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.(2hati+hatj-3hatk)=-4`
B. `vecrxx(-hati+hatj+hatk)=vec0`
C. `vecr.(-hati+hatj+hatk)=0`
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 direction vectors of the lines
The first line is given by:
\[ \vec{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + 2\hat{j} - \hat{k}) \]
From this, we can extract the direction vector \(\vec{b_1}\):
\[ \vec{b_1} = \hat{i} + 2\hat{j} - \hat{k} \]

The second line is given by:
\[ \vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k}) \]
From this, we can extract the direction vector \(\vec{b_2}\):
\[ \vec{b_2} = -\hat{i} + \hat{j} - 2\hat{k} \]

### Step 2: Find a point on the plane
We can choose a point that lies on either of the lines. For simplicity, we can take the point from the first line when \(\lambda = 0\):
\[ \vec{A} = \hat{i} + \hat{j} + 0(\hat{i} + 2\hat{j} - \hat{k}) = \hat{i} + \hat{j} \]
Thus, the position vector of the point is:
\[ \vec{A} = (1, 1, 0) \]

### Step 3: Calculate the normal vector to the plane
To find the normal vector \(\vec{n}\) to the plane, we take 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:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix} \]
Expanding this 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 the minors:
\[ = \hat{i} (2 \cdot -2 - (-1) \cdot 1) - \hat{j} (1 \cdot -2 - (-1) \cdot -1) + \hat{k} (1 \cdot 1 - 2 \cdot -1) \]
\[ = \hat{i} (-4 + 1) - \hat{j} (-2 - 1) + \hat{k} (1 + 2) \]
\[ = -3\hat{i} + 3\hat{j} + 3\hat{k} \]
Thus, we can simplify the normal vector:
\[ \vec{n} = -3(\hat{i} - \hat{j} - \hat{k}) = -3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 4: Write the equation of the plane
The equation of the plane can be expressed in the vector form:
\[ \vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n} \]
Substituting \(\vec{n}\) and \(\vec{A}\):
\[ \vec{r} \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) = (1, 1, 0) \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) \]
Calculating the right-hand side:
\[ = 1 \cdot (-3) + 1 \cdot 3 + 0 \cdot 3 = -3 + 3 + 0 = 0 \]
Thus, the equation of the plane is:
\[ \vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 0 \]

### Final Answer
The vector equation of the plane is:
\[ \vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 0 \]

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