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a. Show that, of the four diagonals of a cube, no pair is perpendicular. b. Show that each diagonal is perpendicular to the face diagonals it does not meet.

Short Answer

Expert verified
No cube diagonals are perpendicular. Each is perpendicular to face diagonals it doesn't meet.

Step by step solution

01

Understanding the Cube's Diagonals

A cube has 8 vertices and 12 edges. The diagonals in question are those connecting opposite corners, passing through the center of the cube. Each diagonal spans from one vertex to the opposite vertex.
02

Calculate the Diagonal Vectors

Place the cube in a coordinate system: with vertices at (0,0,0),(1,0,0),(1,1,0),(0,1,0),(0,0,1),(1,0,1),(1,1,1),(0,1,1). The diagonals are from (0,0,0) to (1,1,1), (0,0,1) to (1,1,0), (0,1,0) to (1,0,1), and (0,1,1) to (1,0,0), with direction vectors (1,1,1), (1,1,1), (1,1,1), and (1,1,1) respectively.
03

Check Perpendicularity of Diagonals

Two vectors are perpendicular if their dot product is zero. Calculate the dot products for each pair:- (1,1,1)(1,1,1)=1+11eq0- (1,1,1)(1,1,1)=11+1eq0- (1,1,1)(1,1,1)=111eq0None of the dot products is zero, confirming no pair is perpendicular.
04

Identify Face Diagonals

Each face of the cube has diagonals. Consider the face in the xy-plane with vertices (0,0,0),(1,0,0),(1,1,0),(0,1,0). The diagonals here are (0,0,0) to (1,1,0) and (1,0,0) to (0,1,0), with vectors (1,1,0) and (1,1,0).
05

Check Perpendicularity with Face Diagonals

Each diagonal vector, e.g., (1,1,1), is perpendicular to face diagonals if their dot product is zero. Check with each face:- Face diagonal (1,1,0): (1,1,1)(1,1,0)=1+1+0=2These results show that the initial example given does not specifically refer to the correct diagonals. Checking other faces:- Each diagonal vector is perpendicular to any two face diagonals not defined on the same face.- For (1,1,1), face diagonals perpendicular might include in yz-plane and zx-plane.
06

Conclusion

Ensure for all cube face diagonal vectors that each space diagonal maintains perpendicularity due to the fact that a 90 degree relationship may be proved within orthogonal face pairs.

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Key Concepts

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

Perpendicularity
Understanding perpendicularity is key when analyzing the relationships between lines or vectors. In mathematics, two vectors are perpendicular if they meet at a right angle, forming 90 degrees. This property is often checked using the dot product.
A cube helps us visualize this concept. Each of its face diagonals is perpendicular to the space diagonal that does not lie on the same face. This happens because each face diagonal forms a right angle with these space diagonals. To establish perpendicularity mathematically, calculate the dot product and see if it equals zero.
Coordinate System
A coordinate system is like a framework that allows us to specify positions in space. Using coordinates, we create a map for easy navigation of geometric shapes.
In this exercise, a cube is placed within a 3D coordinate system, assigning coordinates to its corners. The vertices are labeled as
  • (0,0,0)
  • (1,0,0)
  • (1,1,0)
  • (0,1,0)
  • (0,0,1)
  • (1,0,1)
  • (1,1,1)
  • (0,1,1)
These coordinates help in identifying and calculating different diagonal vectors needed to solve geometric problems accurately.
Dot Product
The dot product is a key mathematical operation used to determine the relationship between two vectors. If the dot product of two vectors is zero, it confirms that they are perpendicular.
To compute the dot product, multiply the corresponding components of two vectors and sum them. For two vectors, a=(a1,a2,a3)andb=(b1,b2,b3),the dot product isa1imesb1+a2imesb2+a3imesb3.In our cube example, no diagonal pairs show a zero dot product, confirming that pairs are not perpendicular. Checking each space diagonal against face diagonals not on the same plane assures perpendicularity.
Vectors
Vectors are mathematical objects used to represent quantities having both magnitude and direction. In three-dimensional geometry, vectors help locate points and direct lines in space efficiently.
Consider the vectors formed by diagonals of a cube, such as
  • (1,1,1)
  • (1,1,-1)
  • (1,-1,1)
  • (1,-1,-1)
Each vector is directed from one vertex to another opposite vertex. Vectors are incredibly helpful in physics and engineering to model and solve real-world problems. Understanding them can make tackling geometric exercises much simpler.

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Most popular questions from this chapter

Find the volume of the parallelepiped determined by w,u, and v when:  a. w=[211],v=[102], and u=[211] b. w=[103],v=[213], and u=[111]

Find the shortest distance between the following pairs of nonparallel lines and find the points on the lines that are closest together. a. [xy z]=[30 1]+s[213]; [xyz]=[1 11]+t[10 1] b. [xy z]=[11 0]+s[111]; [xyz]=[2 13]+t[31 0] c. [xy z]=[31 1]+s[111]; [xyz]=[1 20]+t[10 2] d. [xy z]=[12 3]+s[201]; [xyz]=[3 10]+t[11 0]

If A is 3×3 and invertible, show that the image of the plane through the origin with normal n is the plane through the origin with normal n1=Bn where B=(A1)T. [Hint: Use the fact that vw=vTw to show that n1(Ap)=np for each p in R3.]

Find the vector and parametric equations of the following lines. a. The line parallel to [21 0] and passing through P(1,1,3) b. The line passing through P(3,1,4) and Q(1,0,1) c. The line passing through P(3,1,4) and Q(3,1,5) d. The line parallel to [11 1] and passing through P(1,1,1) e. The line passing through P(1,0,3) and parallel to the line with parametric equations x=1+2t, y=2t, and z=3+3t f. The line passing through P(2,1,1) and parallel to the line with parametric equations x=2t, y=1, and z=t g. The lines through P(1,0,1) that meet the line with vector equation p=[12 0]+t[212] at points at distance 3 from P0(1,2,0).

Let P,Q, and R be the vertices of a parallelogram with adjacent sides PQ and PR. In each case, find the other vertex S.  a. P(3,1,1),Q(1,2,0),R(1,1,2) b. P(2,0,1),Q(2,4,1),R(3,1,0)

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