Problem 108 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The plane containing the line \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}\) and parallel to the line \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{4}\) passes through the point: [Online April 11, 2014] (a) \((1,-2,5)\) (b) \((1,0,5)\) (c) \((0,3,-5)\) (d) \((-1,-3,0)\)

Step-by-Step Solution

Answer

(1, -2, 5) is the correct point.

Step 1: Understand the problem

We need to find which point lies on the plane that contains the given line and is parallel to another line. A plane containing a line must include the point on the line and a direction parallel to another line.

Step 2: Determine the point on the line

The line equation given is \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}\). The point on the line is \((1,2,3)\). This is derived directly when each part of the fraction equals zero.

Step 3: Find the direction ratios

The direction ratios for the line \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}\) are (1, 2, 3). This indicates the direction of the line.

Step 4: Analyze the second line

The second line is given by \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{4}\) with direction ratios (1, 1, 4). The plane is parallel to this line so these direction ratios form a direction vector in the plane.

Step 5: Form the plane equation

We can now use the point \((1,2,3)\) and the direction vectors (1, 2, 3) and (1, 1, 4) to form the equation of the plane. The normal to the plane comes from the cross product of these direction vectors.

Step 6: Calculate cross product

The cross product of the two direction vectors is \(\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & 1 & 4\end{vmatrix}\), which equals \(-5\hat{i} + (-1)\hat{j} + (-1)\hat{k}\). This is the normal to the plane.

Step 7: Equation of the Plane

The equation of the plane is \(-5(x-1) - (y-2) - (z-3) = 0\), or simplifying to \-5x - y - z = -5 - 2 - 3\, or \-5x - y - z = -10\.

Step 8: Test each point

Substitute each option into the plane equation \-5x - y - z = -10\. For \((1, -2, 5)\), it becomes \(-5(1) - (-2) - 5 = -10\) which holds true. The equation doesn't hold for other points.

Key Concepts

Plane Equation

In coordinate geometry, the equation of a plane is a fundamental concept. A plane can be thought of as an infinite flat surface extending in all directions in three-dimensional space. A plane equation is defined using a normal vector; a vector perpendicular to the plane, and a point through which the plane passes. The general form of the plane equation is:
\[ ax + by + cz = d \]where

• \(a, b, c\) are the coefficients of the normal vector
• \((x, y, z)\) are the coordinates of any point on the plane.
• \(d\) is determined by substituting a known point on the plane into the equation.

In this exercise, we derived a plane equation using a given line and making it parallel to another line. We used the cross product of two vectors to find the normal vector, crucial for forming the plane equation.

Line in Space

A line in three-dimensional space is determined by its direction and any single point through which it passes. When dealing with exercises in coordinate geometry, understanding the representation of a line is key.
One common form for representing a line is the **symmetric form**:
\[ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \]where

• \((x_1, y_1, z_1)\) is a known point on the line,
• \((a, b, c)\) are the direction ratios, indicating the line's direction in space.

In our exercise, the line is given in symmetric form, making it easier to identify a point, which in this case was \((1, 2, 3)\). This point was used to help determine the plane.

Direction Ratios

Direction ratios are crucial in identifying the orientation of a line in space. They are a set of three numbers, \((a, b, c)\), proportional to the direction vector of the line.
These ratios show how the line changes in x, y,andz directions.For a line expressed in symmetric form\[ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \]The values \(a, b, c\) are involved with how the line 'travels' through space.
Direction ratios help in constructing the line's representation, and in this problem, they influenced the plane description as both lines each provided a unique set of direction ratios, \((1, 2, 3)\) for our line, and \((1, 1, 4)\) for the parallel line.

Cross Product

The cross product is a vector operation used in geometry to find a vector that is perpendicular to two other vectors. The magnitude of this perpendicular vector is proportional to the area of the parallelogram that the vectors span.
Mathematically, the cross product of two vectors, \(\mathbf{A} = (a_1, a_2, a_3)\) and \(\mathbf{B} = (b_1, b_2, b_3)\), is shown by \[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix}= (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}\]
For this problem, by taking the cross product of direction vectors \((1, 2, 3)\) and \((1, 1, 4)\), we found the normal vector \((-5, -1, -1)\), which was crucial in determining the plane equation. This operation is vital in geometry for finding perpendicular vectors and defining planes.