizing where and how the line meets the plane in space.

Parametric Equations

Parametric equations are essential for describing geometric objects using a parameter, typically \( t \). Instead of expressing variables in terms of each other, we express them separately using the third variable.
These equations provide a smooth transition over a line or curve. For the intersection problem, the parametric line allows us to express \( x \), \( y \), and \( z \) directly:

• \( x = 3t + 2 \)
• \( y = 2t - 1 \)
• \( z = -t + 1 \)

This makes it simpler to substitute back into the plane equations to find exact intersection points. Parametric equations are versatile and widely utilized in 3D geometry for similar calculations.

Distance Formula

The distance formula is a key tool to measure the straight-line distance between two points in a coordinate system. In three-dimensional space, the formula is an extension of the Pythagorean theorem: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
This approach is crucial when calculating distances in geometry problems. We find the distance between point \( P(-1, -3, 2) \) and point \( Q(5, 1, 0) \) by plugging into this formula.
This results in the distance \( PQ = 2\sqrt{14} \), which illustrates how efficient and reliable this formula can be when solving 3D geometry tasks.

Coordinate Geometry

Coordinate geometry, or analytic geometry, combines algebra and geometry to understand spatial configurations through numerical representation. This method allows for precise calculations of geometric properties such as distances, angles, and intersections.
The intersection and distance calculations in this exercise demonstrate the power of coordinate geometry to solve complex spatial problems using algebraic methods.
Through this framework, one can visualize and compute properties of lines, planes, and other geometric figures accurately, making it indispensable for topics like intersection of lines and planes in space.
Mastering basics of coordinate geometry enables students to tackle advanced problems with confidence and efficiency.