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Prove that if two diagonals of a rectangular parallelepiped are perpendicular, then its dimensions are congruent to the sides of a right triangle, and vice versa. \mathrm{~cm}$.

Short Answer

Expert verified
Diagonals are perpendicular if and only if the dimensions form a right triangle.

Step by step solution

01

Understand the Problem

In a rectangular parallelepiped (a 3D box), we need to prove that if the diagonals are perpendicular, the dimensions form a right triangle. Conversely, if the dimensions form a right triangle, show that this implies the diagonals are perpendicular.
02

Express Diagonals in Terms of Dimensions

Let the dimensions of the parallelepiped be \(a\), \(b\), and \(c\). The length of a diagonal, which connects opposing corners, is given by the formula \(d = \sqrt{a^2 + b^2 + c^2}\).
03

Perpendicular Diagonals Condition

For two diagonals to be perpendicular, the dot product of their vectors must be zero. If one diagonal vector has components \([a, b, c]\) and another \([-a, -b, c]\), their dot product is \(-a^2 - b^2 + c^2\). Set this equal to zero to establish the relation governing perpendicular diagonals.
04

Establish the Condition for Right Triangle

From the equation \(-a^2 - b^2 + c^2 = 0\), it simplifies to \(a^2 + b^2 = c^2\), which is the Pythagorean theorem. This means \(a\), \(b\), and \(c\) form a right triangle where \(c\) is the hypotenuse.
05

Reverse Implication

Conversely, if \(a\), \(b\), and \(c\) satisfy \(a^2 + b^2 = c^2\), then substituting back will result in the diagonals being perpendicular, confirming the converse.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diagonals Perpendicularity
In a rectangular parallelepiped, which is essentially a 3D rectangle or a box, the diagonals are long lines running from one corner to the opposite corner. For these diagonals to be perpendicular, their orientation in space should form a right angle, specifically 90 degrees.

To test this, we use the idea of a dot product from vector mathematics. A dot product of zero indicates perpendicular vectors. If two diagonal vectors within the parallelepiped are \[ \vec{D_1} = [a, b, c] \] and \[ \vec{D_2} = [-a, -b, c] \], their dot product is computed as \[ \vec{D_1} \cdot \vec{D_2} = -a^2 - b^2 + c^2. \]

This mathematical condition of the diagonals gives us fascinating insight into the dimensional relations of the box. If the dot product equals zero, the dimensions are intertwined in a way that they must form a right triangle.
Right Triangle
The connection between the perpendicularity of diagonals and right triangles in a parallelepiped lies in geometry's fundamental principles. A right triangle is characterized by one of its angles being 90 degrees.

In the context of a 3D box, the dimensions \(a, b,\) and \(c\) reflect the three sides of a potential right triangle. This condition arises when the lengths of the box follow the Pythagorean theorem, such that \[ a^2 + b^2 = c^2. \] Here, \(c\) serves as the hypotenuse, the longest side of the triangle.

This interrelationship means the dimensions not only define the parallel sides of the parallelepiped but also the right triangle's structure. Only when these dimensions satisfy the Pythagorean relationship, do we achieve perpendicular diagonals.
Pythagorean Theorem
The Pythagorean theorem plays a crucial role in analyzing the conditions set in a rectangular parallelepiped. This ancient principle, which states that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides, is written as \[ a^2 + b^2 = c^2. \]

Applying this theorem to a parallelepiped, the dimensions \(a, b,\) and \(c\) are used to assess the perpendicularity of diagonals. Liberating the same logic, if dimension \(c\) equates to the calculated hypotenuse based on sides \(a\) and \(b\), then the diagonals remain perpendicular to each other. Thus, ensuring the connection between the box's dimensions and the mathematic structure of a right triangle.
  • The diagonals remain perpendicular if and only if this theorem holds true.
  • This relationship gives rise to many geometric and spatial insights in 3D objects.
3D Geometry
3D geometry expands our understanding of space beyond the flat surfaces we encounter in typical 2D scenarios. In a rectangular parallelepiped, we encounter edges, faces, and vital diagonals that cross its expanse.

Understanding the perpendicular nature of diagonals requires translating concepts from 2D into three-dimensional space, linking vector mathematics and geometric properties such as length, angle, and space relations.

The diagonals do not just serve as static lines but encompass the broader spatial awareness of the parallelepiped itself. They highlight how different dimensions like height, width, and depth align to form complex geometric concepts in our tangible world. This deeper exploration of 3D objects reveals intricate relationships like the one between perpendicular diagonals and right triangle dimensions, uncovering the balanced harmonies in geometric structures.

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Most popular questions from this chapter

Describe all ways to superimpose a regular tetrahedron onto itself by rotations, and show that there are 12 such rotations (including the trivial one).

Lateral surface area of prisms. Theorem. The lateral surface area of a prism is equal to the product of a lateral edge and the perimeter of a perpendicular cross section. By a perpendicular cross section (Figure 51) of a prism, we mean the polygon abcde obtained by intersecting all lateral faces of the prism by a plane perpendicular to the lateral edges. Sides of this polygon are perpendicular to the lateral edges \((\S \S 31,20)\). The lateral surface area of the prism is equal to the sum of areas of parallelograms. In each of them, a lateral edge can be considered as the base, and one of the sides of the perpendicular cross section as the altitude. Therefore the lateral surface area is equal to \(A A^{\prime} \cdot a b+\) \(B B^{\prime}+b c+C C^{\prime} \cdot c d+D D^{\prime} \cdot d e+E E^{\prime} \cdot e a=A A^{\prime} \cdot(a b+b c+c d+d e+e a)\). Corollary. The lateral surface area of a right prism is equal to the product of the perimeter of the base and the altitude, because lateral edges of such a prism are congruent to the altitude, and its base can be considered as the perpendicular cross section.

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