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Chapter 11: Problem 5

Show that the points \((1,-1,1),(5,-5,4),(5,0,8)\) and \((1,4,5)\) form a rhombus.

Short Answer

Expert verified
Question: Prove that the points A(1, -1, 1), B(5, -5, 1), C(5, 0, 8), and D(1, 4, 5) form a rhombus. Solution: To prove that these points form a rhombus, we showed that: 1. All sides have equal length, and we found that each side has a length of 5. 2. The diagonals are perpendicular to each other, which was verified by calculating the dot product and finding it to be 0. 3. The diagonals bisect each other, which was proven by finding the same midpoint for both diagonals. Therefore, the given points form a rhombus.

Step by step solution

01

Find the lengths of all sides using the distance formula

We will use the distance formula to find the lengths of all four sides: Distance formula: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\) Length of Side 1 (AB): \(d_{AB}=\sqrt{(5-1)^2+(-5-(-1))^2+(4-1)^2}=\sqrt{16+16+9}=5\) Length of Side 2 (BC): \(d_{BC}=\sqrt{(5-5)^2+(0-(-5))^2+(8-4)^2}=5\) Length of Side 3 (CD): \(d_{CD}=\sqrt{(1-5)^2+(4-0)^2+(5-8)^2}=5\) Length of Side 4 (DA): \(d_{DA}=\sqrt{(1-1)^2+(-1-4)^2+(1-5)^2}=5\) Since all sides have the same length, it is a parallelogram.
02

Find the lengths of diagonals using the distance formula

Now, we will calculate the lengths of both diagonals: Length of Diagonal 1 (AC): \(d_{AC}=\sqrt{(1-5)^2+(-1-0)^2+(1-8)^2}=5\sqrt{2}\) Length of Diagonal 2 (BD): \(d_{BD}=\sqrt{(1-5)^2+(-1+5)^2+(1-4)^2}=5\sqrt{2}\)
03

Find the dot product for diagonals to show they are perpendicular

In order to show that the diagonals are perpendicular to each other, we will calculate their dot product. If the dot product is zero, then the diagonals are perpendicular. Vector AC: \(\overrightarrow{AC} = (1-5, -1-0, 1-8) = (-4, -1, -7)\) Vector BD: \(\overrightarrow{BD} = (1-5, -1+5, 1-4) = (-4, 4, -3)\) Dot product: \(\overrightarrow{AC} \cdot \overrightarrow{BD} = (-4)(-4) + (-1)(4) + (-7)(-3) = 16-4+21 = 0\) Since the dot product is zero, the diagonals are perpendicular to each other.
04

Find the point of intersection and show it bisects both diagonals

Next, we need to find the midpoint of both diagonals and show that they are the same point. Midpoint formula: \(M = \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)\) Midpoint of Diagonal 1 (AC): \(M_{AC}=\left(\frac{1+5}{2},\frac{-1+0}{2},\frac{1+8}{2}\right)=\left(3, -\frac{1}{2}, \frac{9}{2}\right)\) Midpoint of Diagonal 2 (BD): \(M_{BD}=\left(\frac{1+5}{2},\frac{-1+5}{2},\frac{1+4}{2}\right)=\left(3, 2, \frac{5}{2}\right)\) Since the midpoints of both diagonals are the same point, the diagonals bisect each other. Thus, we have shown that the given points form a rhombus.

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Key Concepts

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

Understanding the Distance Formula
The distance formula is a fundamental tool in analytical geometry, particularly when dealing with the coordinates of points in space. It's derived from the Pythagorean theorem and allows us to calculate the straight-line distance between two points, \textbf{A} with coordinates \textbf{(x1, y1, z1)} and \textbf{B} with coordinates \textbf{(x2, y2, z2)}. In three dimensions, the formula is: \[\begin{equation}d = \textbf{AB} = \/{[(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2}\end{equation}\]

When Would You Use the Distance Formula?

  • To determine the length of sides in 3D geometric figures.
  • As a step in proving shapes, like rhombuses, where equal side lengths are a defining property.
  • To assess whether points lie on a circle, sphere, or other geometric shapes.
This is crucial because in problems like forming a rhombus, confirming equal lengths of sides verifies that the shape meets specific criteria of a rhombus: all sides being of equal length.
Dot Product and Perpendicularity
The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is central in determining the angle between two vectors, which in turn is essential in verifying perpendicularity between those vectors.\[\begin{equation}\vec{A} \cdot \vec{B} = A1 * B1 + A2 * B2 + A3 * B3\end{equation}\]

Vector Perpendicularity

  • If the dot product of two vectors is zero, \textbf{AB}, the vectors are perpendicular to each other.
  • This zero-product result is key in identifying right angles within geometric figures.
  • For a rhombus, the diagonals are perpendicular; showing this fact often involves the dot product.
Applying this concept in our problem, the vectors representing the diagonals of the supposed rhombus had a dot product equal to zero, which was a necessary step in proving that the figure is, in fact, a rhombus.
Parallelogram Properties in a Rhombus
A rhombus is a type of parallelogram with its own unique properties. The defining characteristics of a parallelogram apply to rhombuses as well. For instance, opposite sides are equal in length, and opposite angles are equal. However, a rhombus has the additional property that all sides are of equal length. Another critical property shared with parallelograms is that both pairs of opposite sides are parallel to each other.

How Does a Parallelogram Relate to a Rhombus?

  • All rhombuses are parallelograms, but not all parallelograms are rhombuses.
  • Identifying a parallelogram as a rhombus involves ensuring all sides are equal, which is where the distance formula plays a pivotal role.
Observing parallelogram properties in the context of the rhombus helps validate that the sides and angles of the given points conform to the rhombus's definitions.
Bisection of Diagonals
In the geometry of rhombuses, another essential trait is the bisection of diagonals. This means each diagonal cuts the other into two equal parts at the center of the figure. The point which they bisect at is the midpoint of both diagonals and, consequently, becomes the center of the rhombus.

Midpoint Calculation

  • The midpoint of a line segment can be determined using the midpoint formula: \(M = (\frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2})\).
  • This calculation is fundamental in proving that the diagonals of a rhombus bisect each other, an essential step in confirming that a set of points in space describes a rhombus.
In the context of the problem, the calculation of midpoints for both diagonals returning the same point assures us that the diagonals bisect each other and that the figure formed is undeniably a rhombus.

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Most popular questions from this chapter

A variable line in two adjacent positions has direction cosines \((I, m, n)\), \((I+\delta l, m+\delta m, n+\delta n)\). Prove that the small angle \(\delta \theta\) between two positives is given by \(\delta^{2} \theta=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\).

Find the ratio in which the coordinate planes divide the line joining the points \((-2,4,7)\) and \((3,-5,8)\)

Show that the points \(P(3,2,-4), Q(9,8,-10)\) and \(R(5,4,-6)\) are collinear. Find the ratio in which \(R\) divides \(P O\)

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Show that the line joining the points \((1,2,3)\) and \((1,5,7)\) is parallel to the line joining the points \((-4,3,-6)\) and \((2,9,2)\).
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