If a makes the same angle `alpha`, with the lines `vec( r )=lambda(hati+hatj+hatk),vec( r )=mu hati` and `vec( r )=v hatj` then `sec^(2)alpha` is equal to

Options

A. `9+4sqrt(3)`
B. `4+9sqrt(3)`
C. `9-4sqrt(3)`
D. none of these

Answer

To solve the problem, we need to find the value of \( \sec^2 \alpha \) given that a vector \( \vec{A} \) makes the same angle \( \alpha \) with the lines represented by the vectors \( \vec{r_1} = \lambda (\hat{i} + \hat{j} + \hat{k}) \), \( \vec{r_2} = \mu \hat{i} \), and \( \vec{r_3} = v \hat{j} \).

### Step 1: Identify Direction Vectors
The direction vectors for the lines are:
- For \( \vec{r_1} \): \( \vec{d_1} = \hat{i} + \hat{j} + \hat{k} \)
- For \( \vec{r_2} \): \( \vec{d_2} = \hat{i} \)
- For \( \vec{r_3} \): \( \vec{d_3} = \hat{j} \)

### Step 2: Calculate Magnitude of Direction Vectors
The magnitudes of the direction vectors are:
- \( |\vec{d_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \)
- \( |\vec{d_2}| = 1 \)
- \( |\vec{d_3}| = 1 \)

### Step 3: Set Up Dot Products
Since \( \vec{A} \) makes the same angle \( \alpha \) with each direction vector, we can express the dot products as follows:
1. \( \vec{A} \cdot \vec{d_1} = |\vec{A}| |\vec{d_1}| \cos \alpha \)
2. \( \vec{A} \cdot \vec{d_2} = |\vec{A}| |\vec{d_2}| \cos \alpha \)
3. \( \vec{A} \cdot \vec{d_3} = |\vec{A}| |\vec{d_3}| \cos \alpha \)

### Step 4: Express Dot Products
Let \( \vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \).
- For \( \vec{d_1} \):
\[ \vec{A} \cdot \vec{d_1} = a_1 + a_2 + a_3 = |\vec{A}| \sqrt{3} \cos \alpha \]
- For \( \vec{d_2} \):
\[ \vec{A} \cdot \vec{d_2} = a_1 = |\vec{A}| \cos \alpha \]
- For \( \vec{d_3} \):
\[ \vec{A} \cdot \vec{d_3} = a_2 = |\vec{A}| \cos \alpha \]

### Step 5: Relate the Components
From the equations:
1. \( a_1 + a_2 + a_3 = |\vec{A}| \sqrt{3} \cos \alpha \)
2. \( a_1 = |\vec{A}| \cos \alpha \)
3. \( a_2 = |\vec{A}| \cos \alpha \)

Substituting \( a_1 \) and \( a_2 \) into the first equation:
\[ |\vec{A}| \cos \alpha + |\vec{A}| \cos \alpha + a_3 = |\vec{A}| \sqrt{3} \cos \alpha \]
This simplifies to:
\[ 2 |\vec{A}| \cos \alpha + a_3 = |\vec{A}| \sqrt{3} \cos \alpha \]
Thus,
\[ a_3 = |\vec{A}| \sqrt{3} \cos \alpha - 2 |\vec{A}| \cos \alpha = |\vec{A}| (\sqrt{3} - 2) \cos \alpha \]

### Step 6: Calculate Magnitude of \( \vec{A} \)
The magnitude of \( \vec{A} \) is given by:
\[ |\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
Substituting \( a_1 \), \( a_2 \), and \( a_3 \):
\[ |\vec{A}| = \sqrt{(|\vec{A}| \cos \alpha)^2 + (|\vec{A}| \cos \alpha)^2 + (|\vec{A}| (\sqrt{3} - 2) \cos \alpha)^2} \]
This simplifies to:
\[ |\vec{A}| = |\vec{A}| \cos \alpha \sqrt{2 + (\sqrt{3} - 2)^2} \]
Calculating \( (\sqrt{3} - 2)^2 = 3 - 4\sqrt{3} + 4 = 7 - 4\sqrt{3} \):
\[ |\vec{A}| = |\vec{A}| \cos \alpha \sqrt{2 + 7 - 4\sqrt{3}} = |\vec{A}| \cos \alpha \sqrt{9 - 4\sqrt{3}} \]

### Step 7: Find \( \sec^2 \alpha \)
Now, we know:
\[ |\vec{A}|^2 = (|\vec{A}| \cos \alpha)^2 (9 - 4\sqrt{3}) \]
Dividing both sides by \( (|\vec{A}| \cos \alpha)^2 \):
\[ \sec^2 \alpha = \frac{|\vec{A}|^2}{(|\vec{A}| \cos \alpha)^2} = 9 - 4\sqrt{3} \]

### Final Answer
Thus, the value of \( \sec^2 \alpha \) is:
\[ \sec^2 \alpha = 9 - 4\sqrt{3} \]