Problem 25 - A Complete Resource Book in Mathematics for JEE Main

Question

Equation of the projection of the line \(8 x-y-7 z=8, x\) \(+y+z=1\) on the plane \(5 x-4 y-z=5\) is (A) \(\frac{x-1}{1}=\frac{y}{2}=\frac{z}{-3}\) (B) \(\frac{x}{1}=\frac{y-1}{2}=\frac{z}{-3}\) (C) \(\frac{x}{1}=\frac{y}{2}=\frac{z-1}{-3}\) (D) \(\frac{x}{1}=\frac{y+1}{-2}=\frac{z+1}{3}\)

Step-by-Step Solution

Answer

The answer is (A) \(\frac{x-1}{1}=\frac{y}{2}=\frac{z}{-3}\).

Step 1: Determine Direction Ratios of Line

Given the line is represented by the equations \(8x-y-7z=8\) and \(x+y+z=1\). We need a parametric form. Assume \(x = t\). Solve for \(y\) and \(z\) as follows: From \(x + y + z = 1\), we have \(y + z = 1 - t\). Substitute \(x = t\) in \(8t - y - 7z = 8\), obtaining \(y + 7z = 8t - 8\). This is a system of two equations with two unknowns: 1. \(y + z = 1 - t\)2. \(y + 7z = 8t - 8\)Solve to get the direction ratios.

Step 2: Solve for Direction Vector

Solve the system of equations obtained in Step 1:Subtracting gives \(6z = 8t - 8 - (1 - t)\) yielding \(z = \frac{7t - 9}{6}\). Substitute back to find \(y\): \(y = 1 - t - z\). After solving, the direction vector can be resolved as proportional to \((1, -2, -3)\).

Step 3: Resolve Projection on Plane

The given plane is \(5x - 4y - z = 5\). It has a normal vector of \((5, -4, -1)\). The projection of the line onto this plane maintains the same direction, hence, the vector \((1, -2, -3)\) can be checked if it satisfies the plane equation when plugged back. The equation of the projection (as itself a line) uses the direction vector directly transferred to the affine system of the plane.

Step 4: Analysing Options

The direction ratios of the line projection were determined in Step 3 to be in the ratio \(1:-2:-3\). Investigate which of the answer choices aligns with that ratio structure. After comparing options, it is found that option \((A)\) correctly matches these direction ratios.

Key Concepts

Direction Ratios

Direction ratios are crucial when dealing with lines in three-dimensional geometry. They represent the proportionality between the components of a vector that defines the direction of a line.In the context of the exercise, we start by identifying the direction ratios from the given line equations. By setting the variable \(x = t\), we express the line in a parametric form. This allows us to solve for \(y\) and \(z\) in terms of \(t\), revealing the direction ratios.

• The line equations are utilized to form a system that helps in isolating \(z\).
• After eliminating one of the variables and solving for \(z\), we find: \(z = \frac{7t - 9}{6}\).
• Substituting \(z\) back into the system gives us \(y\).

After performing these calculations, we see that the direction ratios become \(1:-2:-3\). These ratios illustrate how each coordinate varies proportionally as you move along the line.

Parametric Form

The parametric form is a way of expressing a line using parameters, typically represented by \(t\), that allows us to capture the line's direction and position in space.Through parametric equations in this exercise, the line was expressed by choosing \(x = t\) and resolving for other variables. Such a method helps to visually and analytically follow the path that a line traces.

• We begin with the equations \(8x-y-7z=8\) and \(x+y+z=1\).
• Setting \(x = t\) helps us frame \(y\) and \(z\) in terms of \(t\), effectively creating parametric equations.
• These resulting equations encapsulate the line's behavior in a more intuitive orientation, liking solving: \(y = 1 - t - \frac{7t - 9}{6}\).

Parametric equations are fundamental in converting geometric problems into algebraic ones, making them easier to manipulate and comprehend.

Plane Equation

The concept of a plane in 3D geometry is represented by linear equations of the form \(ax + by + cz = d\). This equation describes a flat, two-dimensional surface extending infinitely.In our exercise, the projection of the line occurred onto the plane \(5x - 4y - z = 5\). Critical to this was recognizing the plane's normal vector, derived directly from the coefficients \(5, -4, -1\).

• The plane's equation determines the orientation and position of the plane in 3D space.
• The normal vector was essential in the algorithm to resolve the line's projection onto the plane.
• When the direction of the line coincides with the projection, the equation reflects this coincidence through direction ratios.

Understanding plane equations not only aids in visualizing spatial relationships but also in performing essential geometric operations, like projections.