Problem 28 - A Complete Resource Book in Mathematics for JEE Main

Question

The perpendicular distance of a corner of a unit cube form a diagonal not passing through it is (A) \(\sqrt{6}\) (B) \(\frac{\sqrt{6}}{3}\) (C) \(\frac{3}{\sqrt{6}}\) (D) none of these

Step-by-Step Solution

Answer

Option (B) \(\frac{\sqrt{6}}{3}\) is the correct answer.

Step 1: Understanding the Problem

We need to find the perpendicular distance from a vertex of a unit cube to a space diagonal that does not pass through that vertex. A unit cube has vertices at (0, 0, 0) through (1, 1, 1). A space diagonal would go from (0, 0, 0) to (1, 1, 1).

Step 2: Finding the Relevant Points

To find the perpendicular distance, consider the vertex (1, 0, 0), and the space diagonal which passes through points (0,0,0) and (1,1,1). The point (1, 0, 0) is not part of this diagonal.

Step 3: Equation of the Diagonal

The direction vector of the diagonal from (0,0,0) to (1,1,1) is (1,1,1). The equation in parametric form is \( (t, t, t) \).

Step 4: Finding Orthogonal Vector

The vector from (1, 0, 0) to a point on the line can be written as \( (t-1, t, t) \). For it to be perpendicular, dot product with the line's direction \( (1,1,1) \) should be zero: \( (t-1) + t + t = 0 \).

Step 5: Solving for t

Solve \( 3t - 1 = 0 \), giving \( t = \frac{1}{3} \). Hence, the closest point is \( (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) \).

Step 6: Calculating the Distance

The distance between the vertex (1, 0, 0) and the point \( \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right) \) is found using the distance formula: \[ \sqrt{\left( 1 - \frac{1}{3} \right)^2 + \left( 0 - \frac{1}{3} \right)^2 + \left( 0 - \frac{1}{3} \right)^2} \]. Simplifying, we find: \[ \sqrt{\left( \frac{2}{3} \right)^2 + \left( \frac{1}{3} \right)^2 + \left( \frac{1}{3} \right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{6}{9}} = \frac{\sqrt{6}}{3} \].

Key Concepts

Distance in Space

Understanding the distance in space involves the calculation of the shortest path between two points in a three-dimensional setting. This is crucial in geometry problems as it enables us to solve questions involving spatial structures like cubes, spheres, and other 3D figures.
To achieve this, we use the distance formula, derived from the Pythagorean Theorem. In 3D, the position of a point is expressed as \(x, y, z\). If you have two points, \(x_1, y_1, z_1\) and \(x_2, y_2, z_2\), the distance \(d\) between them is calculated as:

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\).

This equation gives the straight-line distance or the "as-the-crow-flies" distance.
Practical applications of this concept range from virtual simulations to real-world navigation where determining the most direct route is essential.

Diagonal of a Cube

A cube is a three-dimensional object with six equal square faces. Identifying the diagonal of a cube assists in many geometry-related problems.
A cube's diagonal can be categorized into two types:

• Face diagonal: This is the diagonal that cuts across the square face of the cube. Utilizing the Pythagorean Theorem, its length is \(\sqrt{2} \,a\) where \(a\) is the side of the cube.
• Space diagonal: This extends from one corner of the cube to the opposite corner across the cube. Its length is \(\sqrt{3} \,a\). For a unit cube (where \(a = 1\)), the space diagonal is \(\sqrt{3}\).

Understanding these diagonals is essential in math exercises that demand precise calculations involving symmetrical properties and angles.

Vector Mathematics

In geometry and physics, vector mathematics plays a significant role in solving complex spatial problems. Vectors allow us to describe both magnitude and direction. This means vectors are vital when dealing with distances and directions in space.
For example, the direction vector of a line passing through two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\) can be represented as \(\overrightarrow{AB} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\). This helps in various calculations like determining if two lines are perpendicular, parallel, or finding the shortest distance between them.
Vectors are often combined using operations like addition, subtraction, and dot product. The dot product, in particular, is instrumental in calculating perpendicularity, as two vectors are perpendicular if their dot product is zero. This underpinning allows one to explore more intricate aspects of spatial geometry.

Perpendicular Distance Calculation

Perpendicular distance calculation involves determining the shortest path from a point to a line or plane, standing at a right angle to it. It is a frequent requirement in vector mathematics and geometry problems.
To find the perpendicular distance from a point to a line in space:

• Identify the point \(P \, (x_1, y_1, z_1)\) and a line \((x, y, z) = (x_0 + at, y_0 + bt, z_0 + ct)\) where \(t\) is a parameter.
• The direction vector of the line is \(d = (a, b, c)\).
• Compute \(\overrightarrow{PQ}\) from \(P\) to any point \(Q\) on the line. The perpendicular distance is minimized when \(\overrightarrow{PQ}\) is orthogonal to \(d\).
• Solve \(\overrightarrow{PQ} \cdot d = 0\) to find the exact point on the line, then use the distance formula to calculate the perpendicular distance.

This principle assists in a wide range of applications from academic problems to engineering designs where precise measurements are critical.