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Chapter 1: Problem 4

One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors \(\mathbf{b}=(1,2,4)\) and \(\mathbf{c}=(4,2,1)\) by evaluating their scalar product.

Short Answer

Expert verified
The angle between the vectors is \( \cos^{-1}(\frac{4}{7}) \).

Step by step solution

01

Write the formula for the scalar product

The scalar product (or dot product) of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). This can also be expressed in terms of the magnitudes of the vectors and the cosine of the angle between them: \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta \).
02

Calculate the scalar product of \(\mathbf{b}\) and \(\mathbf{c}\)

The scalar product of \( \mathbf{b} = (1,2,4) \) and \( \mathbf{c} = (4,2,1) \) is calculated as follows: \( \mathbf{b} \cdot \mathbf{c} = 1 \times 4 + 2 \times 2 + 4 \times 1 = 4 + 4 + 4 = 12 \).
03

Find the magnitudes of \(\mathbf{b}\) and \(\mathbf{c}\)

The magnitude of \( \mathbf{b} \) is \( \|\mathbf{b}\| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21} \). The magnitude of \( \mathbf{c} \) is \( \|\mathbf{c}\| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{21} \).
04

Solve for the cosine of the angle

Using the relationship \( \mathbf{b} \cdot \mathbf{c} = \|\mathbf{b}\|\|\mathbf{c}\|\cos\theta \), substitute the known values: \( 12 = (\sqrt{21})(\sqrt{21})\cos\theta \). This simplifies to \( 12 = 21\cos\theta \). Solving for \( \cos\theta \), we get \( \cos\theta = \frac{12}{21} = \frac{4}{7} \).
05

Calculate the angle \(\theta\)

To find the angle \( \theta \), take the inverse cosine: \( \theta = \cos^{-1}(\frac{4}{7}) \). Use a calculator to find the angle in degrees or radians.

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Key Concepts

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

Angle Between Vectors
Understanding the angle between vectors is a fundamental concept in vector mathematics. It helps us understand the orientation of one vector relative to another. The angle between vectors can determine whether vectors are aligned, perpendicular, or somewhere in between. To find this angle, we first need to calculate the scalar product (also known as the dot product) of the two vectors.
This dot product relates directly to the cosine of the angle. This is because the formula for the scalar product can be expressed as:
  • \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta \)
Here, \( \theta \) is the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \). To find \( \theta \), we simply rearrange the formula to solve for \( \cos\theta \) and then determine the angle using inverse trigonometric functions.
Vector Magnitudes
The magnitude of a vector is a crucial element when working with vectors. It's essentially the length of the vector, measured in the multidimensional space in which it resides. To determine vector magnitudes, we use the Euclidean distance formula.
For a vector like \( \mathbf{b} = (1, 2, 4) \), its magnitude \( \|\mathbf{b}\| \) is calculated as:
  • \( \|\mathbf{b}\| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21} \)
Similarly, for vector \( \mathbf{c} = (4, 2, 1) \):
  • \( \|\mathbf{c}\| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{21} \)
By computing magnitudes, we lay the groundwork needed to use the scalar product formula involving the cosine of the angle between vectors.
Cosine of Angle
The cosine of the angle between two vectors comes into play when we express the scalar product formula in terms of vector magnitudes and the angle itself.
Using the provided scalar product equation \( \mathbf{b} \cdot \mathbf{c} = \|\mathbf{b}\|\|\mathbf{c}\|\cos\theta \), we can solve for \( \cos\theta \). After calculating the magnitudes and the scalar product, we substitute them into the equation:
  • \( 12 = (\sqrt{21})(\sqrt{21})\cos\theta \)
  • Simplifying gives us \( 12 = 21\cos\theta \).
  • Solving for \( \cos\theta \) yields \( \cos\theta = \frac{12}{21} = \frac{4}{7} \).
The value of \( \cos\theta \) tells us how vectors are oriented relative to each other and is crucial for finding the angle.
Dot Product Calculation
Calculating the dot product of vectors is a straightforward yet powerful process. The dot product gives us both a measure of the vectors' alignment and a way to solve for the angle between them. To calculate this feature for vectors \( \mathbf{b} = (1, 2, 4) \) and \( \mathbf{c} = (4, 2, 1) \), follow these steps:
  • Multiply corresponding components together and sum them up:
  • \( \mathbf{b} \cdot \mathbf{c} = (1 \times 4) + (2 \times 2) + (4 \times 1) = 4 + 4 + 4 = 12 \)
The resulting scalar, 12, gives us a key intermediary piece allowing us to find the cosine of the angle between the vectors and subsequently the angle itself.

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Most popular questions from this chapter

The position of a moving particle is given as a function of time \(t\) to be $$\mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin (\omega t)+\hat{\mathbf{z}} v_{\mathrm{o}} t$$ where \(b, c, v_{\mathrm{o}}\) and \(\omega\) are constants. Describe the particle's orbit.

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

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 \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

Prove that if \(\mathbf{v}(t)\) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t) .\) Prove the converse that if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t),\) then \(|\mathbf{v}(t)|\) is constant. [Hint: Consider the derivative of \(\mathbf{v}^{2}\).] This is a very handy result. It explains why, in two- dimensional polars, \(d \hat{\mathbf{r}} / d t\) has to be in the direction of \(\hat{\boldsymbol{\phi}}\) and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

A ball is thrown with initial speed \(v_{\mathrm{o}}\) up an inclined plane. The plane is inclined at an angle \(\phi\) above the horizontal, and the ball's initial velocity is at an angle \(\theta\) above the plane. Choose axes with \(x\) measured up the slope, \(y\) normal to the slope, and \(z\) across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance \(R=2 v_{\mathrm{o}}^{2} \sin \theta \cos (\theta+\phi) /\left(g \cos ^{2} \phi\right)\) from its launch point. Show that for given \(v_{\mathrm{o}}\) and \(\phi,\) the maximum possible range up the inclined plane is \(R_{\max }=v_{\mathrm{o}}^{2} /[g(1+\sin \phi)]\)
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