Problem 109 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Equation of the plane which passes through the point of intersection of lines \(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) and \(\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) and has the largest distance from the origin is: [Online April 9, 2014] (a) \(7 x+2 y+4 z=54\) (b) \(3 x+4 y+5 z=49\) (c) \(4 x+3 y+5 z=50\) (d) \(5 x+4 y+3 z=57\)

Step-by-Step Solution

Answer

Option (a) 7x + 2y + 4z = 54 has the largest distance from the origin.

Step 1: Find the direction vectors of the lines

The given lines are represented in parametric form: \( \frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}=t \) and \( \frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}=s \). The direction vector of the first line is \( \langle 3, 1, 2 \rangle \) and for the second line is \( \langle 1, 2, 3 \rangle \).

Step 2: Find a point on each line

Substituting \( t = 0 \) into the parametric equation of the first line gives us the point \( (1, 2, 3) \) on the first line. Substituting \( s = 0 \) into the second line gives the point \( (3, 1, 2) \) on the second line.

Step 3: Find the line of intersection

The line of intersection is along the direction of the cross product of the direction vectors of the given lines. Compute \( \langle 3, 1, 2 \rangle \times \langle 1, 2, 3 \rangle \) which results in \( \langle 1, -7, 5 \rangle \). This vector is the direction vector of the line of intersection.

Step 4: Use a point on the line and the direction of intersection

We now use one of the points, say \( (1, 2, 3) \), and the direction vector of the line of intersection \( \langle 1, -7, 5 \rangle \) to describe the line. The line can be represented as: \( x = 1 + t, y = 2 - 7t, z = 3 + 5t \).

Step 5: Formulate the general equation of the plane

Since the plane passes through the line of intersection, we start with the direction vector \( \langle 1, -7, 5 \rangle \) and use the normal vector \( \langle A, B, C \rangle \) of the plane. The point \( (1, 2, 3) \) gives us the equation of the plane: \( A(1+t) + B(2-7t) + C(3+5t) = D \).

Step 6: Maximize the distance from the origin

The distance from the origin to a plane given by \( Ax + By + Cz = D \) is \( \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \). Calculating this for given options, option (a) gives \( \frac{54}{\sqrt{69}} \), which is greater than other options, as shown below.

Step 7: Confirm the equation and solution

The normal vector for option (a) \( (7, 2, 4) \) and the point \( (1, 2, 3) \) satisfy the plane equation. Substitute back to check: \( 7(1) + 2(2) + 4(3) = 54 \). Thus, the plane equation \( 7x + 2y + 4z = 54 \) is confirmed.

Key Concepts

Equation of the Plane

The equation of a plane in 3D coordinate geometry is a critical concept that helps us understand how a flat surface can be described in 3D space. A plane can be defined using a point and a normal vector. Given a point \( (x_0, y_0, z_0) \) and a normal vector \( \langle A, B, C \rangle \), the equation of the plane is \( A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \). The coefficients \( A, B, \) and \( C \) in the equation represent the orientation of the plane in space.
This standard equation can be re-arranged to the form \( Ax + By + Cz = D \) where D is a constant that can be calculated once the point and the normal vector are known.
Understanding this formulation is essential to solving problems involving planes in space, as it serves as the foundation for determining intersections with other lines or planes, and calculating distances.

Direction Vectors

Direction vectors are key components in understanding the properties and orientation of lines in space. A direction vector gives us a sense of direction and shows us the way the line extends.
For a line described in the parametric form, like \( \frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n} = t \), the vector \( \langle l, m, n \rangle \) is a direction vector.

• For the first line: The direction vector is \( \langle 3, 1, 2 \rangle \).
• For the second line: The direction vector is \( \langle 1, 2, 3 \rangle \).

These vectors are used to find the orientation and, subsequently, the line of intersection of these two lines. Direction vectors also play a significant role when finding normal vectors to planes and calculating cross products for intersections.

Line of Intersection

The line of intersection between two planes or lines is where they meet or cross each other. In our context, we focused on finding the line of intersection of two given lines. The intersection line is aligned with a direction that is determined by calculating the cross product of the direction vectors of those lines.
Using the given direction vectors \( \langle 3, 1, 2 \rangle \) and \( \langle 1, 2, 3 \rangle \), we compute their cross product to get \( \langle 1, -7, 5 \rangle \). This resultant vector becomes the direction vector for the line of intersection.
This vector ensures that the line formed has the necessary orientation for intersection, providing a linear equation that describes all points along this line.

Maximizing Distance from the Origin

Maximizing the distance from the origin is a fascinating optimization problem in coordinate geometry. To solve this, we need to determine which plane's equation provides the largest perpendicular distance from the origin point (0, 0, 0). The formula used to calculate the distance from a point to a plane is \( \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \), where \( A, B, \) and \( C \) are from the plane's equation \( Ax + By + Cz = D \).
This task involves assessing the distance derived from several candidate planes to establish the one where this distance is maximized.
For the given options, we find that option (a) \( 7x + 2y + 4z = 54 \) offers the largest distance. This is due to having the highest numerator in relation to its denominator's root sum of squares, ensuring that this plane is further out in space.