.. (a) Find the scalar product of the vectors A and B given in Exercise 1.360. (b) Find the angle between these two vectors. 1.40 A = 4.00 +7.00 and B-5.00 - 2.00j,
.. (a) Find the scalar product of the vectors A and B given in Exercise 1.360. (b) Find the angle between these two vectors. 1.40 A = 4.00 +7.00 and B-5.00 - 2.00j,
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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1.40 explain me easily way please
![### Problem 1.40
**(a) Find the scalar product of the vectors** \(\mathbf{A}\) **and** \(\mathbf{B}\) **given in Exercise 1.36 (b). Find the angle between these two vectors.**
Given vectors:
\[ \mathbf{A} = 4.00 \hat{i} + 7.00 \hat{j} \]
\[ \mathbf{B} = 5.00 \hat{i} - 2.00 \hat{j} \]
#### Explanation of Diagram
The diagram provided appears to illustrate vectors \(\mathbf{A}\) and \(\mathbf{B}\) in a two-dimensional coordinate system. The x- and y-axes are labeled \(x\) and \(y\) respectively.
Vector \(\mathbf{A}\) is shown extending from the origin to a point (4.00, 7.00) in the first quadrant, while Vector \(\mathbf{B}\) extends from origin to a point (5.00, -2.00), which places it in the fourth quadrant. The angle between the vectors, \(\theta\), is represented but not directly labeled with a specific value.
### Steps to Solve:
1. **Find the Scalar (Dot) Product**:
The scalar product (dot product) of two vectors \(\mathbf{A} \cdot \mathbf{B}\) is calculated as follows:
\[
\mathbf{A \cdot B} = (A_x \cdot B_x) + (A_y \cdot B_y)
\]
where \(A_x\) and \(A_y\) are the components of vector \(\mathbf{A}\), and \(B_x\) and \(B_y\) are the components of vector \(\mathbf{B}\).
2. **Calculate the Magnitudes of the Vectors**:
The magnitude of vector \(\mathbf{A}\) is:
\[
|\mathbf{A}| = \sqrt{A_x^2 + A_y^2}
\]
Similarly, the magnitude of vector \(\mathbf{B}\) is:
\[
|\mathbf{B}| = \sqrt{B_x^2 + B_y^2}
\]
3. **Find the Angle Between the Vectors**:
The angle \(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2Fa3bbaf55-6b77-445d-aa7b-08eaee226573%2Fjt4hq3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 1.40
**(a) Find the scalar product of the vectors** \(\mathbf{A}\) **and** \(\mathbf{B}\) **given in Exercise 1.36 (b). Find the angle between these two vectors.**
Given vectors:
\[ \mathbf{A} = 4.00 \hat{i} + 7.00 \hat{j} \]
\[ \mathbf{B} = 5.00 \hat{i} - 2.00 \hat{j} \]
#### Explanation of Diagram
The diagram provided appears to illustrate vectors \(\mathbf{A}\) and \(\mathbf{B}\) in a two-dimensional coordinate system. The x- and y-axes are labeled \(x\) and \(y\) respectively.
Vector \(\mathbf{A}\) is shown extending from the origin to a point (4.00, 7.00) in the first quadrant, while Vector \(\mathbf{B}\) extends from origin to a point (5.00, -2.00), which places it in the fourth quadrant. The angle between the vectors, \(\theta\), is represented but not directly labeled with a specific value.
### Steps to Solve:
1. **Find the Scalar (Dot) Product**:
The scalar product (dot product) of two vectors \(\mathbf{A} \cdot \mathbf{B}\) is calculated as follows:
\[
\mathbf{A \cdot B} = (A_x \cdot B_x) + (A_y \cdot B_y)
\]
where \(A_x\) and \(A_y\) are the components of vector \(\mathbf{A}\), and \(B_x\) and \(B_y\) are the components of vector \(\mathbf{B}\).
2. **Calculate the Magnitudes of the Vectors**:
The magnitude of vector \(\mathbf{A}\) is:
\[
|\mathbf{A}| = \sqrt{A_x^2 + A_y^2}
\]
Similarly, the magnitude of vector \(\mathbf{B}\) is:
\[
|\mathbf{B}| = \sqrt{B_x^2 + B_y^2}
\]
3. **Find the Angle Between the Vectors**:
The angle \(\
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