The components of a vector along the x- and y- directions are `(n+1)` and 1, respectively. If the coordinate system is rotated by an angle `theta=60^(@)`, then the components change to `n` and 3. The value of `n` is

Options

A. a. 2
B. b. `cos 60^(@)`
C. c. `sin 60^(@)`
D. d. 3.5

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial components of the vector
The initial components of the vector along the x- and y-directions are given as:
- \( V_x = n + 1 \)
- \( V_y = 1 \)

### Step 2: Write the expression for the magnitude of the initial vector
The magnitude of the vector can be calculated using the Pythagorean theorem:
\[ |V| = \sqrt{(V_x)^2 + (V_y)^2} = \sqrt{(n + 1)^2 + 1^2} \]

### Step 3: Identify the new components after rotation
After rotating the coordinate system by an angle \( \theta = 60^\circ \), the new components of the vector are:
- \( V_x' = n \)
- \( V_y' = 3 \)

### Step 4: Write the expression for the magnitude of the new vector
The magnitude of the new vector can also be expressed as:
\[ |V'| = \sqrt{(V_x')^2 + (V_y')^2} = \sqrt{n^2 + 3^2} \]

### Step 5: Set the magnitudes equal to each other
Since the length of the vector does not change with the rotation of the coordinate axes, we can set the two magnitudes equal:
\[ \sqrt{(n + 1)^2 + 1^2} = \sqrt{n^2 + 3^2} \]

### Step 6: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (n + 1)^2 + 1^2 = n^2 + 3^2 \]

### Step 7: Expand both sides
Expanding both sides results in:
\[ (n^2 + 2n + 1) + 1 = n^2 + 9 \]
This simplifies to:
\[ n^2 + 2n + 2 = n^2 + 9 \]

### Step 8: Simplify the equation
Subtract \( n^2 \) from both sides:
\[ 2n + 2 = 9 \]

### Step 9: Solve for \( n \)
Subtract 2 from both sides:
\[ 2n = 7 \]
Now divide by 2:
\[ n = \frac{7}{2} = 3.5 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{3.5} \]

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