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Magazine Chapter 1: Problem 47
Find the angle between each of the following pairs of vectors: $$(a)\vec{A}=-2.00 \hat{\imath}+6.00 \hat{J} \quad\( and \)\quad \vec{B}=2.00 \hat{\imath}-3.00 \hat{J}$$ $$(b) \vec{A}=3.00 \hat{\imath}+5.00 \hat{J} \quad\( and \)\quad \vec{B}=10.00 \hat{\imath}+6.00 \hat{J}$$ $$(c)\vec{A}=-4.00 \hat{\imath}+2.00 \hat{J} \quad\( and \)\quad \vec{B}=7.00 \hat{\imath}+14.00 \hat{J}$$
Short Answer
Expert verified
For (a), find \( \theta \) using the equation; for (b), find \( \theta \) with similar calculations; for (c), \( \theta = 90^{\circ} \).
Step by step solution
01
Understand the Dot Product Formula for Vectors
The dot product of two vectors \( \vec{A} = A_x \hat{\imath} + A_y \hat{J} \) and \( \vec{B} = B_x \hat{\imath} + B_y \hat{J} \) is calculated as \( \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y \). The dot product can also be expressed as \( \vec{A} \cdot \vec{B} = |A||B|\cos(\theta) \), where \( \theta \) is the angle between the vectors.
02
Use Dot Product to Find the Angle Between the Vectors
The angle between two vectors can be found using the formula \( \theta = \cos^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{|A||B|}\right) \). This formula derives from equating the dot product definition \( \vec{A} \cdot \vec{B} \) in two ways: component-wise and in terms of the vectors' magnitudes and angle.
03
Step 3(a): Calculate Dot Product for Vectors \( \vec{A} \) and \( \vec{B} \)
For (a), \( \vec{A} = -2.00 \hat{\imath} + 6.00 \hat{J} \) and \( \vec{B} = 2.00 \hat{\imath} - 3.00 \hat{J} \). The dot product \( \vec{A} \cdot \vec{B} = (-2.00)(2.00) + (6.00)(-3.00) = -4.00 - 18.00 = -22.00 \).
04
Step 4(a): Compute Magnitudes of Vectors
The magnitude of \( \vec{A} \) is \( |A| = \sqrt{(-2.00)^2 + (6.00)^2} = \sqrt{4.00 + 36.00} = \sqrt{40.00} \). \( |B| = \sqrt{(2.00)^2 + (-3.00)^2} = \sqrt{4.00 + 9.00} = \sqrt{13.00} \).
05
Step 5(a): Determine the Angle \( \theta \)
Substitute the dot product and magnitudes into the formula: \( \theta = \cos^{-1}\left(\frac{-22.00}{\sqrt{40.00}\cdot\sqrt{13.00}}\right) \). Calculate \( \theta \) using a calculator.
06
Step 3(b): Calculate Dot Product for Vectors \( \vec{A} \) and \( \vec{B} \)
For (b), \( \vec{A} = 3.00 \hat{\imath} + 5.00 \hat{J} \) and \( \vec{B} = 10.00 \hat{\imath} + 6.00 \hat{J} \). The dot product \( \vec{A} \cdot \vec{B} = (3.00)(10.00) + (5.00)(6.00) = 30.00 + 30.00 = 60.00 \).
07
Step 4(b): Compute Magnitudes of Vectors
The magnitude of \( \vec{A} \) is \( |A| = \sqrt{(3.00)^2 + (5.00)^2} = \sqrt{9.00 + 25.00} = \sqrt{34.00} \). \( |B| = \sqrt{(10.00)^2 + (6.00)^2} = \sqrt{100.00 + 36.00} = \sqrt{136.00} \).
08
Step 5(b): Determine the Angle \( \theta \)
Substitute the dot product and magnitudes into the formula: \( \theta = \cos^{-1}\left(\frac{60.00}{\sqrt{34.00}\cdot\sqrt{136.00}}\right) \). Calculate \( \theta \) using a calculator.
09
Step 3(c): Calculate Dot Product for Vectors \( \vec{A} \) and \( \vec{B} \)
For (c), \( \vec{A} = -4.00 \hat{\imath} + 2.00 \hat{J} \) and \( \vec{B} = 7.00 \hat{\imath} + 14.00 \hat{J} \). The dot product \( \vec{A} \cdot \vec{B} = (-4.00)(7.00) + (2.00)(14.00) = -28.00 + 28.00 = 0.00 \).
10
Step 4(c): Compute Magnitudes of Vectors
The magnitude of \( \vec{A} \) is \( |A| = \sqrt{(-4.00)^2 + (2.00)^2} = \sqrt{16.00 + 4.00} = \sqrt{20.00} \). \( |B| = \sqrt{(7.00)^2 + (14.00)^2} = \sqrt{49.00 + 196.00} = \sqrt{245.00} \).
11
Step 5(c): Determine the Angle \( \theta \)
Since the dot product is zero, \( \theta = \cos^{-1}(0) = 90.00 \text{ degrees} \). This indicates the vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular.
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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 a valuable tool for working with vectors. For two vectors, say \( \vec{A} = A_x \hat{\imath} + A_y \hat{J} \) and \( \vec{B} = B_x \hat{\imath} + B_y \hat{J} \), the dot product is calculated as the sum of the products of their respective components: \( \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y \). This simple multiplication and addition operation helps to find how much one vector extends in the direction of another.
It is not just a mere sum or multiplication; the result gives insight into the angle between two vectors. If the vectors are aligned perfectly, the dot product will be the highest, a positive maximum. If they are perpendicular, or 90 degrees apart, the dot product will be zero, indicating no projection. For vectors in opposite directions, the dot product will be negative.
This makes the dot product particularly useful in physics and engineering when you need to understand directional relationships.
It is not just a mere sum or multiplication; the result gives insight into the angle between two vectors. If the vectors are aligned perfectly, the dot product will be the highest, a positive maximum. If they are perpendicular, or 90 degrees apart, the dot product will be zero, indicating no projection. For vectors in opposite directions, the dot product will be negative.
This makes the dot product particularly useful in physics and engineering when you need to understand directional relationships.
Magnitude of a Vector
Understanding the magnitude of a vector is crucial because it tells us about a vector's length or size. For a vector \( \vec{A} = A_x \hat{\imath} + A_y \hat{J} \), its magnitude \( |\vec{A}| \) is given by \( |\vec{A}| = \sqrt{A_x^2 + A_y^2} \). This is simply the length calculated through the Pythagorean theorem and is very akin to finding the hypotenuse of a right triangle with base \( A_x \) and height \( A_y \).
The magnitude is always non-negative since it represents a distance in geometry and physical contexts such as displacement and force.
For any vector calculations, knowing the magnitude helps in normalizing vectors, which involves scaling them to unit length without changing their direction. This is often done for simplifying computations and analyses in physics and computer graphics.
The magnitude is always non-negative since it represents a distance in geometry and physical contexts such as displacement and force.
For any vector calculations, knowing the magnitude helps in normalizing vectors, which involves scaling them to unit length without changing their direction. This is often done for simplifying computations and analyses in physics and computer graphics.
Perpendicular Vectors
Vectors are considered perpendicular, or orthogonal, if the angle between them is 90 degrees. A unique property of perpendicular vectors is that their dot product is zero, as seen in instance (c) of the original exercise. This is because the cosine of 90 degrees is zero, leading the dot product formula \( \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos(\theta) \) to equal zero.
When working with vectors, identifying perpendicularity can simplify many problems, including those involving projection or decomposition of vectors into components.
In both mathematics and physics, where vectors frequently represent forces or directions, understanding their perpendicular relationships is essential for solving equilibrium equations and optimizing paths.
When working with vectors, identifying perpendicularity can simplify many problems, including those involving projection or decomposition of vectors into components.
In both mathematics and physics, where vectors frequently represent forces or directions, understanding their perpendicular relationships is essential for solving equilibrium equations and optimizing paths.
