Chapter 1: Problem 6
Find the acute angle between the two vectors \(\mathbf{A}=2 \mathbf{a}_{x}+\mathbf{a}_{y}+3 \mathbf{a}_{z}\) and \(\mathbf{B}=\mathbf{a}_{x}-3 \mathbf{a}_{y}+2 \mathbf{a}_{z}\) by using the definition of \((a)\) the dot product; \((b)\) the cross product.
Short Answer
Expert verified
Answer: The acute angle between the two vectors is approximately \(1.21\) radians or \(69.3^\circ\).
Step by step solution
01
(a) Using the Dot Product
To find the angle between the two vectors using the dot product, we'll follow these steps:
1. Calculate the dot product of the two vectors (\(\mathbf{A} \cdot \mathbf{B}\)).
2. Calculate the magnitudes of both vectors (\(|\mathbf{A}|\) and \(|\mathbf{B}|\)).
3. Use these values and the dot product formula to find the cosine of the angle between the vectors. The formula is \(\cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\)
4. Calculate the angle in radians and degrees.
02
Calculate the Dot Product
The dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: \(\mathbf{A} \cdot \mathbf{B} = (2)(1) + (1)(-3) + (3)(2) = 2 - 3 + 6 = 5\).
03
Calculate the Magnitudes
The magnitudes of vectors \(\mathbf{A}\) and \(\mathbf{B}\) are given by:
\(|\mathbf{A}| = \sqrt{(2)^2 + (1)^2 + (3)^2} = \sqrt{14}\) and \(|\mathbf{B}| = \sqrt{(1)^2 + (-3)^2 + (2)^2} = \sqrt{14}\).
04
Calculate the Cosine of the Angle
Using the formula for the cosine of the angle between the two vectors, we have: $$\cos(\theta) = \frac{5}{\sqrt{14} \sqrt{14}} = \frac{5}{14}$$
05
Calculate the Angle in Radians and Degrees
To find the angle, we take the inverse cosine (also known as arccos) of the value found above: $$\theta = \arccos \left(\frac{5}{14}\right) \approx 1.21 \text{ radians}$$ To convert to degrees, use the formula: $$\theta = \frac{180}{\pi} \times 1.21 \approx 69.3^\circ$$.
06
(b) Using the Cross Product
To find the angle between the two vectors using the cross product, we'll follow these steps:
1. Calculate the cross product of the two vectors (\(\mathbf{A} \times \mathbf{B}\)).
2. Calculate the magnitudes of the cross product vector (\(|\mathbf{A} \times \mathbf{B}|\)).
3. Use these values and the cross product formula to find the sine of the angle between the vectors. The formula is \(\sin\theta = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}||\mathbf{B}|}\)
4. Calculate the angle in radians and degrees using this sine value.
07
Calculate the Cross Product
The cross product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: $$\mathbf{A} \times \mathbf{B} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} (1)(2)-(-3)(3) \\ (3)(1)-(2)(2) \\ (2)(-3)-(1)(1) \end{pmatrix} = \begin{pmatrix} 11 \\ -1 \\ -7 \end{pmatrix} $$
08
Calculate the Magnitude of the Cross Product Vector
The magnitude of the cross product vector is given by: $$|\mathbf{A} \times \mathbf{B}| = \sqrt{(11)^2 + (-1)^2 + (-7)^2} = \sqrt{141}$$
09
Calculate the Sine of the Angle
Using the formula for the sine of the angle between the two vectors, we have: $$\sin(\theta) = \frac{\sqrt{141}}{\sqrt{14} \sqrt{14}} = \frac{\sqrt{141}}{14} $$
10
Calculate the Angle in Radians and Degrees
To find the angle, we take the inverse sine (also known as arcsin) of the value found above: $$\theta' = \arcsin \left(\frac{\sqrt{141}}{14}\right) \approx 1.21 \text{ radians}$$ To convert to degrees, use the formula: $$\theta' = \frac{180}{\pi} \times 1.21 \approx 69.3^\circ$$.
Both methods yield the same result: the acute angle between the two vectors is approximately \(1.21\) radians or \(69.3^\circ\).
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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, also known as the scalar product, is a fundamental concept in vector algebra. It's an easy way to determine the relationship between two vectors as it gives insight into the angle between them. The dot product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is defined as the sum of the products of their corresponding components:
- Formula: \( \mathbf{A} \cdot \mathbf{B} = A_1B_1 + A_2B_2 + A_3B_3 \).
- In this specific exercise, the dot product was calculated as \( 5 \).
- \( \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|} \).
Cross Product
The cross product, unlike the dot product, results in another vector instead of a scalar. It's crucial for finding a vector that is perpendicular to the initial two vectors. The cross product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is calculated by:
- Formula: \( \mathbf{A} \times \mathbf{B} = (A_2B_3 - A_3B_2, A_3B_1 - A_1B_3, A_1B_2 - A_2B_1) \).
- In the given exercise: \( \mathbf{A} \times \mathbf{B} = (11, -1, -7) \).
- \( \sin(\theta) = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}||\mathbf{B}|} \).
Angle Between Vectors
Determining the angle between vectors is an important task in vector algebra. Whether you are using the dot or cross product, the angle between two vectors \( \theta \) can reveal much about their spatial relationship:
- Acute angles indicate vectors pointing more towards each other.
- Obtuse angles indicate vectors diverging away.
Vector Magnitudes
Vector magnitudes give us an idea of the length or size of a vector in space. They are an essential part of vector operations, especially when finding angles or performing normalization:
- The magnitude of a vector \( \mathbf{A} = (A_1, A_2, A_3) \) is calculated using: \( |\mathbf{A}| = \sqrt{A_1^2 + A_2^2 + A_3^2} \).
- For the vectors in the exercise, both \( \mathbf{A} \) and \( \mathbf{B} \) have magnitudes of \( \sqrt{14} \).