Problem 15
Question
In the following problems, the magnitudes \(A\) and \(B\) of two perpendicular vectors are given. Find the resultant and the angle that it makes with \(B\) $$A=1.25 \quad B=2.07$$
Step-by-Step Solution
Verified Answer
The magnitude of the resultant vector is approximately 2.39, and the angle it makes with vector B is about 31.0 degrees.
1Step 1: Identify the Components
Since the vectors are perpendicular, we can consider one to be along the 'x' axis and the other along the 'y' axis. For the sake of this problem, let's assume vector A is along the x-axis and vector B is along the y-axis.
2Step 2: Calculating the Resultant Vector
The resultant vector (R) can be found using the Pythagorean theorem because the vectors are perpendicular: \[ R = \sqrt{A^2 + B^2} \] where A is the magnitude of the first vector and B is the magnitude of the second vector.
3Step 3: Find the Angle With Vector B
To find the angle (\( \theta \) that the resultant makes with vector B (along the y-axis), we use the tangent function: \[ \tan(\theta) = \frac{A}{B} \] Solve for \( \theta \) using the arctangent function: \[ \theta = \arctan\left(\frac{A}{B}\right) \]
4Step 4: Compute the Result
Insert the given values for A and B into the equations: \[ R = \sqrt{(1.25)^2 + (2.07)^2} \] and \[ \theta = \arctan\left(\frac{1.25}{2.07}\right) \] Calculate the numerical values to get the magnitude of the resultant vector and the angle with respect to vector B.
Key Concepts
Pythagorean TheoremVector ComponentsArctangent FunctionAngle of Vectors
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, particularly useful when working with right triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In mathematical terms, if we label these sides as 'a' and 'b', and the hypotenuse as 'c', the theorem is expressed as:
\[ c^2 = a^2 + b^2 \]
This theorem is incredibly valuable when calculating the resultant of two perpendicular vectors. Vectors, which are quantities having both magnitude and direction, can be represented as directed line segments in either two- or three-dimensional space. When two vectors are perpendicular, akin to the sides 'a' and 'b' of a right triangle, their resultant is analogous to the hypotenuse 'c'. Therefore, to find the magnitude of the resultant vector, we apply the Pythagorean theorem, taking the magnitudes of the two vectors as the sides of a right triangle.
\[ c^2 = a^2 + b^2 \]
This theorem is incredibly valuable when calculating the resultant of two perpendicular vectors. Vectors, which are quantities having both magnitude and direction, can be represented as directed line segments in either two- or three-dimensional space. When two vectors are perpendicular, akin to the sides 'a' and 'b' of a right triangle, their resultant is analogous to the hypotenuse 'c'. Therefore, to find the magnitude of the resultant vector, we apply the Pythagorean theorem, taking the magnitudes of the two vectors as the sides of a right triangle.
Vector Components
Vectors are best understood when broken down into their components. The components of a vector represent its effect in each direction of the chosen coordinate system. For example, in a two-dimensional space, any vector can be decomposed into 'x' (horizontal) and 'y' (vertical) components.
In the context of our exercise, vector components are crucial to finding the resultant vector. Since the vectors A and B are perpendicular to each other, we can treat A as the x-component and B as the y-component. The components make the calculation simpler because they allow us to apply the Pythagorean theorem directly.
Once the components are identified, it's straightforward to calculate the magnitude and direction of the resultant vector through the process of vector addition, where the sum of the individual components in their respective axes gives the resultant vector's corresponding components.
In the context of our exercise, vector components are crucial to finding the resultant vector. Since the vectors A and B are perpendicular to each other, we can treat A as the x-component and B as the y-component. The components make the calculation simpler because they allow us to apply the Pythagorean theorem directly.
Once the components are identified, it's straightforward to calculate the magnitude and direction of the resultant vector through the process of vector addition, where the sum of the individual components in their respective axes gives the resultant vector's corresponding components.
Arctangent Function
The arctangent function, often denoted as 'arctan', is the inverse of the tangent function commonly used in trigonometry. It is the function that provides an angle whose tangent is a given number.
When expressing the relationship between two sides of a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side. Mathematically, for an angle \( \theta \):
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For our problem, the angle of interest is the one that the resultant vector makes with vector B. Since we have the lengths of the opposite (vector A) and the adjacent (vector B), we can find this angle by using the arctangent function:
\[ \theta = \arctan\left(\frac{A}{B}\right) \]
This inverse function allows us to compute the angle when given the ratio of the sides of the triangle formed by the vectors, facilitating the transition from the ratio measurement (tangent) to the angle measurement.
When expressing the relationship between two sides of a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side. Mathematically, for an angle \( \theta \):
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For our problem, the angle of interest is the one that the resultant vector makes with vector B. Since we have the lengths of the opposite (vector A) and the adjacent (vector B), we can find this angle by using the arctangent function:
\[ \theta = \arctan\left(\frac{A}{B}\right) \]
This inverse function allows us to compute the angle when given the ratio of the sides of the triangle formed by the vectors, facilitating the transition from the ratio measurement (tangent) to the angle measurement.
Angle of Vectors
Understanding the angle between vectors is a crucial concept in vector math, as it describes the orientation of one vector with respect to another. In the case of perpendicular vectors, this angle is 90 degrees or \( \frac{\pi}{2} \) radians. However, when determining the angle that a resultant vector makes with one of its components, we must use trigonometry.
For our problem, we seek the angle \( \theta \) between the resultant vector and vector B. After calculating the magnitude of the resultant using the Pythagorean theorem, we turn our attention to finding this angle. As mentioned earlier, this angle can be found using the arctangent function, which computes the angle using the ratio of the two vector magnitudes (A being the opposite side and B the adjacent side in our right triangle).
This angle is significant because it provides insight into the vector's direction, which is as important as its magnitude when dealing with vector quantities in physics and engineering.
For our problem, we seek the angle \( \theta \) between the resultant vector and vector B. After calculating the magnitude of the resultant using the Pythagorean theorem, we turn our attention to finding this angle. As mentioned earlier, this angle can be found using the arctangent function, which computes the angle using the ratio of the two vector magnitudes (A being the opposite side and B the adjacent side in our right triangle).
This angle is significant because it provides insight into the vector's direction, which is as important as its magnitude when dealing with vector quantities in physics and engineering.
Other exercises in this chapter
Problem 14
Evaluate the following, giving your answer in decimal degrees to three significant digits. $$\cos ^{-1} 0.229$$
View solution Problem 14
Sketch each right triangle and find all missing parts. Work to three significant digits and express the angles in decimal degrees. $$a=63.9 \quad b=84.3$$
View solution Problem 15
After leaving port, a ship holds a course \(\mathrm{N} 46^{\circ} 12^{\prime} \mathrm{E}\) for 225 mi. Find how far north and how far east of the port the ship
View solution Problem 15
Evaluate the following, giving your answer in decimal degrees to three significant digits. $$\arctan 4.26$$
View solution