Math Problem Statement

Find vector C⃗ such that A⃗ +B⃗ +C⃗ =0⃗ . Write your answer in component form. Express your answer in meters in terms of the unit vectors i^ and j^ . Use the 'unit vector' button to denote unit vectors in your answer.

Find vector C⃗ such that A⃗ +B⃗ +C⃗ =0⃗ . Write your answer in component form. Express your answer in meters in terms of the unit vectors i^ and j^.

Solution

We are tasked with finding the vector C\vec{C} such that:

A+B+C=0\vec{A} + \vec{B} + \vec{C} = \vec{0}

To do this, we'll first determine the components of vectors A\vec{A} and B\vec{B}, and then solve for C\vec{C}.

Step 1: Resolve vector A\vec{A}

From the image, vector A\vec{A} has a magnitude of 4 m and is at an angle of 60° above the positive x-axis. To find the components of A\vec{A}:

Ax=Acos(60)=4cos(60)=40.5=2mA_x = A \cdot \cos(60^\circ) = 4 \cdot \cos(60^\circ) = 4 \cdot 0.5 = 2 \, \text{m} Ay=Asin(60)=4sin(60)=432=23m3.46mA_y = A \cdot \sin(60^\circ) = 4 \cdot \sin(60^\circ) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m} \approx 3.46 \, \text{m}

Thus, the components of A\vec{A} are:

A=2i^+3.46j^\vec{A} = 2 \hat{i} + 3.46 \hat{j}

Step 2: Resolve vector B\vec{B}

Vector B\vec{B} has a magnitude of 3 m and is at an angle of 20° below the positive x-axis. To find the components of B\vec{B}:

Bx=Bcos(20)=3cos(20)30.94=2.82mB_x = B \cdot \cos(20^\circ) = 3 \cdot \cos(20^\circ) \approx 3 \cdot 0.94 = 2.82 \, \text{m} By=Bsin(20)=3sin(20)30.34=1.03mB_y = B \cdot \sin(20^\circ) = 3 \cdot \sin(20^\circ) \approx 3 \cdot 0.34 = 1.03 \, \text{m}

Since the y-component of B\vec{B} is pointing downward (negative y-direction), we take:

By=1.03mB_y = -1.03 \, \text{m}

Thus, the components of B\vec{B} are:

B=2.82i^1.03j^\vec{B} = 2.82 \hat{i} - 1.03 \hat{j}

Step 3: Solve for vector C\vec{C}

Using the equation A+B+C=0\vec{A} + \vec{B} + \vec{C} = \vec{0}, we can solve for C\vec{C}:

C=(A+B)\vec{C} = -(\vec{A} + \vec{B})

First, sum the components of A\vec{A} and B\vec{B}:

A+B=(2+2.82)i^+(3.461.03)j^\vec{A} + \vec{B} = (2 + 2.82)\hat{i} + (3.46 - 1.03)\hat{j} A+B=4.82i^+2.43j^\vec{A} + \vec{B} = 4.82 \hat{i} + 2.43 \hat{j}

Now, the components of C\vec{C} are:

C=(4.82i^+2.43j^)\vec{C} = -(4.82 \hat{i} + 2.43 \hat{j}) C=4.82i^2.43j^\vec{C} = -4.82 \hat{i} - 2.43 \hat{j}

Final Answer:

The vector C\vec{C} in component form is:

C=4.82i^2.43j^m\vec{C} = -4.82 \hat{i} - 2.43 \hat{j} \, \text{m}

Would you like further details or explanations?

Here are five related questions to expand on this topic:

  1. How would the solution change if vector A\vec{A} had a different angle?
  2. How can you calculate the magnitude of vector C\vec{C}?
  3. What is the geometric interpretation of the sum of vectors A+B+C=0\vec{A} + \vec{B} + \vec{C} = 0?
  4. How do you apply vector addition in 3D space?
  5. What are the alternative methods for resolving vectors into components?

Tip: Always pay attention to the direction (sign) of vector components when resolving them along axes!

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Unit Vectors

Formulas

A_x = A * cos(θ)
A_y = A * sin(θ)
B_x = B * cos(θ)
B_y = B * sin(θ)
C⃗ = -(A⃗ + B⃗)

Theorems

Vector Resolution into Components
Trigonometric Ratios

Suitable Grade Level

Grades 10-12