Problem 62 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+\) \(3 y-\alpha z+\beta=0 .\) Then \((\alpha, \beta)\) equals [2009] (A) \((6,-17)\) (B) \((-6,7)\) (C) \((5,-15)\) (D) \((-5,15)\)

Step-by-Step Solution

Answer

The correct pair is \((-6, 7)\), option B.

Step 1: Identify the Direction Ratios of the Line

The given line equations are \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\). The direction ratios of the line are \((3, -5, 2)\).

Step 2: Determine a Point on the Line

From the given symmetric form of the line, we can extract the point \((2, 1, -2)\) which lies on the line.

Step 3: Substitute the Point into the Plane Equation

Substitute the point \((2, 1, -2)\) into the plane equation \(x + 3y - \alpha z + \beta = 0\). This yields \(2 + 3(1) + 2\alpha + \beta = 0\). Simplifying, we get \(5 + 2\alpha + \beta = 0\).

Step 4: Check the Perpendicularity Condition

For the line to lie on the plane, its direction ratios \((3, -5, 2)\) must be perpendicular to the normal to the plane, \((1, 3, -\alpha)\). This condition is met if their dot product is zero: \(3\cdot 1 - 5\cdot 3 + 2(-\alpha) = 0\). Simplifying, \(3 - 15 - 2\alpha = 0\), we find \(\alpha = -6\).

Step 5: Solve for \(\beta\) Using \(\alpha\)

Substitute \(\alpha = -6\) in the equation from Step 3: \(5 + 2(-6) + \beta = 0\). This simplifies to \(5 - 12 + \beta = 0\) or \(\beta = 7\).

Step 6: Verify the Solution

The values \((\alpha, \beta) = (-6, 7)\) satisfy both the line's condition on the plane and the perpendicularity condition. Hence, these are the correct values.

Key Concepts

Direction Ratios

Understanding the concept of direction ratios is fundamental in three-dimensional geometry, especially when dealing with lines in space. Direction ratios are a set of three numbers that are proportional to the direction cosines of a line. For a given line in 3D, direction ratios help in defining its orientation.
In our exercise, the line is expressed in symmetric form: \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\). Here, the coefficients of \(x, y, z\) in the denominators \((3, -5, 2)\) are the direction ratios of the line. This means that the line extends in the direction defined by this vector.
Direction ratios are crucial when checking the relationship between a line and a plane, such as establishing perpendicularity.

Symmetric Form of a Line

The symmetric form of a line is a useful representation in three-dimensional geometry. It describes a line by giving a particular point on the line and direction ratios. This form is given by \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\), where \((x_1, y_1, z_1)\) is a point on the line and \((a, b, c)\) are its direction ratios.
In the problem, the symmetric form \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\) indicates that the line passes through the point \((2, 1, -2)\) with direction ratios \((3, -5, 2)\). This form makes it easy to extract both a reference point on the line and its direction in space. It's particularly helpful in calculations involving intersections with planes.

Plane Equation

The general equation of a plane in three-dimensional geometry is described as \(ax + by + cz + d = 0\), where \((a, b, c)\) represents the normal vector to the plane. The problem gives a specific plane equation \(x + 3y - \alpha z + \beta = 0\). Here, the coefficients \((1, 3, -\alpha)\) are key in determining the relationship of the plane to other geometric elements, such as lines.
To find whether a line lies in a given plane, you can substitute the coordinates of a point on the line into the plane equation, which must be satisfied. If the equation holds true, the point, and therefore the line, lies on the plane. This method was used in the exercise to deduce values for \(\beta\), confirming the line's presence in the plane.

Perpendicularity Condition

The perpendicularity condition is pivotal when analyzing the geometrical relationship between a line and a plane. For a line to lie within a plane, its direction ratios must be perpendicular to the normal vector of the plane. This condition is mathematically expressed through the dot product, which should equal zero.
In this exercise, the direction ratios of the line are \((3, -5, 2)\) and the normal to the plane is given by \((1, 3, -\alpha)\). The dot product, \(3 \,\cdot 1 - 5 \,\cdot 3 + 2(-\alpha)\), equaling zero, establishes \(\alpha = -6\). This illustrates how a simple relationship such as perpendicularity can simplify complex spatial problems.