Problem 66 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the line, \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\) meets the plane, \(x+2 y+3 z=15\) at a point \(P\), then the distance of \(P\) from the origin is: [April 09 2019I](a) \(\sqrt{5} / 2\) (b) \(2 \sqrt{5}\) (c) \(9 / 2\) (d) \(7 / 2\)

Step-by-Step Solution

Answer

The distance of \(P\) from the origin is \(9/2\).

Step 1: Parameterize the line equation

The given line equation is \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4} = t\), where \(t\) is a parameter. Using this parameter, we can write the equations for \(x\), \(y\), and \(z\) as: \(x = 2t + 1\), \(y = 3t - 1\), and \(z = 4t + 2\).

Step 2: Substitute line equations into the plane equation

Substitute \(x = 2t + 1\), \(y = 3t - 1\), and \(z = 4t + 2\) into the plane equation \(x + 2y + 3z = 15\). This gives: \((2t + 1) + 2(3t - 1) + 3(4t + 2) = 15\).

Step 3: Simplify and solve for t

Simplify the equation from Step 2: \(2t + 1 + 6t - 2 + 12t + 6 = 15\), leading to \(20t + 5 = 15\). Solve for \(t\) by subtracting 5 from both sides to get \(20t = 10\). Then, divide both sides by 20 to find \(t = 1/2\).

Step 4: Find the coordinates of point P

Substitute \(t = 1/2\) back into the parameterized equations: \(x = 2(1/2) + 1 = 2\), \(y = 3(1/2) - 1 = 1/2\), and \(z = 4(1/2) + 2 = 4\). Therefore, point \(P\) is \((2, 1/2, 4)\).

Step 5: Calculate the distance from the origin

The distance from the origin \((0,0,0)\) to the point \((2, 1/2, 4)\) is calculated using the distance formula: \(\sqrt{(2-0)^2 + (1/2-0)^2 + (4-0)^2 } = \sqrt{4 + 1/4 + 16} = \sqrt{81/4} = 9/2\).

Key Concepts

Parametric Equations

Understanding parametric equations is essential when dealing with lines in three-dimensional geometry. A parametric equation expresses the coordinates of the points on a line as functions of a single parameter—commonly denoted as \( t \).
For the line given by \( \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4} \), the parameter \( t \) represents a way of moving along the line. By rewriting this into the form of \( x = 2t + 1 \), \( y = 3t - 1 \), and \( z = 4t + 2 \), we describe how each coordinate changes as \( t \) varies.

• This expression allows us to easily substitute and calculate specific points on the line.
• The parameter \( t \) essentially dictates the position along the line, from negative infinity to positive infinity.

Intersection of Line and Plane

Finding where a line intersects a plane is about determining the common point they share. This involves substituting the parametric equations of the line into the plane's equation.
Given the plane equation \( x+2y+3z=15 \) and the parametric forms \( x = 2t + 1 \), \( y = 3t - 1 \), and \( z = 4t + 2 \), we plug these into the plane's equation to check for the values of \( t \) that satisfy it: \( (2t + 1) + 2(3t - 1) + 3(4t + 2) = 15. \)

• This substitution converts the equation from three variables \( x, y, z \) to a single variable \( t \), which simplifies the process immensely.
• Solving this equation gives us \( t = 1/2 \), which can then be used to find the exact point of intersection \( P \).

Through this, we find \( P \) to be the point \( (2, 1/2, 4) \) where the line intersects the plane.

Distance Formula

The distance formula is a robust tool for calculating the shortest distance between two points in space. For points in three-dimensional geometry, it's expressed as: \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
This formula derives from the Pythagorean Theorem and generalizes it to 3D.
When we need the distance from the origin \((0,0,0)\) to a point \((2, 1/2, 4)\), we substitute in:

• \( x_1 = 0, y_1 = 0, z_1 = 0\)
• \( x_2 = 2, y_2 = 1/2, z_2 = 4\)

The computation becomes \( \sqrt{2^2 + (1/2)^2 + 4^2} = \sqrt{4 + 1/4 + 16} = \sqrt{81/4} = 9/2 \). This distance is what we found as the result in the exercise solution.

Coordinate Geometry

Coordinate geometry, also known as Cartesian geometry, uses algebra to model geometric spaces. It allows us to work with coordinates and equations to express geometric relationships and solve problems. The intersection of a line and a plane or the calculation of distance exemplifies its usefulness.
In the context of our problem:

• The plane \( x+2y+3z=15 \) describes a flat two-dimensional surface within a three-dimensional space.
• The line given by the parametric equations is another geometric entity that can be represented in this space.

By using coordinate geometry, we can calculate where these geometric forms intersect and measure distances directly based on their coordinates, linking algebraic methods with geometric understanding.