Find the vector equation of the plane in which the lines `vecr=hati+hatj+lambda(hati+2hatj-hatk)` and `vecr=(hati+hatj)+mu(-hati+hatj-2hatk)` lie.

Options

A. `vecr.(hati+hatj+hatk)=0`
B. `vecr.(hati-hatj-hatk)=0`
C. `vecr.(hati+hatj+hatk)=3`
D. none of these

Explanation

To find the vector equation of the plane in which the given lines lie, we can follow these steps:

### Step 1: Identify the lines
The two lines are given in vector form:
1. \( \vec{r} = \hat{i} + \hat{j} + \lambda (\hat{i} + 2\hat{j} - \hat{k}) \)
2. \( \vec{r} = (\hat{i} + \hat{j}) + \mu (-\hat{i} + \hat{j} - 2\hat{k}) \)

### Step 2: Extract direction vectors and points
From the first line:
- Point \( A \) (when \( \lambda = 0 \)): \( \vec{A} = \hat{i} + \hat{j} \)
- Direction vector \( \vec{b_1} = \hat{i} + 2\hat{j} - \hat{k} \)

From the second line:
- Point \( B \) (when \( \mu = 0 \)): \( \vec{B} = \hat{i} + \hat{j} \)
- Direction vector \( \vec{b_2} = -\hat{i} + \hat{j} - 2\hat{k} \)

### Step 3: Find the normal vector to the plane
The normal vector \( \vec{n} \) to the plane can be found using the cross product of the two direction vectors \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{n} = \vec{b_1} \times \vec{b_2} \]

Calculating the cross product:
\[ \vec{b_1} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} \]

Using the determinant:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ -1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} \]

Calculating each of the minors:
1. \( \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-1)(1) = -4 + 1 = -3 \)
2. \( \begin{vmatrix} 1 & -1 \\ -1 & -2 \end{vmatrix} = (1)(-2) - (-1)(-1) = -2 - 1 = -3 \)
3. \( \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = (1)(1) - (2)(-1) = 1 + 2 = 3 \)

Putting it all together:
\[ \vec{n} = -3\hat{i} + 3\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 4: Write the equation of the plane
Using the point \( \vec{A} = \hat{i} + \hat{j} \) and the normal vector \( \vec{n} \):
The equation of the plane can be written as:
\[ \vec{n} \cdot (\vec{r} - \vec{A}) = 0 \]
Substituting \( \vec{n} \) and \( \vec{A} \):
\[ (-3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (\vec{r} - (\hat{i} + \hat{j})) = 0 \]

Expanding this:
\[ -3(x - 1) + 3(y - 1) + 3(z - 0) = 0 \]
This simplifies to:
\[ -3x + 3y + 3z = 0 \quad \text{or} \quad x - y - z = 0 \]

### Final Answer
The vector equation of the plane is:
\[ x - y - z = 0 \]