The value(s) of `mu` for which the straight lines `vecr=3hati-2hatj-4hatk+lambda_(1)(hati-hatj+mu hatk)` and `vec r=5hati-2hatj+hatk+lambda_(2)(hati+mu hatj+2hatk)` are coplanar is/are :

Options

A. `(5+sqrt(33))/(4)`
B. `(-5+sqrt(33))/(4)`
C. `(5-sqrt(33))/(4)`
D. `(-5-sqrt(33))/(4)`

Answer

To find the value(s) of \( \mu \) for which the given lines are coplanar, we will follow these steps:

### Step 1: Identify the lines in vector form
The equations of the lines are given as:
1. \( \vec{r_1} = 3\hat{i} - 2\hat{j} - 4\hat{k} + \lambda_1(\hat{i} - \hat{j} + \mu \hat{k}) \)
2. \( \vec{r_2} = 5\hat{i} - 2\hat{j} + \hat{k} + \lambda_2(\hat{i} + \mu \hat{j} + 2\hat{k}) \)

### Step 2: Identify points and direction vectors
From the equations, we can identify:
- For line 1:
- Point \( A_1 = 3\hat{i} - 2\hat{j} - 4\hat{k} \)
- Direction vector \( \vec{B_1} = \hat{i} - \hat{j} + \mu \hat{k} \)

- For line 2:
- Point \( A_2 = 5\hat{i} - 2\hat{j} + \hat{k} \)
- Direction vector \( \vec{B_2} = \hat{i} + \mu \hat{j} + 2\hat{k} \)

### Step 3: Calculate \( A_2 - A_1 \)
We compute:
\[ A_2 - A_1 = (5\hat{i} - 2\hat{j} + \hat{k}) - (3\hat{i} - 2\hat{j} - 4\hat{k}) \]
Calculating this gives:
\[ A_2 - A_1 = (5 - 3)\hat{i} + (-2 + 2)\hat{j} + (1 + 4)\hat{k} = 2\hat{i} + 0\hat{j} + 5\hat{k} = 2\hat{i} + 5\hat{k} \]

### Step 4: Calculate \( \vec{B_1} \times \vec{B_2} \)
Next, we calculate the cross product:
\[ \vec{B_1} = \hat{i} - \hat{j} + \mu \hat{k} \]
\[ \vec{B_2} = \hat{i} + \mu \hat{j} + 2\hat{k} \]
The cross product \( \vec{B_1} \times \vec{B_2} \) is calculated using the determinant:
\[ \vec{B_1} \times \vec{B_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & \mu \\ 1 & \mu & 2 \end{vmatrix} \]
Calculating this determinant gives:
\[ \hat{i} \left((-1)(2) - (\mu)(\mu)\right) - \hat{j} \left((1)(2) - (1)(\mu)\right) + \hat{k} \left((1)(\mu) - (-1)(1)\right) \]
This simplifies to:
\[ \hat{i}(-2 - \mu^2) - \hat{j}(2 - \mu) + \hat{k}(\mu + 1) \]
Thus,
\[ \vec{B_1} \times \vec{B_2} = (-2 - \mu^2)\hat{i} - (2 - \mu)\hat{j} + (\mu + 1)\hat{k} \]

### Step 5: Use the coplanarity condition
The lines are coplanar if:
\[ (A_2 - A_1) \cdot (\vec{B_1} \times \vec{B_2}) = 0 \]
Substituting \( A_2 - A_1 = 2\hat{i} + 5\hat{k} \):
\[ (2\hat{i} + 5\hat{k}) \cdot \left((-2 - \mu^2)\hat{i} - (2 - \mu)\hat{j} + (\mu + 1)\hat{k}\right) = 0 \]
Calculating the dot product:
\[ 2(-2 - \mu^2) + 5(\mu + 1) = 0 \]
This expands to:
\[ -4 - 2\mu^2 + 5\mu + 5 = 0 \implies -2\mu^2 + 5\mu + 1 = 0 \]
Multiplying through by -1 gives:
\[ 2\mu^2 - 5\mu - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ \mu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = -5, c = -1 \):
\[ \mu = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \mu = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4} \]
Thus, the values of \( \mu \) are:
\[ \mu = \frac{5 + \sqrt{33}}{4} \quad \text{and} \quad \mu = \frac{5 - \sqrt{33}}{4} \]

### Final Answer
The values of \( \mu \) for which the lines are coplanar are:
\[ \mu = \frac{5 + \sqrt{33}}{4} \quad \text{and} \quad \mu = \frac{5 - \sqrt{33}}{4} \]