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a. Let \(P_{1}, P_{2}, P_{3}, P_{4}, P_{5},\) and \(P_{6}\) be six points equally spaced on a circle with centre \(C\). Show that $$ \overrightarrow{C P}_{1}+\overrightarrow{C P}_{2}+\overrightarrow{C P}_{3}+\overrightarrow{C P}_{4}+\overrightarrow{C P}_{5}+\overrightarrow{C P}_{6}=\mathbf{0} $$ b. Show that the conclusion in part (a) holds for any even set of points evenly spaced on the circle. c. Show that the conclusion in part (a) holds for three points. d. Do you think it works for any finite set of points evenly spaced around the circle?

Short Answer

Expert verified
The vector sum is zero for evenly spaced points as roots of unity sum to zero.

Step by step solution

01

Understanding the Problem

In this problem, we will demonstrate that the vector sum of radial vectors from the center to evenly spaced points on a circle results in the zero vector. This involves examining both specific and general cases, initially for six points.
02

Conceptual Approach with Complex Numbers

We represent the situation using the complex plane where the center of the circle is the origin. The position of each point on the circle can be expressed as a complex number with modulus (radius) equal and arguments separated by equal central angles.
03

Writing Vectors as Complex Numbers

Let's place the circle on the complex plane with radius 1 for simplicity (this does not affect the generality of our result). The points are represented as \( e^{i\theta}, e^{i(\theta + \pi/3)}, e^{i(\theta + 2\pi/3)}, e^{i(\theta + \pi)}, e^{i(\theta + 4\pi/3)}, e^{i(\theta + 5\pi/3)} \).
04

Summing the Complex Numbers

The vector sum corresponds to the sum of complex numbers: \[ z_1 + z_2 + z_3 + z_4 + z_5 + z_6 = e^{i\theta} \left( 1 + e^{i\pi/3} + e^{i2\pi/3} + e^{i\pi} + e^{i4\pi/3} + e^{i5\pi/3} \right) \]. The terms inside the parentheses form a geometric series.
05

Calculating the Geometric Series

The series \( 1 + e^{i\pi/3} + e^{i2\pi/3} + e^{i\pi} + e^{i4\pi/3} + e^{i5\pi/3} \) is a sum of the sixth roots of unity, excluding 1. The roots of unity sum to zero, which makes this series sum to zero.
06

Conclusion for Six Points

Since the inner sum is zero, the entire expression simplifies to zero: \[ e^{i\theta}(0) = 0 \]. This confirms that the vector sum is indeed zero.
07

Generalizing for Even Number of Points

For any even number \(2k\) of points, they can be expressed as the sum of the roots of unity \(z_j = e^{i2\pi j/(2k)}\), where the sum is \(\sum_{j=0}^{2k-1} z_j = 0\). This holds for any even \(2k\).
08

Checking for Three Points

For three points, we have vectors \( e^{i\theta}, e^{i(\theta + 2\pi/3)}, e^{i(\theta + 4\pi/3)} \). These correspond to third roots of unity, which also sum to zero: \( 1 + \omega + \omega^2 = 0 \).
09

Evaluating for Any Finite Set

For any finite set of points evenly spaced on a circle, they can be represented by the roots of unity \( \sum_{j=0}^{n-1} e^{i2\pi j/n} \), which sum to zero by properties of the roots of unity.

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Key Concepts

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

Roots of Unity
Roots of unity are an interesting concept, especially in the context of complex numbers and circular symmetry. Simply put, the roots of unity are complex numbers that, when raised to a certain power, equal 1. If you have an equation like \(z^n = 1\), the solutions for \(z\) are the \(n\)-th roots of unity. They are often represented as points on the complex plane. Each root is evenly spaced around the unit circle, much like clock hands pointing to each hour in a twelve-hour cycle.

Consider these points: \(1, e^{i2\pi/n}, e^{i4\pi/n}, \, \ldots, e^{i2(n-1)\pi/n}\). They exhibit fascinating properties and sum to zero because they are symmetrically distributed.
In practical terms, when you add all the \(n\)-th roots of unity together, their vector sum is zero. This happens due to the symmetrical nature of how the vectors point outward from the origin and balance each other out.
Complex Numbers
Complex numbers are a powerful tool for visualizing and solving geometry problems involving circles. These numbers have two parts: a real part and an imaginary part, expressed as \(a + bi\). On the complex plane, the number forms a point where \(a\) is the x-coordinate and \(b\) is the y-coordinate.

What's amazing is how these numbers naturally connect to rotations and magnitudes. When used to represent points on a circle, complex numbers can simplify many problems. Instead of thinking in terms of angles alone, think about each point as a complex number with a certain magnitude and direction (angle).
In our problem, converting vectors into complex numbers allowed us to easily handle the sum of multiple vectors by merely adding corresponding complex numbers. This method taps into the power of exponential notation: \( e^{i \theta}\), whereby a point on a circle with radius 1 can be represented as a unit complex number. Thus, transformations or rotations become simple multiplications.
Geometric Series
A geometric series is a series of numbers where each term is a constant multiple of the previous term, expressed as \(a, ar, ar^2, ar^3, \ldots\). It's incredibly useful, especially when dealing with problems involving roots of unity or evenly spaced points on a circle.

In the context of this problem, roots of unity form a geometric series. When computing the sum of vectors (complex numbers), this series emerges as \(1 + e^{i\pi/3} + e^{i2\pi/3} + e^{i\pi} + e^{i4\pi/3} + e^{i5\pi/3}\) for six points. The sum of this series equals zero due to the properties of geometric series and the symmetrical distribution of the points.
Knowing how to calculate and work with these series allows you to easily find sums and simplifies the problem-solving process when working with symmetric or repeating patterns.
Vector Sum
The vector sum is a fundamental concept in both physics and mathematics. It's the result of adding two or more vectors together. Vectors have both magnitude and direction, making them ideal for representing forces, velocities, or positions.

In problems involving circular symmetry, like ours, vector sums help prove symmetries. If you take vectors from the center of the circle to each point, these vectors add up not just in terms of their length, but also their direction.
The crucial part here is symmetry: vectors spread evenly around a circle (like clock hands) will naturally point in such a way that they cancel each other out. Thus, when you sum vectors pointing to evenly spaced points around a circle, you find they form a closed shape, resulting in a net vector sum of zero. This provides a neat way to demonstrate balanced forces or structures in circular patterns.

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Most popular questions from this chapter

Find the vector and parametric equations of the following lines. a. The line parallel to \(\left[\begin{array}{r}2 \\ -1 \\\ 0\end{array}\right]\) 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 \(\left[\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right]\) and passing through \(P(1,1,1)\) e. The line passing through \(P(1,0,-3)\) and parallel to the line with parametric equations \(x=-1+2 t\), \(y=2-t,\) and \(z=3+3 t\) f. The line passing through \(P(2,-1,1)\) and parallel to the line with parametric equations \(x=2-t,\) \(y=1,\) and \(z=t\) \(\mathrm{g}\). The lines through \(P(1,0,1)\) that meet the line with vector equation \(\mathbf{p}=\left[\begin{array}{l}1 \\ 2 \\\ 0\end{array}\right]+t\left[\begin{array}{r}2 \\ -1 \\ 2\end{array}\right]\) at points at distance 3 from \(P_{0}(1,2,0)\).

If a plane contains two distinct points \(P_{1}\) and \(P_{2}\), show that it contains every point on the line through \(P_{1}\) and \(P_{2}\).

Prove that the triangle inequality \(\|\mathbf{u}+\mathbf{v}\| \leq\|\mathbf{u}\|+\|\mathbf{v}\|\) holds for all vectors \(\mathbf{u}\) and \(\mathbf{v}\). [Hint: Consider the triangle with \(\mathbf{u}\) and \(\mathbf{v}\) as two sides. \(]\)

In each case, find \(\overrightarrow{P Q}\) and \(\|\overrightarrow{P Q}\| .\) a. \(P(1,-1,3), Q(3,1,0)\) b. \(P(2,0,1), Q(1,-1,6)\) c. \(P(1,0,1), Q(1,0,-3)\) d. \(P(1,-1,2), Q(1,-1,2)\) e. \(P(1,0,-3), Q(-1,0,3)\) f. \(P(3,-1,6), Q(1,1,4)\)

Let \(P, Q, R,\) and \(S\) be four points, \(n o t\) all on one plane, as in the diagram. Show that the volume of the pyramid they determine is $$\frac{1}{6}|\overrightarrow{P Q} \cdot(\overrightarrow{P R} \times \overrightarrow{P S})|$$

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