Problem 49 - 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

The shortest distance is 1.

Step 1: Identify line representations

The first line is given in parametric form: \(x=1, y=-1-t, z=t\). The second line is the intersection of the planes \(x+y+z+1=0\) and \(2x-y+z+3=0\). To find the intersection line, solve for combinations of these equations.

Step 2: Find direction vectors

The direction vector of the first line is \(\mathbf{a} = (0, -1, 1)\). For the second line, the direction vector is a cross-product of the normal vectors of the planes: \(\mathbf{b} = (1, 1, 1)\) and \(\mathbf{n} = (2, -1, 1)\). Calculate \(\mathbf{d} = \mathbf{b} \times \mathbf{n}\).

Step 3: Calculate the cross product

Perform the cross product \(\mathbf{d} = (1, 1, 1) \times (2, -1, 1)\), resulting in \(\mathbf{d} = (2, 1, -3)\), which is the direction vector of the line intersection.

Step 4: Find a point on the second line

Assume \(z = 0\) for simplicity in solving the simultaneous equations. Thus, \(x + y = -1\) and \(2x - y = -3\). Solving these gives \(x = -1, y = 0\). Therefore, point \((-1, 0, 0)\) lies on the intersection line.

Step 5: Determine the formula for shortest distance

Use the formula for shortest distance between two skew lines: \(D = \frac{|(\mathbf{c_1} - \mathbf{c_2}) \cdot (abla_1 \times abla_2)|}{|abla_1 \times abla_2|}\), where \(\mathbf{c_1} = (1, -1, 0)\) for the first line and \(\mathbf{c_2} = (-1, 0, 0)\) for the second line.

Step 6: Compute the shortest distance

Calculate \(abla_1 \times abla_2 = \mathbf{a} \times \mathbf{d} = (0, -1, 1) \times (2, 1, -3)\), which results in \(-(-1\times -3 + 1\times 2)i + (1\times 2 - 0\times -3)j + (0\times 1 - -1\times 2)k = (-1, 2, 2)\). Then compute \( (\mathbf{c_1} - \mathbf{c_2}) = (2, -1, 0)\), and \((2,-1,0) \cdot (-1, 2, 2) = -2+(-2)= -4\). Thus, \(D = \frac{| -4 |}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{4}{3}\)."

Key Concepts

Skew lines

In three-dimensional geometry, "skew lines" are lines that do not intersect and are not parallel. Unlike parallel lines that do not intersect because they maintain the same distance, skew lines do not intersect because they are situated in different planes. Skew lines are a unique feature in three dimensions because in two dimensions, lines that do not intersect are either parallel or they do intersect at some point.

To visualize skew lines, imagine one line running along the floor of a room, and another line running along the ceiling that does not run parallel to the floor line. Despite neither being parallel nor converging, they maintain a consistent spatial relationship due to their positioning in different planes. This characteristic is essential when determining the spatial structure of systems and forms a basis for calculating the shortest distance between such lines.

Shortest distance between lines

The shortest distance between skew lines is crucial in geometry and can be determined mathematically. For two skew lines, the shortest distance is defined as the line segment perpendicular to both. This distance is not immediately apparent and requires a systematic calculation.

To find this distance, we typically use the formula:

\[ D = \frac{|(\mathbf{c_1} - \mathbf{c_2}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|}{|\mathbf{a_1} \times \mathbf{a_2}|} \]

Where:

• \(\mathbf{c_1}\) and \(\mathbf{c_2}\) are points on the respective lines.
• \(\mathbf{a_1}\) and \(\mathbf{a_2}\) are direction vectors of the lines.

The cross product \((\mathbf{a_1} \times \mathbf{a_2})\) provides a direction perpendicular to both lines, forming the basis for the computation. Calculating this allows for the measurement of the shortest distance along that direction. Understanding this concept allows for accurate spatial reasoning in various applications, from architecture to physics.

Direction vectors

Direction vectors are foundational in defining the orientation of lines in space. A direction vector gives a sense of the line's trajectory, pointing from one point on the line to another.

In finding the shortest distance between skew lines, direction vectors are indispensable. They are represented as a set of coordinates that describe how the line advances through the spatial dimensions. For example, the direction vector \((v_1, v_2, v_3)\) indicates a line's movement along the axes.

For the line presented in parametric form: \(x = a, y = b + tx, z = c + tz\), the direction vector is often \((0, x, z)\). Calculating a direction vector for lines requires understanding the line equations, which in turn acts as a guide in creating or analyzing spatial systems. Cross products of direction vectors are particularly useful in deriving perpendicular alignment essential for computing shortest distances between skew and other types of lines.