The equation of the plane contaiing the lines `vecr=veca_(1)+lamda vecb` and `vecr=veca_(2)+muvecb` is

Options

A. `vecr ( vec a_(1)- vec a _(2)) xx vecb = [ veca _(1) veca_(2)vecb]`
B. `vecr . (veca _(2)-veca _(1)) [ veca _(1) veca _(2) vecb ]`
C. `vecr. (veca _(1) - veca_(2)) xx vecb [veca _(2) veca_(1) vecb]`
D. None of these

Answer

To find the equation of the plane containing the lines given by the vector equations \(\vec{r} = \vec{a_1} + \lambda \vec{b}\) and \(\vec{r} = \vec{a_2} + \mu \vec{b}\), we can follow these steps:

### Step 1: Identify the Direction Vector and Points
The two lines are given in parametric form. The direction vector for both lines is \(\vec{b}\). The points through which these lines pass are \(\vec{a_1}\) and \(\vec{a_2}\).

### Step 2: Find the Vector Between the Two Points
The vector between the points \(\vec{a_1}\) and \(\vec{a_2}\) is given by:
\[ \vec{BA} = \vec{a_1} - \vec{a_2} \]

### Step 3: Determine the Normal Vector to the Plane
The normal vector \(\vec{n}\) to the plane can be found using the cross product of the direction vector \(\vec{b}\) and the vector \(\vec{BA}\):
\[ \vec{n} = \vec{b} \times (\vec{a_1} - \vec{a_2}) \]

### Step 4: Use the Point-Normal Form of the Plane Equation
The equation of the plane can be expressed in the point-normal form:
\[ \vec{n} \cdot (\vec{r} - \vec{a_1}) = 0 \]
Substituting \(\vec{n}\) gives:
\[ (\vec{b} \times (\vec{a_1} - \vec{a_2})) \cdot (\vec{r} - \vec{a_1}) = 0 \]

### Step 5: Expand the Equation
Expanding this equation, we have:
\[ \vec{r} \cdot (\vec{b} \times (\vec{a_1} - \vec{a_2})) - \vec{a_1} \cdot (\vec{b} \times (\vec{a_1} - \vec{a_2})) = 0 \]

### Step 6: Rearranging the Equation
This can be rearranged to give the final equation of the plane:
\[ \vec{r} \cdot (\vec{b} \times (\vec{a_1} - \vec{a_2})) = \vec{a_1} \cdot (\vec{b} \times (\vec{a_1} - \vec{a_2})) \]

### Final Equation
Thus, the equation of the plane containing the two lines is:
\[ \vec{r} \cdot (\vec{b} \times (\vec{a_1} - \vec{a_2})) = \text{constant} \]