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 values of `mu` for which the given straight lines are coplanar, we will follow these steps:

### Step 1: Identify the position and direction vectors of the lines
The first line is given by:
\[ \vec{r_1} = 3\hat{i} - 2\hat{j} - 4\hat{k} + \lambda_1(\hat{i} - \hat{j} + \mu \hat{k}) \]
From this, we can identify:
- Position vector \( \vec{l_1} = 3\hat{i} - 2\hat{j} - 4\hat{k} \)
- Direction vector \( \vec{m_1} = \hat{i} - \hat{j} + \mu \hat{k} \)

The second line is given by:
\[ \vec{r_2} = 5\hat{i} - 2\hat{j} + \hat{k} + \lambda_2(\hat{i} + \mu \hat{j} + 2\hat{k}) \]
From this, we can identify:
- Position vector \( \vec{l_2} = 5\hat{i} - 2\hat{j} + \hat{k} \)
- Direction vector \( \vec{m_2} = \hat{i} + \mu \hat{j} + 2\hat{k} \)

### Step 2: Find the vector \( \vec{l_2} - \vec{l_1} \)
Calculating \( \vec{l_2} - \vec{l_1} \):
\[ \vec{l_2} - \vec{l_1} = (5\hat{i} - 2\hat{j} + \hat{k}) - (3\hat{i} - 2\hat{j} - 4\hat{k}) = (5 - 3)\hat{i} + (0)\hat{j} + (1 + 4)\hat{k} = 2\hat{i} + 5\hat{k} \]

### Step 3: Compute the cross product \( \vec{m_1} \times \vec{m_2} \)
Using the determinant method to find \( \vec{m_1} \times \vec{m_2} \):
\[ \vec{m_1} = \begin{pmatrix} 1 \\ -1 \\ \mu \end{pmatrix}, \quad \vec{m_2} = \begin{pmatrix} 1 \\ \mu \\ 2 \end{pmatrix} \]
The cross product is given by the determinant:
\[ \vec{m_1} \times \vec{m_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & \mu \\ 1 & \mu & 2 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i}((-1)(2) - (\mu)(\mu)) - \hat{j}((1)(2) - (1)(\mu)) + \hat{k}((1)(\mu) - (1)(-1)) \]
\[ = \hat{i}(-2 - \mu^2) - \hat{j}(2 - \mu) + \hat{k}(\mu + 1) \]
Thus, we have:
\[ \vec{m_1} \times \vec{m_2} = (-\mu^2 - 2)\hat{i} + (\mu - 2)\hat{j} + (\mu + 1)\hat{k} \]

### Step 4: Use the coplanarity condition
The lines are coplanar if:
\[ (\vec{l_2} - \vec{l_1}) \cdot (\vec{m_1} \times \vec{m_2}) = 0 \]
Substituting the values:
\[ (2\hat{i} + 5\hat{k}) \cdot ((-\mu^2 - 2)\hat{i} + (\mu - 2)\hat{j} + (\mu + 1)\hat{k}) = 0 \]
Calculating the dot product:
\[ 2(-\mu^2 - 2) + 5(\mu + 1) = 0 \]
Expanding this:
\[ -2\mu^2 - 4 + 5\mu + 5 = 0 \]
Rearranging gives:
\[ 2\mu^2 - 5\mu - 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( \mu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -5, c = -1 \):
\[ \mu = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4} \]

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