Two vectors `vec(a)` and `vec(b)` are at an angle of `60^(@)` with each other . Their resultant makes an angle of `45^(@)` with `vec(a)` If `|vec(b)|=2`unit , then `|vec(a)|` is

Options

A. `sqrt(3)`
B. `sqrt(3)-1`
C. `sqrt(3)+1`
D. `sqrt(3)//2`

Explanation

To solve the problem step by step, we will use the information given about the vectors and their angles.

### Step 1: Understand the given information
We have two vectors, \( \vec{a} \) and \( \vec{b} \), with the following properties:
- The angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \).
- The magnitude of \( \vec{b} \) is \( | \vec{b} | = 2 \) units.
- The resultant vector \( \vec{R} \) makes an angle of \( 45^\circ \) with \( \vec{a} \).

### Step 2: Use the formula for the resultant of two vectors
The magnitude of the resultant \( \vec{R} \) of two vectors \( \vec{a} \) and \( \vec{b} \) can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos(60^\circ)} \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we can simplify this to:
\[ |\vec{R}| = \sqrt{|\vec{a}|^2 + 2^2 + 2 |\vec{a}| \cdot 2 \cdot \frac{1}{2}} \]
This simplifies to:
\[ |\vec{R}| = \sqrt{|\vec{a}|^2 + 4 + 2 |\vec{a}|} \]

### Step 3: Use the angle of the resultant
The angle \( \theta \) that the resultant makes with \( \vec{a} \) can be expressed using the tangent function:
\[ \tan(45^\circ) = \frac{|\vec{b}| \sin(60^\circ)}{|\vec{a}| + |\vec{b}| \cos(60^\circ)} \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ 1 = \frac{2 \cdot \frac{\sqrt{3}}{2}}{|\vec{a}| + 2 \cdot \frac{1}{2}} \]
This simplifies to:
\[ 1 = \frac{\sqrt{3}}{|\vec{a}| + 1} \]

### Step 4: Solve for \( |\vec{a}| \)
Cross-multiplying gives us:
\[ |\vec{a}| + 1 = \sqrt{3} \]
Thus, we can isolate \( |\vec{a}| \):
\[ |\vec{a}| = \sqrt{3} - 1 \]

### Conclusion
The magnitude of vector \( \vec{a} \) is:
\[ |\vec{a}| = \sqrt{3} - 1 \]