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Chapter 1: Problem 13

For the following exercises, find the measure of the angle between the three- dimensional vectors a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly. $$ \mathbf{a}=\langle 3,-1,2\rangle, \mathbf{b}=\langle 1,-1,-2\rangle $$

Short Answer

Expert verified
The angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{2} \) radians.

Step by step solution

01

Dot Product of Vectors

To find the measure of the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), we first calculate their dot product. The dot product is given by \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]Substituting the given vector components:\[ 3 \times 1 + (-1) \times (-1) + 2 \times (-2) = 3 + 1 - 4 = 0 \]
02

Calculate Magnitudes of Vectors

Next, compute the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \). The magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is:\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]For \( \mathbf{a} \):\[ \|\mathbf{a}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]For \( \mathbf{b} \):\[ \|\mathbf{b}\| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
03

Calculating Cosine of the Angle

Use the dot product and magnitudes to find the cosine of the angle \( \theta \) between the vectors:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \times \|\mathbf{b}\|} \]Substitute the values:\[ \cos \theta = \frac{0}{\sqrt{14} \times \sqrt{6}} = 0 \]
04

Calculate the Angle

The angle \( \theta \) whose cosine is 0 is \( \frac{\pi}{2} \) radians, as the cosine of \( \frac{\pi}{2} \) is exactly zero. Thus, the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{2} \) radians.

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Key Concepts

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

Understanding the Dot Product
The dot product is a fundamental operation when working with vectors. It provides insights into the angle and relationship between two vectors. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is calculated as: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \] It's a scalar, meaning the result is a single number. This result can inform you about the vectors' directions:
  • If the dot product is positive, the angle between the vectors is acute.
  • If it is zero, the vectors are perpendicular or orthogonal.
  • A negative result indicates an obtuse angle.
In our exercise, with \( \mathbf{a} \cdot \mathbf{b} = 0 \), the vectors are orthogonal, confirming a right angle.
Calculating Vector Magnitude
A vector's magnitude indicates its length in space. To compute the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), we use: \[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
  • Magnitude of \( \mathbf{a} \): For vector \( \mathbf{a} = \langle 3, -1, 2 \rangle \), the magnitude is \( \|\mathbf{a}\| = \sqrt{14} \).
  • Magnitude of \( \mathbf{b} \): The vector \( \mathbf{b} = \langle 1, -1, -2 \rangle \) has a magnitude of \( \|\mathbf{b}\| = \sqrt{6} \).
Understanding magnitudes helps when calculating angles between vectors since they normalize the space direction by outlining its "length."
Cosine of the Angle Between Vectors
Once you have the dot product and magnitudes, you can compute the cosine of the angle \( \theta \) between vectors using this formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \times \|\mathbf{b}\|} \] This formula relates vector lengths and the dot product, allowing you to find \( \theta \). It's equivalent to defining how the direction of one vector aligns with another in terms of the angle. For our vectors, since the dot product is zero: \[ \cos \theta = \frac{0}{\sqrt{14} \times \sqrt{6}} = 0 \] Thus, \( \theta \) corresponds to an angle where cosine equals zero, namely \( \frac{\pi}{2} \).
Using Radian Measure
Radian measure is an essential concept in trigonometry and calculus for quantifying angles. It's based on the radius of a circle. One full rotation around a circle is \( 2\pi \) radians, equating to 360 degrees. This measurement simplifies mathematical formulas and operations involving angles. To convert angles from radians to degrees, use: \[ \text{Degrees} = \text{Radians} \times \frac{180}{\pi} \] In the given exercise, the angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{2} \) radians. This quantity signifies a 90-degree angle, showing the vectors are perpendicular.

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