Math Problem Statement

In the sum \(\vec{A} + \vec{B} = \vec{C}\), vector \(\vec{A}\) has a magnitude of 13.8 m and is angled 36.7° counterclockwise from the +x direction, and vector \(\vec{C}\) has a magnitude of 13.7 m and is angled 24.7° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of \(\vec{B}\)? State your angle as a positive number.

Solution

The problem you shared asks to find the magnitude and angle of vector B in the vector equation:

A+B=C\vec{A} + \vec{B} = \vec{C}

Given Data:

  • Magnitude of A\vec{A} = 13.8 m, and its angle = 36.7° (counterclockwise from the +x direction)
  • Magnitude of C\vec{C} = 13.7 m, and its angle = 24.7° (counterclockwise from the -x direction)

We need to:

  1. Find the magnitude of B\vec{B}.
  2. Find the angle of B\vec{B}.

Solution Approach:

We will break A\vec{A} and C\vec{C} into their components and use the vector equation to solve for B\vec{B}.

Step 1: Resolve vectors A\vec{A} and C\vec{C} into components

  • For A\vec{A}, the angle is given as 36.7° from the +x axis: Ax=Acos(36.7)A_x = A \cdot \cos(36.7^\circ) Ay=Asin(36.7)A_y = A \cdot \sin(36.7^\circ) Substituting A=13.8mA = 13.8 \, \text{m}: Ax=13.8cos(36.7)A_x = 13.8 \cdot \cos(36.7^\circ) Ay=13.8sin(36.7)A_y = 13.8 \cdot \sin(36.7^\circ)

  • For C\vec{C}, 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°: Cx=Ccos(155.3)C_x = C \cdot \cos(155.3^\circ) Cy=Csin(155.3)C_y = C \cdot \sin(155.3^\circ) Substituting C=13.7mC = 13.7 \, \text{m}: Cx=13.7cos(155.3)C_x = 13.7 \cdot \cos(155.3^\circ) Cy=13.7sin(155.3)C_y = 13.7 \cdot \sin(155.3^\circ)

Step 2: Use the vector sum equation

The components of B\vec{B} are found using:

Bx=CxAxB_x = C_x - A_x By=CyAyB_y = C_y - A_y

Step 3: Find the magnitude and angle of B\vec{B}

  • Magnitude of B\vec{B}: B=Bx2+By2B = \sqrt{B_x^2 + B_y^2}
  • Angle of B\vec{B}: θB=tan1(ByBx)\theta_B = \tan^{-1}\left(\frac{B_y}{B_x}\right)

I'll now calculate the values.The magnitude and angle of vector B\vec{B} are as follows:

  • Magnitude of B\vec{B}: 23.65m23.65 \, \text{m}
  • Angle of B\vec{B}: 173.88-173.88^\circ (relative to the +x axis)

Since the problem asks for the angle as a positive number, we can express the angle as:

θB=360173.88=186.12\theta_B = 360^\circ - 173.88^\circ = 186.12^\circ

Summary:

  • (a) Magnitude of B\vec{B}: 23.65m23.65 \, \text{m}
  • (b) Angle of B\vec{B}: 186.12186.12^\circ

Would you like further clarification or details on the steps?

Here are some related questions to expand your understanding:

  1. How are vector components related to their magnitudes and angles?
  2. Why do we resolve vectors into x and y components to solve vector addition problems?
  3. What happens to the angle when a vector points in the negative x-direction?
  4. Can you use the Pythagorean theorem for vector addition in all cases?
  5. 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