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Chapter 7: Problem 86

Find the smallest positive angle between the vectors \(\langle- 3,5\rangle\) and \((1,6)\)

Short Answer

Expert verified
The smallest positive angle between the vectors is approximately 42.07掳.

Step by step solution

01

- Find the Dot Product of the Vectors

The dot product of vectors \(\textbf{a} = \langle -3, 5\rangle\) and \(\textbf{b} = (1, 6)\) is given by \(\textbf{a} \cdot \textbf{b} = a_1b_1 + a_2b_2\). Substitute the given values: \ \(\textbf{a} \cdot \textbf{b} = (-3)(1) + (5)(6) = -3 + 30 = 27\).
02

- Compute Magnitudes of Each Vector

The magnitude of vector \(\textbf{a} = \langle -3, 5\rangle\) is given by \(\|\textbf{a}\| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}\). The magnitude of vector \(\textbf{b} = (1, 6)\) is \(\|\textbf{b}\| = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}\).
03

- Use the Dot Product and Magnitudes to Find Cosine of the Angle

The cosine of the angle \(\theta\) between the two vectors is given by \(\textbf{a} \cdot \textbf{b} = \|\textbf{a}\| \|\textbf{b}\| \cos \theta\). Therefore, \(\cos \theta = \frac{\textbf{a} \cdot \textbf{b}}{\|\textbf{a}\| \|\textbf{b}\|} = \frac{27}{\sqrt{34} \sqrt{37}} = \frac{27}{\sqrt{1258}}\).
04

- Solve for the Angle \(\theta\)

Use inverse cosine to find \(\theta\): \(\theta = \cos^{-1}(\frac{27}{\sqrt{1258}})\). Using a calculator, \(\theta \approx 42.07 \degree\).

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Key Concepts

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

Dot Product
The dot product is an essential operation when dealing with vectors. It combines two vectors to produce a single scalar number. For two vectors \(\textbf{a} \) and \(\textbf{b}\), the dot product is calculated by multiplying the corresponding components of each vector and then summing these products. For the vectors \(\textbf{a} = \langle-3, 5\rangle\) and \(\textbf{b} = (1, 6)\), performing these operations gives us:
\(\textbf{a} \cdot \textbf{b} = (-3) \times 1 + 5 \times 6 = -3 + 30 = 27\).
Notice how simply multiplying and adding the components provides us with a scalar value of 27. This scalar result is a measure of how much one vector goes in the direction of another.
Magnitude of Vectors
Understanding the magnitude of a vector is crucial in vector mathematics. The magnitude, often seen as the length or size of the vector, is found using the Pythagorean theorem.
For a vector \(\textbf{a} = \langle a_1, a_2 \rangle\), the magnitude is: \[ \|\textbf{a}\| = \sqrt{a_1^2 + a_2^2} \. \]
Specifically, for \(\textbf{a} = \langle-3, 5\rangle\), we get: \[ \|\textbf{a}\| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}\. \]
Similarly, for \(\textbf{b} = (1, 6)\), the magnitude is: \[ \|\textbf{b}\| = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}\. \]
Magnitudes help us understand the vectors' scale and direction.
Inverse Cosine
The inverse cosine function, denoted as \(\text{cos}^{-1}\) or \(\arccos\), is used to find the angle when the cosine value is provided.
To find the smallest positive angle between two vectors using the dot product and magnitudes, we use the formula:
\[ \cos \theta = \frac{\textbf{a} \cdot \textbf{b}}{\|\textbf{a}\| \|\textbf{b}\|} \. \]
With the calculated values of \(\textbf{a} \cdot \textbf{b} = 27\), \(\frac{27}{\sqrt{34} \sqrt{37}} = \approx \frac{27}{35.46}\).
Thus, solving for the angle \(\theta\): \[ \theta = \cos^{-1}(\frac{27}{\sqrt{1258}}) \approx 42.07\degree. \]
The inverse cosine gives the angle corresponding to the calculated cosine value.

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Most popular questions from this chapter

Find \(z_{1} z_{2}\) and \(z_{1} / z_{2}\) for each pair of complex numbers, using trigonometric form. Write the answer in the form \(a+b i\). $$z_{1}=3-4 i, z_{2}=-1+3 i$$

Write each vector as a linear combination of the unit vectors i and \(\mathbf{j}\). $$\langle- 3, \sqrt{2}\rangle$$

Determine whether each pair of vectors is parallel, perpendicular, or neither. $$\langle- 2,3\rangle,\langle 6,4\rangle$$

Explain why the Pythagorean theorem is a special case of the law of cosines.

Given that \(\mathbf{A}=\langle 3,1\rangle\) and \(\mathbf{B}=\langle- 2,3\rangle,\) find the magnitude and direction angle for each of the following vectors. $$\mathbf{B}+\mathbf{A}$$
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