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Chapter 1: Problem 8

A triangle is defined by the vertices of three vectors \(\mathbf{A}, \mathrm{B}\), and \(\mathrm{C}\) that extend from the origin. In terms of \(A, B\), and \(C\) show that the vector sum of the successive sides of the triangle \((A B+B C+C A)\) is zero.

Short Answer

Expert verified
The vector sum \(AB + BC + CA\) is zero.

Step by step solution

01

Understanding Vectors

Visualize the triangle formed by the points where vectors \(\mathbf{A}\), \(\textbf{B}\), and \(\textbf{C}\) terminate. The vectors represent the sides of a triangle.
02

Determine Vector Directions

The vector \(\textbf{AB}\) is the direction from point \(A\) to \(B\), so it is defined as \(\textbf{B} - \textbf{A}\). Similarly, \(\textbf{BC} = \textbf{C} - \textbf{B}\) and \(\textbf{CA} = \textbf{A} - \textbf{C}\).
03

Calculate Vector Sum

Substitute the vectors into the expression: \[ \textbf{AB} + \textbf{BC} + \textbf{CA} = (\textbf{B} - \textbf{A}) + (\textbf{C} - \textbf{B}) + (\textbf{A} - \textbf{C}) \]
04

Simplify the Expression

Combine the terms: \[ \textbf{B} - \textbf{A} + \textbf{C} - \textbf{B} + \textbf{A} - \textbf{C} \]Cancel out the terms: \[ \textbf{B} - \textbf{B} + \textbf{C} - \textbf{C} + \textbf{A} - \textbf{A} = \textbf{0} \]Thus, the vector sum is zero.

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Key Concepts

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

triangle geometry
To understand the problem, it's important to visualize the triangle within vector space. Imagine vectors \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \) originating from a common point, usually the origin, and extending outwards. These vectors are not just lines; they form a bounded closed shape known as a triangle. The points where each vector ends, actually define the three vertices of our geometric triangle.

Now, consider walking along these vectors from one to the next to form a loop: you start from the tip of vector \( \mathbf{A} \), head towards \( \mathbf{B} \), then towards \( \mathbf{C} \), and finally come back to \( \mathbf{A} \). This loop signifies that the vector journey forms a closed path, resulting in no net displacement, i.e., the sum of vectors around the closed triangle is zero.
vector representation
Vectors in this context are tools that help us describe the sides of the triangle in terms of direction and magnitude. A vector is usually represented as an arrow, having both direction and magnitude, making it perfect to symbolize the triangle sides.

When we talk about vectors \( \mathbf{AB} \), \( \mathbf{BC} \), and \( \mathbf{CA} \), we mean the directional displacement from one point to another.
  • Vector \( \mathbf{AB} \) is the vector from point \( A \) to \( B \), represented mathematically as \( \mathbf{B} - \mathbf{A} \).
  • Similarly, \( \mathbf{BC} = \mathbf{C} - \mathbf{B} \), indicating movement from \( B \) to \( C \).
  • Lastly, \( \mathbf{CA} = \mathbf{A} - \mathbf{C} \) describes the return journey from \( C \) to \( A \).
Each of these ties back to the original position and orientation of the corresponding vectors, showing how they connect points in their trajectory.
algebraic simplification
This concept is crucial to find out why the sum of these vector sides equates to zero. Once we represent each vector mathematically, we substitute these into the summed expression as given in the problem: \( \mathbf{AB} + \mathbf{BC} + \mathbf{CA} = (\mathbf{B} - \mathbf{A}) + (\mathbf{C} - \mathbf{B}) + (\mathbf{A} - \mathbf{C}) \).

By writing it clearly in expanded form, observe how each vector part balances another:
  • \( \mathbf{B} - \mathbf{B} = \mathbf{0} \)
  • \( \mathbf{C} - \mathbf{C} = \mathbf{0} \)
  • \( \mathbf{A} - \mathbf{A} = \mathbf{0} \)
The overall outcome of these operations shows perfect cancellation, meaning that the vectors sum up to \( \mathbf{0} \). This algebraic reduction reinforces why the total displacement in this closed loop is indeed zero, explaining how the mathematical terms coincide with physical intuition.

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Most popular questions from this chapter

The origin of Cartesian coordinates is at the Earth's center. The moon is on the \(z\) -axis, a fixed distance \(R\) away (center-to-center distance). The tidal force exerted by the moon on a particle at the earth's surface (point \(x, y, z)\) is given by $$ F_{x}=-G M m \frac{x}{R^{3}}, \quad F_{y}=-G M m \frac{y}{R^{3}}, \quad F_{z}=+2 G M m \frac{z}{R^{3}} . $$ Find the potential that yields this tidal force. $$ \text { ANS, }-\frac{G M m}{R^{3}}\left(z^{2}-\frac{1}{2} x^{2}-\frac{1}{2} y^{2}\right) $$ In terms of the Legendre polynomials of Chapter 12 this becomes $$ -\frac{G M m}{R^{3}} r^{2} P_{2}(\cos \theta) $$

The vertices of parallelogram \(A B C D\) are \((1,0,0),(2,-1,0),(0,-1,1)\), and \((-1,0,1)\) in order. Calculate the vector areas of triangle \(A B D\) and of triangle \(B C D\). Are the two vector areas equal? $$ \text { ANS. } \operatorname{Area}_{A B D}=-\frac{1}{2}(\hat{\mathbf{x}}+\hat{\mathbf{y}}+2 \hat{\mathbf{z}}) \text { . } $$

If \(\mathbf{x}^{\prime} \cdot \mathbf{y}=\delta_{x y},(\mathbf{x}, \mathbf{y}=\mathbf{a}, \mathbf{b}, \mathbf{c})\), prove that $$ \mathbf{a}^{\prime}=\frac{\mathbf{b} \times \mathbf{c}}{\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}} $$ (This is the converse of Problem 1.5.15.)

Find the work done going around a unit circle in the \(x y\) -plane: (a) counterclockwise from 0 to \(\pi\), (b) clockwise from 0 to \(-\pi\), doing work against a force field given by $$ \mathbf{F}=\frac{-\hat{\mathbf{x}} y}{x^{2}+y^{2}}+\frac{\hat{\mathbf{y}} x}{x^{2}+y^{2}} $$ Note that the work done depends on the path.

Show that \(\nabla(u v)=v \nabla u+u \nabla v\), where \(u\) and \(v\) are differentiable scalar functions of \(x, y\), and \(z\). (a) Show that a necessary and sufficient condition that \(u(x, y, z)\) and \(v(x, y, z)\) are related by some function \(f(u, v)=0\) is that \((\nabla u) \times(\nabla v)=0\). (b) If \(u=u(x, y)\) and \(v=v(x, y)\), show that the condition \((\nabla u) \times(\nabla v)=0\) leads to the two-dimensional Jacobian $$ J\left(\frac{u, v}{x, y}\right)=\left|\begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right|=0 . $$ The functions \(u\) and \(v\) are assumed differentiable.
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