Problem 78 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

On which of the following lines lies the point of inter-section of the line, \(\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}\) and the plane, \(x+y+z=2 ? \quad\) [Jan. 10, 2019 (II)] (a) \(\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}\) (b) \(\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}\) (c) \(\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}\) (d) \(\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}\)

Step-by-Step Solution

Answer

The intersection point lies on line (b).

Step 1: Parameterize the Line

The line given by \( \frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1} \) can be parameterized as \( x = 4 + 2t, \ y = 5 + 2t, \ z = 3 + t \) where \( t \) is a parameter.

Step 2: Substitute in the Plane Equation

Substitute the parameterized coordinates into the plane equation \( x + y + z = 2 \). Doing this gives \( (4 + 2t) + (5 + 2t) + (3 + t) = 2 \).

Step 3: Solve for the Parameter

Simplify the equation from Step 2: \( 12 + 5t = 2 \). Solving for \( t \) gives \( 5t = 2 - 12 = -10 \) so \( t = -2 \).

Step 4: Find Point of Intersection

Substitute \( t = -2 \) back into the parameterized coordinates: \( x = 4 + 2(-2) = 0 \), \( y = 5 + 2(-2) = 1 \), \( z = 3 + (-2) = 1 \). Thus, the point of intersection is \( (0, 1, 1) \).

Step 5: Check Intersection with Options

Determine which line contains the point \( (0, 1, 1) \) by substituting it into each option and checking consistency. For option (b) \( \frac{x-4}{1} = \frac{y-5}{1} = \frac{z-5}{-1} \), substitute \( x=0, y=1, z=1 \) and check: \( \frac{0-4}{1} = -4 \), \( \frac{1-5}{1} = -4 \), \( \frac{1-5}{-1} = -4 \). All are equal, thus the point lies on this line.

Key Concepts

Line and Plane Intersection

In coordinate geometry, understanding the intersection of lines and planes is a crucial concept. When a line intersects a plane, their meeting point is called the point of intersection. To determine this point, parametric equations for the line are used, as they provide a way to express each coordinate in terms of a parameter. In the given exercise, we have a line expressed by the parametric equation

• x = 4 + 2t
• y = 5 + 2t
• z = 3 + t

We also have a plane defined by the equation x + y + z = 2.

To find their intersection, you substitute the parametric expressions into the plane's equation and solve for the parameter t. This procedure gives us the specific point on the line that lies on the plane as well.

Finding the intersection point is important as it can help to solve real-world problems involving surfaces and directions, such as determining where a road (line) might pass through a mountain (plane). Using algebraic operations, you can precisely locate this intersection, ensuring both mathematical and logical accuracy.

Parametric Equations

Parametric equations are a powerful tool in coordinate geometry. They allow us to express geometrical entities, like lines, using parameters. A parametric equation of a line expresses each coordinate, such as x, y, and z, in terms of one or more parameters. This method simplifies the examination of lines and their interactions with other forms, like planes.

In the exercise given, the line is parameterized as:

• x = 4 + 2t
• y = 5 + 2t
• z = 3 + t

Here, the parameter t represents any real number, and altering t traces various points on the line.

Parametrization is integral to solving geometrical problems as it breaks down layers of complexity. Instead of dealing with separate instances, parametric equations offer a unified approach to represent and solve equations involving curves, surfaces, and their intersections.

Points of Intersection

Finding the points of intersection between lines and planes is a common problem in geometry and trigonometry. These points provide significant insight into spatial relationships and shape orientations. When a line meets a plane, they intersect at a single point, unless the line is coplanar with the plane, in which case they share all points along the line.

The procedure for finding this intersection involves substituting the parametric coordinates of a line into the equation of a plane, simplifying, and solving. In this exercise, by setting:

• Parameter values in line: x = 4 + 2t, y = 5 + 2t, z = 3 + t
• Into plane: x + y + z = 2

Solving for t helps locate the exact intersection point.

Practically, points of intersection may represent project goals like route planning, collision detection in simulation software, or even artwork conception in 3D modeling. Clearly understanding how to calculate these points is essential to mastering spatial reasoning and applying it in varied engineering and scientific contexts.