What is the angle between two vector forces of equal magnitude such that their resultant is one-third of either of the original forces?

Options

A. `cos^(-1)(-17/18)`
B. `cos^(-1)(-1/3)`
C. `45^(@)`
D. `120^(@)`

Explanation

To find the angle between two vector forces of equal magnitude such that their resultant is one-third of either of the original forces, we can follow these steps:

### Step 1: Define the Forces
Let the magnitude of each force be \( A \). Therefore, we have:
- Force 1: \( \vec{F_1} = A \)
- Force 2: \( \vec{F_2} = A \)

### Step 2: Define the Resultant Force
According to the problem, the resultant force \( \vec{R} \) is given as:
\[ R = \frac{A}{3} \]

### Step 3: Use the Formula for Resultant of Two Forces
The magnitude of the resultant \( R \) of two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) is given by the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]

### Step 4: Substitute the Values
Substituting \( F_1 = A \), \( F_2 = A \), and \( R = \frac{A}{3} \) into the formula gives:
\[ \frac{A}{3} = \sqrt{A^2 + A^2 + 2A \cdot A \cos \theta} \]
This simplifies to:
\[ \frac{A}{3} = \sqrt{2A^2 + 2A^2 \cos \theta} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root yields:
\[ \left(\frac{A}{3}\right)^2 = 2A^2(1 + \cos \theta) \]
This simplifies to:
\[ \frac{A^2}{9} = 2A^2(1 + \cos \theta) \]

### Step 6: Divide by \( A^2 \)
Assuming \( A \neq 0 \), we can divide both sides by \( A^2 \):
\[ \frac{1}{9} = 2(1 + \cos \theta) \]

### Step 7: Solve for \( \cos \theta \)
Rearranging gives:
\[ 1 + \cos \theta = \frac{1}{18} \]
Thus,
\[ \cos \theta = \frac{1}{18} - 1 = -\frac{17}{18} \]

### Step 8: Find the Angle \( \theta \)
Now, we can find \( \theta \) using the inverse cosine function:
\[ \theta = \cos^{-1}\left(-\frac{17}{18}\right) \]

### Final Answer
The angle \( \theta \) between the two vector forces is:
\[ \theta = \cos^{-1}\left(-\frac{17}{18}\right) \]

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