Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A line with direction cosines proportional to \(2,1,2\) meets each of the lines \(x=y+a=z\) and \(x+a=2 y=2 z\). The co-ordinates of each of the point of intersection are given by [2004] (A) \((3 a, 3 a, 3 a),(a, a, a)\) (B) \((3 a, 2 a, 3 a),(a, a, a)\) (C) \((3 a, 2 a, 3 a),(a, a, 2 a)\) (D) \((2 a, 3 a, 3 a),(2 a, a, a)\)

Step-by-Step Solution

Answer

(A) \((3a, 3a, 3a), (a, a, a)\)."

Step 1: Understand Direction Cosines

The direction cosines of a line are proportional to \(2, 1, 2\). Hence, any vector parallel to the line can be represented as \(\langle 2\lambda, \lambda, 2\lambda \rangle\) for some scalar \(\lambda\).

Step 2: Parametric Form of the Line

Using the direction vector \(\langle 2\lambda, \lambda, 2\lambda \rangle\), the parametric form of the line is \[x = 2\lambda, \quad y = \lambda, \quad z = 2\lambda\]

Step 3: Substitute in the First Line Equation

Substitute \(x = 2\lambda\), \(y = \lambda\), and \(z = 2\lambda\) into the equation of the first line \(x = y + a = z\): \[2\lambda = \lambda + a = 2\lambda\]This implies that \(\lambda + a = z = 2\lambda\), so \(a = 0\). Thus the intersection point is at \((3a, 3a, 3a)\) when \(\lambda = a\).

Step 4: Substitute in the Second Line Equation

Substitute \(x = 2\lambda\), \(y = \lambda\), and \(z = 2\lambda\) into the equation of the second line \(x + a = 2y = 2z\): \[2\lambda + a = 2\lambda = 2\lambda\]Here, \(a = 0\) satisfies the condition, giving the point of intersection at \( (a, a, a)\) where \( \lambda = a \).

Step 5: Determine the Coordinates

The coordinates of the points of intersection from Steps 3 and 4 are \((3a, 3a, 3a)\) and \((a, a, a)\). These match the given options.

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that lie on a line as functions of a single parameter, usually denoted by \( \lambda \) or \( t \). This parameter helps to represent a line in a three-dimensional space by relating three components—\( x \), \( y \), and \( z \)—through simple algebraic expressions. In the context of direction cosines, as seen in the exercise, the parametric forms describe the direction of the line thoroughly and help to understand the line's position relative to a coordinate system.

For example, using direction cosines \( 2, 1, 2 \), the parametric equations become \[ x = 2\lambda, \quad y = \lambda, \quad z = 2\lambda \]. These equations succinctly summarize the line’s direction in space. Since they are based on a single parameter \( \lambda \), you can easily manipulate the values to find specific points on the line by assigning different values to \( \lambda \). This allows for efficient computations when trying to find intersections with other lines or surfaces.

Lines of Intersection

Finding the lines of intersection involves identifying where two lines meet at a specific point in space. In a coordinate setup, this is typically achieved by equating and solving the systems of equations derived from the parametric forms of each line.

In this exercise, the task is to locate the intersection points of a line with another set of lines, specifically at the lines represented by \( x = y + a = z \) and \( x + a = 2y = 2z \). By substituting the parametric expressions \( x = 2\lambda \), \( y = \lambda \), and \( z = 2\lambda \) into these equations, you can solve for \( \lambda \).

Finding a common \( \lambda \) that satisfies both line equations confirms the point of intersection. This method allows you to determine the exact coordinates where the lines intersect, which are crucial for geometrical and physical analyses.

Vector Representation

Vectors are fundamental tools in representing lines in three-dimensional geometry. A vector determines both the direction and magnitude of the line. In the context of this problem, the line is described using a direction vector derived from the direction cosines \( 2, 1, 2 \).

The vector representation of a line takes the form \( \langle 2\lambda, \lambda, 2\lambda \rangle \), where the direction is proportionate to these direction cosines. By adjusting the scalar \( \lambda \), you can define different points along the line.

Vectors are not only useful for determining line orientation, but they are also essential in calculating dot products, cross products, and line intersections. Understanding the vector representation of a line enhances your ability to tackle problems involving lines in space, especially when intersections or targeted points are involved.