The equation of the plane containing the line `vecr= veca + k vecb` and perpendicular to the plane `vecr . vecn =q` is :

Options

A. `(vecr -vecb). (vecn xx veca) =0`
B. `(vecr -veca ).{vecn x (veca xx vecb)}=0`
C. `(vecr -veca). (vecn xx vecb)=0`
D. `(vecr-vecb) .{vecn xx (veca xx vecb)}=0`

Answer

To find the equation of the plane that contains the line given by the vector equation \(\vec{r} = \vec{a} + k\vec{b}\) and is perpendicular to the plane defined by \(\vec{r} \cdot \vec{n} = q\), we can follow these steps:

### Step 1: Identify the point on the line
The line is given by \(\vec{r} = \vec{a} + k\vec{b}\). Here, \(\vec{a}\) is a position vector of a point on the line. This point will also lie on the required plane.

**Hint:** The point \(\vec{a}\) is essential because it provides a specific location through which the plane passes.

### Step 2: Determine the direction vectors
The direction vector of the line is \(\vec{b}\). The normal vector of the plane defined by \(\vec{r} \cdot \vec{n} = q\) is \(\vec{n}\). Since the required plane contains the line, both \(\vec{b}\) and \(\vec{n}\) are direction vectors lying in the required plane.

**Hint:** Remember that the normal vector of the required plane can be found using the cross product of the two direction vectors.

### Step 3: Find the normal vector of the required plane
To find the normal vector \(\vec{n_1}\) of the required plane, we take the cross product of \(\vec{n}\) and \(\vec{b}\):
\[ \vec{n_1} = \vec{n} \times \vec{b} \]

**Hint:** The cross product gives a vector that is perpendicular to both \(\vec{n}\) and \(\vec{b}\), which is what we need for the normal of the required plane.

### Step 4: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ \vec{r} \cdot \vec{n_1} = \vec{a} \cdot \vec{n_1} \]
Substituting \(\vec{n_1}\) into the equation, we have:
\[ \vec{r} \cdot (\vec{n} \times \vec{b}) = \vec{a} \cdot (\vec{n} \times \vec{b}) \]

**Hint:** This equation represents all points \(\vec{r}\) that lie in the required plane.

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ \vec{r} \cdot (\vec{n} \times \vec{b}) - \vec{a} \cdot (\vec{n} \times \vec{b}) = 0 \]
This can be expressed as:
\[ \vec{n} \times \vec{b} \cdot (\vec{r} - \vec{a}) = 0 \]

**Hint:** This form emphasizes that the vector \(\vec{r} - \vec{a}\) is perpendicular to the normal vector \(\vec{n} \times \vec{b}\).

### Conclusion
Thus, the equation of the required plane is:
\[ \vec{r} \cdot (\vec{n} \times \vec{b}) = \vec{a} \cdot (\vec{n} \times \vec{b}) \]
or equivalently,
\[ \vec{r} - \vec{a} \cdot (\vec{n} \times \vec{b}) = 0 \]

### Final Answer
The equation of the plane containing the line and perpendicular to the given plane is:
\[ \vec{r} \cdot (\vec{n} \times \vec{b}) = \vec{a} \cdot (\vec{n} \times \vec{b}) \]