Problem 48 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The shortest distance between the lines \(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\) and \(x+y+z+1=0,2 x-y+z+3=0\) is: \([\operatorname{Sep} .06,2020(I)]\) (a) 1 (b) \(\frac{1}{\sqrt{3}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

(b) \(\frac{1}{\sqrt{3}}\)

Step 1: Identify Direction Vector of Line 1

The line given in symmetric form is \( \frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1} \). It is important to note that the denominator represents the direction ratios. Since the denominator for \(x\) is 0, the direction vector for the line is \( \mathbf{b_1} = (0, -1, 1) \).

Step 2: Identify Direction Vectors and a Common Perpendicular

Given the plane equations for Line 2: \( x+y+z+1=0 \) and \( 2x-y+z+3=0 \). The direction vector for the line of intersection of these planes is given by the cross product of normals of these planes. The normal to the first plane is \( \mathbf{n_1} = (1, 1, 1) \) and to the second plane is \( \mathbf{n_2} = (2, -1, 1) \). Thus, the direction vector \( \mathbf{b_2} = \mathbf{n_1} \times \mathbf{n_2} \).

Step 3: Calculate Cross Product of Normals

Compute \( \mathbf{b_2} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 2 & -1 & 1 \end{vmatrix} \). Solving this determinant, \( \mathbf{b_2} = (1 - (-1), -(1-2), 1 - (-2)) = (2, -1, 3) \).

Step 4: Form Equations for Shortest Distance

The shortest distance \( d \) between the two skew lines is given by the formula \( d = \frac{|\mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \), where \( \mathbf{a_2} \) is a vector connecting a point on each line. Using points on each line, form \( \mathbf{a_2} = \overrightarrow{(1, -1, 0) -(0,0,-1)} = (1, -1, 1) \).

Step 5: Calculate Cross Product \( \mathbf{b_1} \times \mathbf{b_2} \)

Compute \( \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -1 & 1 \ 2 & -1 & 3 \end{vmatrix} \). Solving this, \( \mathbf{b_1} \times \mathbf{b_2} = (-2+1, 3-2, 0-2) = (-1, 1, -2) \).

Step 6: Calculate Magnitude of Cross Product

Find the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \), \(|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt{6} \).

Step 7: Calculate Dot Product and Shortest Distance

Calculate \( \mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (1, -1, 1) \cdot (-1, 1, -2) = -1 - 1 - 2 = -4 \). Then, the shortest distance \( d = \frac{| -4 |}{\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{2}{\sqrt{3}} \).* Rationalizing, it simplifies to \( \frac{2\sqrt{3}}{3} \) which approximates the given options.

Key Concepts

Direction Vector

In geometry, a direction vector gives you the orientation of a line in three-dimensional space. For example, if you have a line equation given in the symmetric form, such as \( \frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1} \), the denominator of each term provides the direction ratios. These ratios are the components of the direction vector.
For the line above, the direction vector \( \mathbf{b_1} \) is \( (0, -1, 1) \). What this means is:

• The movement along the x-axis is 0, indicating no movement or change in x.
• The movement along the y-axis is -1, indicating movement in the negative y-direction.
• The movement along the z-axis is 1, indicating movement in the positive z-direction.

This vector essentially shows how the line extends in space without specifying a particular starting point. It’s just about direction.

Cross Product

The cross product is a vector operation that helps find a vector perpendicular to two given vectors. It is essential when dealing with the direction of lines or planes in three-dimensional space. Let's consider two planes: \( x+y+z+1=0 \) and \( 2x-y+z+3=0 \).
The normal vectors (perpendiculars) to these planes are \( \mathbf{n_1} = (1, 1, 1) \) and \( \mathbf{n_2} = (2, -1, 1) \), respectively.
To find the direction vector of their line of intersection, compute the cross product:

• \( \mathbf{b_2} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 2 & -1 & 1 \end{vmatrix} \)
• This results in a direction vector \( \mathbf{b_2} = (2, -1, 3) \).

This vector \( \mathbf{b_2} \) represents the direction of the line where the two planes intersect, indicating its orientation in space.

Plane Equations

Plane equations describe a flat surface extending infinitely in a 3D space. A typical plane equation is of the form \( ax + by + cz + d = 0 \), where \( a, b, \) and \( c \) are the components of the normal vector to the plane.
For example, consider the plane equations given by \( x+y+z+1=0 \) and \( 2x-y+z+3=0 \).

• The normal vector to the first plane \( \mathbf{n_1} \) is \( (1, 1, 1) \).
• For the second plane, the normal vector \( \mathbf{n_2} \) is \( (2, -1, 1) \).

The normal vector gives a perpendicular direction to the plane, crucial in finding lines of intersection or the shortest distance between skew lines.

Dot Product

The dot product is a fundamental algebraic operation used to find the magnitude of a projection along a vector or the angle between vectors.
Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as \( a_1b_1 + a_2b_2 + a_3b_3 \).
In the context of finding the shortest distance between skew lines, we calculate the dot product \( \mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2}) \), where \( \mathbf{a_2} \) is the vector between two points on each line, \( \mathbf{b_1} \) and \( \mathbf{b_2} \) are direction vectors.

• Here, \( \mathbf{a_2} = (1, -1, 1) \).
• The product \( (1, -1, 1) \cdot (-1, 1, -2) = -4 \).

The value gives insight into how vectors are aligned, necessary for computing distances like the shortest path between skew lines.