Math Problem Statement
Solution
The problem you shared asks to find the magnitude and angle of vector B in the vector equation:
Given Data:
- Magnitude of = 13.8 m, and its angle = 36.7° (counterclockwise from the +x direction)
- Magnitude of = 13.7 m, and its angle = 24.7° (counterclockwise from the -x direction)
We need to:
- Find the magnitude of .
- Find the angle of .
Solution Approach:
We will break and into their components and use the vector equation to solve for .
Step 1: Resolve vectors and into components
-
For , the angle is given as 36.7° from the +x axis: Substituting :
-
For , the angle is given as 24.7° counterclockwise from the -x axis, so its angle relative to the +x axis is 180° - 24.7° = 155.3°: Substituting :
Step 2: Use the vector sum equation
The components of are found using:
Step 3: Find the magnitude and angle of
- Magnitude of :
- Angle of :
I'll now calculate the values.The magnitude and angle of vector are as follows:
- Magnitude of :
- Angle of : (relative to the +x axis)
Since the problem asks for the angle as a positive number, we can express the angle as:
Summary:
- (a) Magnitude of :
- (b) Angle of :
Would you like further clarification or details on the steps?
Here are some related questions to expand your understanding:
- How are vector components related to their magnitudes and angles?
- Why do we resolve vectors into x and y components to solve vector addition problems?
- What happens to the angle when a vector points in the negative x-direction?
- Can you use the Pythagorean theorem for vector addition in all cases?
- How would this process change if the vectors were in 3D space?
Tip: Always double-check angle references (from +x or -x) when resolving vector directions.
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Math Problem Analysis
Mathematical Concepts
Vector Addition
Trigonometry
Pythagorean Theorem
Inverse Tangent
Formulas
A_x = A * cos(θ)
A_y = A * sin(θ)
C_x = C * cos(θ)
C_y = C * sin(θ)
B_x = C_x - A_x
B_y = C_y - A_y
B = sqrt(B_x^2 + B_y^2)
θ_B = tan^{-1}(B_y / B_x)
Theorems
Pythagorean Theorem
Trigonometric Identities
Suitable Grade Level
College/University Level
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