Two vectors having equal magnitudes A make an angle `theta` with each other. Find the magnitude and direction of the resultant.

Options

A. `2A cos"" (theta)/(2),` along bisector
B. `A cos"" (theta)/(2),"at" 45^(@)` from on a vector
C. `2A cos"" (theta)/(2),` along bisector
D. `A cos"" (theta)/(2),` along bisector

Explanation

To find the magnitude and direction of the resultant of two vectors of equal magnitude \( A \) that make an angle \( \theta \) with each other, we can follow these steps:

### Step 1: Understand the Vectors
Let the two vectors be represented as \( \vec{A_1} \) and \( \vec{A_2} \), both having a magnitude \( A \) and an angle \( \theta \) between them.

### Step 2: Use the Law of Cosines
The magnitude of the resultant vector \( R \) can be calculated using the formula:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos(\theta)} \]
This simplifies to:
\[ R = \sqrt{2A^2(1 + \cos(\theta))} \]

### Step 3: Simplify the Resultant Magnitude
Factoring out \( A^2 \):
\[ R = \sqrt{2A^2(1 + \cos(\theta))} = A\sqrt{2(1 + \cos(\theta))} \]

### Step 4: Use the Cosine Identity
Using the identity \( 1 + \cos(\theta) = 2\cos^2(\frac{\theta}{2}) \):
\[ R = A\sqrt{2 \cdot 2\cos^2(\frac{\theta}{2})} = A \cdot 2 \cdot \cos(\frac{\theta}{2}) = 2A \cos(\frac{\theta}{2}) \]

### Step 5: Find the Direction of the Resultant
To find the direction of the resultant vector, we can use the formula for the angle \( \alpha \) that the resultant makes with one of the vectors. The tangent of this angle can be given by:
\[ \tan(\alpha) = \frac{A \sin(\theta)}{A + A \cos(\theta)} = \frac{\sin(\theta)}{1 + \cos(\theta)} \]

### Step 6: Use the Sine and Cosine Identity
Using the identity \( \sin(\theta) = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}) \) and \( 1 + \cos(\theta) = 2\cos^2(\frac{\theta}{2}) \):
\[ \tan(\alpha) = \frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{2\cos^2(\frac{\theta}{2})} = \tan(\frac{\theta}{2}) \]

### Step 7: Conclude the Direction
Thus, we find that:
\[ \alpha = \frac{\theta}{2} \]

### Final Result
The magnitude of the resultant vector \( R \) is:
\[ R = 2A \cos\left(\frac{\theta}{2}\right) \]
And the direction of the resultant vector is:
\[ \alpha = \frac{\theta}{2} \]