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Chapter 1: Problem 9

In elementary trigonometry, you probably learned the law of cosines for a triangle of sides \(a, b,\) and \(c,\) that \(c^{2}=a^{2}+b^{2}-2 a b \cos \theta,\) where \(\theta\) is the angle between the sides \(a\) and \(b\). Show that the law of cosines is an immediate consequence of the identity \((\mathbf{a}+\mathbf{b})^{2}=a^{2}+b^{2}+2 \mathbf{a} \cdot \mathbf{b}\)

Short Answer

Expert verified
By expanding the identity and substituting the dot product formula, the law of cosines is verified.

Step by step solution

01

Expanding the Given Identity

The identity given is \((\mathbf{a} + \mathbf{b})^2 = a^2 + b^2 + 2 \mathbf{a} \cdot \mathbf{b}\).This identity holds for any vectors \(\mathbf{a}\) and \(\mathbf{b}\). Expanding the left side results in\((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}\),which confirms the identity with \(a^2 = \mathbf{a} \cdot \mathbf{a}\) and \(b^2 = \mathbf{b} \cdot \mathbf{b}\).
02

Relating Vectors to Triangle Sides

Consider a triangle where the sides are represented as vectors. Let \(\mathbf{c} = \mathbf{a} + \mathbf{b}\) be the vector from one vertex to another. This means that the square of the magnitude of \(\mathbf{c}\) is\[ c^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}). \]Substituting the identity from Step 1,\[ c^2 = a^2 + b^2 + 2 \mathbf{a} \cdot \mathbf{b}. \]
03

Incorporating the Angle Between Vectors

Using the dot product relationship, we know that\( \mathbf{a} \cdot \mathbf{b} = a b \cos \theta \),where \(\theta\) is the angle between the vectors \(\mathbf{a}\) and \(\mathbf{b}\).Substituting into the equation from Step 2, we get\[ c^2 = a^2 + b^2 + 2 a b \cos \theta. \]
04

Deriving the Law of Cosines Form

Manipulate the equation from Step 3 to reach the form given in the law of cosines:\[ c^2 = a^2 + b^2 + 2 a b \cos(\theta), \]which can be rewritten as\[ c^2 = a^2 + b^2 - 2 a b \cos(180掳 - \theta). \]Since \(\cos(180掳 - \theta) = -\cos(\theta)\), this simplifies back to\[ c^2 = a^2 + b^2 - 2 a b \cos \theta. \]The law of cosines is thus confirmed.

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Key Concepts

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

Trigonometry
Trigonometry is the branch of mathematics that deals with the study of triangles, particularly right triangles, and the relationships between their angles and sides. One of the most powerful tools in trigonometry is the Law of Cosines, which helps to find unknown sides or angles in any triangle, not just right-angled ones. The formula for the Law of Cosines is: \[c^2 = a^2 + b^2 - 2ab \cos(\theta).\]This formula shows how the sides of a triangle relate through angles. When you know two sides of a triangle and the angle between them, the Law of Cosines allows you to solve for the third side. This is especially useful when the triangle is not a right triangle, where simpler tools like the Pythagorean theorem would not apply. Understanding the Law of Cosines involves knowing how angles affect the length of sides across triangles. Familiarize yourself with concepts of angles measured in radians or degrees to fully utilize this law in solving problems.
Vector Algebra
Vector Algebra is a way to understand and solve geometric problems using vectors, which are quantities that have both magnitude and direction. In the context of the Law of Cosines, vectors can represent the sides of a triangle. Think of each side of a triangle as a vector. If vectors \( \mathbf{a} \) and \( \mathbf{b} \) represent two sides, then the vector \( \mathbf{c} \) can represent the resultant side, calculated using vector addition, such as \( \mathbf{c} = \mathbf{a} + \mathbf{b} \).Vectors make it possible to calculate the magnitude and direction of a resultant vector using principles of vector addition and subtraction, which directly ties back to the Law of Cosines. Vector Algebra involves operations like addition, subtraction, and scalar multiplication, all of which lay the foundation to use vectors in describing sides of triangles. The intersection of these vectors at particular angles will provide different magnitudes for the results, hence connecting to how the Law of Cosines predicts the sides of triangles. Knowing how vectors behave, through vector addition or orientation, gives a powerful way to compute side lengths of triangles beyond basic geometry.
Dot Product
The Dot Product, also known as the scalar product, is a critical tool in understanding the relationship between vectors. It computes a single number from two vectors, providing insights into the angle between them. Mathematically, the dot product is defined as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \), where \( \theta \) is the angle between the two vectors.In terms of the Law of Cosines, the dot product connects the vectors representing the sides of a triangle. By substituting \( \mathbf{a} \cdot \mathbf{b} = ab \cos(\theta) \) into the equation for \( c^2 \), we derive how dots on the sides reflect the angle's influence between them. This makes the dot product an invaluable part of vector calculation, helping us relate vector algebra to geometric concepts like the angles in a triangle.The dot product helps us understand projection: how one vector projects onto another, which translates to how one side of the triangle influences another based on the angle. This deepens comprehension of how the Law of Cosines bridges between abstract vector concepts and concrete trigonometric applications. Understanding dot products enriches problem-solving by offering insights into how triangles' sides and angles interrelate through vector properties.

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Most popular questions from this chapter

You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle \(\theta\) with the horizontal. Choose your origin at one of the two corners that touch the floor, the \(x\) axis pointing along the bottom edge of the board, the \(y\) axis pointing up the slope, and the \(z\) axis normal to the board. You now kick a frictionless puck that is resting at \(O\) so that it slides across the board with initial velocity \(\left(v_{\mathrm{ox}}, v_{\mathrm{oy}}, 0\right) .\) Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from \(O\) when it does so.

Conservation laws, such as conservation of momentum, often give a surprising amount of information about the possible outcome of an experiment. Here is perhaps the simplest example: Two objects of masses \(m_{1}\) and \(m_{2}\) are subject to no external forces. Object 1 is traveling with velocity \(\mathbf{v}\) when it collides with the stationary object \(2 .\) The two objects stick together and move off with common velocity \(\mathbf{v}^{\prime}\). Use conservation of momentum to find \(\mathbf{v}^{\prime}\) in terms of \(\mathbf{v}, m_{1},\) and \(m_{2}\)

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure \(1.8 .\) The electric and magnetic fields at a point \(\mathbf{r}_{1}\) due to a charge \(q_{2}\) at \(\mathbf{r}_{2}\) moving with constant velocity \(\mathbf{v}_{2}\) (with \(v_{2} \ll c\) ) are \(^{15}\) $$\mathbf{E}\left(\mathbf{r}_{1}\right)=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \quad \text { and } \quad \mathbf{B}\left(\mathbf{r}_{1}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}}$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2}\) is the vector pointing from \(\mathbf{r}_{2}\) to \(\mathbf{r}_{1}\). (The first of these you should recognize as Coulomb's law.) If \(\mathbf{F}_{12}^{\mathrm{el}}\) and \(\mathbf{F}_{12}^{\text {mag }}\) denote the electric and magnetic forces on a charge \(q_{1}\) at \(\mathbf{r}_{1}\) with velocity \(\mathbf{v}_{1},\) show that \(F_{12}^{\operatorname{mag}} \leq\left(v_{1} v_{2} / c^{2}\right) F_{12}^{\mathrm{el}} .\) This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin \(O\) and the three edges that emanate from \(O\) defined by vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) Show that the volume of the parallelepiped is \(|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\)

In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]
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