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The Triangle Inequality Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in the plane. a. Use the Triangle Rule for adding vectors to explain why \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}| .\) This result is known as the Triangle Inequality. b. Under what conditions is \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}| ?\)

Short Answer

Expert verified
Answer: The Triangle Inequality for vectors states that the magnitude of the sum of two vectors is always less than or equal to the sum of the magnitudes of those two vectors, i.e., \(|\mathbf{u}+\mathbf{v}| \leq |\mathbf{u}| + |\mathbf{v}|\). The equality holds when the two vectors are collinear and have the same direction, meaning they form a straight line.

Step by step solution

01

Understanding the Triangle Rule for Adding Vectors

The Triangle Rule for adding vectors states that if we have two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), we can represent them as sides of a triangle. The sum of the vectors, \(\mathbf{u}+\mathbf{v}\), is represented by the resultant vector, which is the third side of the triangle.
02

Apply the Cosine Law

In this step, we use the Cosine Law from trigonometry to find the relationship between the magnitudes of the vectors. Let \(\theta\) be the angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\). Then we can write the Cosine Law as: \(|\mathbf{u}+\mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2|\mathbf{u}||\mathbf{v}|\cos\theta\).
03

Derive the Triangle Inequality

Now, we know \(\cos\theta\) has a range of \(-1 \leq \cos\theta \leq 1\). The minimum value of \(-2|\mathbf{u}||\mathbf{v}|\cos\theta\) is \(-2|\mathbf{u}||\mathbf{v}|\), and the maximum value is \(2|\mathbf{u}||\mathbf{v}|\). Therefore, \(|\mathbf{u}|^2 + |\mathbf{v}|^2 - 2|\mathbf{u}||\mathbf{v}| \leq |\mathbf{u}+\mathbf{v}|^2 \leq |\mathbf{u}|^2 + |\mathbf{v}|^2 + 2|\mathbf{u}||\mathbf{v}|\). Then, taking square roots of the inequality, we can conclude that the Triangle Inequality holds: \(|\mathbf{u}+\mathbf{v}| \leq |\mathbf{u}| + |\mathbf{v}|\).
04

Determine the Conditions for Equality

The equality \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}|\) holds when \(-2|\mathbf{u}||\mathbf{v}|\cos\theta = 2|\mathbf{u}||\mathbf{v}|\), which means that \(\cos\theta = 1\) or \(\theta = 0\). Thus, the condition for equality in the Triangle Inequality is when the two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are collinear and have the same direction (i.e., they form a straight line).

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Key Concepts

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

Vectors
Vectors are fundamental mathematical objects that have both direction and magnitude. They are typically represented as arrows in a plane or space, where:
  • The length of the arrow indicates the magnitude.
  • The direction of the arrow indicates the direction of the vector.

Vectors are crucial in physics and engineering because they can represent various quantities such as velocity, force, and displacement. To add vectors, you can use the Triangle Rule, which involves placing the tail of the next vector at the head of the previous one. The resulting vector (or resultant) is then drawn from the tail of the first vector to the head of the last vector. This method visually explains why adding vectors in this way forms a triangle.
Cosine Law
The Cosine Law is a helpful theorem in trigonometry, particularly useful in finding the magnitude of a vector sum. For two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), forming a triangle with angle \(\theta\) between them, the Cosine Law is given by:\[|\mathbf{u}+\mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2|\mathbf{u}||\mathbf{v}|\cos\theta\]
This equation allows you to relate the magnitudes of the vectors to the magnitude of their resultant vector and the angle between them. It essentially acts as a bridge, helping you to utilize the angle information to compute the resultant magnitude in non-right triangles. The Cosine Law reduces to the Pythagorean theorem when the angle \(\theta\) is \(90^\circ\), since \(\cos90^\circ = 0\).
Collinear
To say that two vectors are collinear means they lie on the same or a parallel line. When vectors are collinear, one is a scalar multiple of the other. This is central to understanding when equality in the Triangle Inequality holds, as it requires the vectors to be collinear and pointing in the same direction.
  • When two vectors have an angle of \(\theta = 0\), they are perfectly aligned, resulting in the cosine of the angle, \(\cos\theta = 1\), maximizing their sum.
  • Conversely, if \(\theta = 180^\circ\), the vectors are opposite in direction, which minimizes their overlap in sum.

Therefore, vectors in a straight line with no deviation, either overlapping or extending their line in the same direction, reach the maximum possible sum in magnitude.
Magnitude
The magnitude of a vector measures its size or length. In mathematical terms, it's typically represented with absolute value bars: \(|\mathbf{v}|\). It represents how far the vector's reach extends from the origin point to its endpoint in space.
  • For a vector \(\mathbf{v} = (v_x, v_y)\), its magnitude is calculated using the formula \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\).
  • Magnitude is always a non-negative quantity, as it represents a length.

In the context of the Triangle Inequality, the magnitude helps us comprehend and quantify the relationship between the sides of the triangle formed by vectors \(\mathbf{u}\), \(\mathbf{v}\), and their resultant \(\mathbf{u}+\mathbf{v}\). This concept allows us to compare the vector sides and calculate the resultant in geometric terms.

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned} \mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i}+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} \\ &+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k} \end{aligned}$$

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Consider an object moving along the circular trajectory \(\mathbf{r}(t)=\langle A \cos \omega t, A \sin \omega t\rangle,\) where \(A\) and \(\omega\) are constants. a. Over what time interval \([0, T]\) does the object traverse the circle once? b. Find the velocity and speed of the object. Is the velocity constant in either direction or magnitude? Is the speed constant? c. Find the acceleration of the object. d. How are the position and velocity related? How are the position and acceleration related? e. Sketch the position, velocity, and acceleration vectors at four different points on the trajectory with \(A=\omega=1\)

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