Problem 18 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A plane \(\mathrm{P}\) meets the coordinate axes at \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) respectively. The centroid of \(\Delta \mathrm{ABC}\) is given to be \((1,1,2)\). Then the equation of the line through this centroid and perpendicular to the plane \(\mathrm{P}\) is: [Sep. 06, 2020 (II)] (a) \(\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-2}{1}\) (b) \(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-2}{2}\) (c) \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}\) (d) \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\)

Step-by-Step Solution

Answer

Option (c) is the correct answer.

Step 1: Understand the Problem

We need to find the equation of the line passing through the centroid \((1, 1, 2)\) and perpendicular to the plane \(P\), which meets the coordinate axes at points \(A, B, C\). The equation of the line will be in the form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).

Step 2: Use the Condition for the Centroid

The coordinates of the centroid of a triangle \(\Delta ABC\) with vertices at \((x_1, 0, 0)\), \((0, y_1, 0)\), and \((0, 0, z_1)\) is given by \(\left(\frac{x_1}{3}, \frac{y_1}{3}, \frac{z_1}{3}\right)\). Since the centroid is \((1, 1, 2)\), we equate:\[ \frac{x_1}{3} = 1, \quad \frac{y_1}{3} = 1, \quad \frac{z_1}{3} = 2 \]Solving these equations gives us \(x_1 = 3\), \(y_1 = 3\), and \(z_1 = 6\).

Step 3: Determine the Plane Equation P

Since the plane meets the coordinate axes at \((x_1, 0, 0)\), \((0, y_1, 0)\), and \((0, 0, z_1)\), the equation for plane \(P\) is:\[ \frac{x}{x_1} + \frac{y}{y_1} + \frac{z}{z_1} = 1 \]Substituting \(x_1 = 3\), \(y_1 = 3\), and \(z_1 = 6\), we have:\[ \frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1 \]

Step 4: Find the Normal to Plane P

The normal vector to the plane \(\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1\) is given by the coefficients of \(x\), \(y\), and \(z\):\[ \langle 1/3, 1/3, 1/6 \rangle \]By multiplying each component by \(6\) to clear the fractions, we find the direction vector for the line:\[ \langle 2, 2, 1 \rangle \]

Step 5: Write the Equation of the Line

Using the direction vector \(\langle 2, 2, 1 \rangle\) and passing through the point \((1, 1, 2)\), the equation of the line is:\[ \frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1} \]

Step 6: Select the Correct Option

Comparing the obtained line equation with the given options, we choose:(c) \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}\)

Key Concepts

Centroid of a Triangle

In 3D geometry, finding the centroid of a triangle is quite similar to the process used in 2D but with an added dimension. The centroid is a crucial concept as it represents a kind of "average" or "balance point" of the vertices of the triangle. For a triangle \((\Delta ABC)\), with vertices \((x_1, 0, 0), (0, y_1, 0), (0, 0, z_1)\), the centroid \((G)\) is found using the formula for each coordinate:
\(G = \left(\frac{x_1}{3}, \frac{y_1}{3}, \frac{z_1}{3}\right)\)

This means to find the centroid, you simply take the average of the x-coordinates, the y-coordinates, and the z-coordinates of the triangle's vertices. In the exercise, this method showed that \(x_1 = 3\), \(y_1 = 3\), \(z_1 = 6\). Thus the triangle's centroid is \( (1,1,2) \). Understanding how to calculate the centroid helps us solve many problems related to balance and symmetry in shapes.

Equation of a Plane

The equation of a plane in 3D space is a way to represent a flat surface that extends infinitely in the space in which it resides. When the plane intersects the coordinate axes at specific points, it can be expressed in intercept form. For a plane that meets the coordinate axes at points \((x_1, 0, 0), (0, y_1, 0), (0, 0, z_1)\), its equation can be written as:
\(\frac{x}{x_1} + \frac{y}{y_1} + \frac{z}{z_1} = 1\)

In the given problem, the plane cuts the axes at points \((3, 0, 0), (0, 3, 0), (0, 0, 6)\), so the equation becomes:
\(\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1\)

This form of a plane equation is particularly handy because it gives an immediate sense of how the plane is oriented in space, showing the relationships between the distances the plane is from each of the axes.

Line Perpendicular to a Plane

Finding a line perpendicular to a plane involves pinpointing a line that intersects the plane at a right angle. In 3D geometry, this line is often defined using the normal vector of the plane. The normal vector is simply the vector orthogonal to the plane's surface. For the plane \(\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1\), its normal vector is \(<1/3, 1/3, 1/6>\). Multiplying this vector by 6 to avoid fractions, we get the direction vector \(<2, 2, 1>\).

This direction vector is crucial because it tells us the direction in which the line moves away from the plane. If the line needs to pass through a specific point, such as the centroid \(1, 1, 2)\), we can then write the equation of the line as:

\(\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}\)

This equation concisely represents the line in terms of its relation to a known point and its direction in space. Recognizing how to derive and interpret this line is fundamental for solving problems involving perpendicular lines to planes.