If the position vectors of the point A and B are `3hat(i)+hat(j)+2hat(k) and hat(i)-2hat(j)-4hat(k)` respectively. Then the eqaution of the plane through B and perpendicular to AB is

Options

A. `2x+3y+6z+28=0`
B. `2x+3y+6z=28`
C. `2x-3y+6z+28=0`
D. `3x-2y+6z=28`

Answer

To find the equation of the plane through point B and perpendicular to the line segment AB, we can follow these steps:

### Step 1: Identify the position vectors of points A and B.
The position vectors are given as:
- Point A: \( \vec{A} = 3\hat{i} + \hat{j} + 2\hat{k} \)
- Point B: \( \vec{B} = \hat{i} - 2\hat{j} - 4\hat{k} \)

### Step 2: Find the vector AB.
The vector \( \vec{AB} \) can be found by subtracting the position vector of B from that of A:
\[ \vec{AB} = \vec{A} - \vec{B} = (3\hat{i} + \hat{j} + 2\hat{k}) - (\hat{i} - 2\hat{j} - 4\hat{k}) \]
Calculating this gives:
\[ \vec{AB} = (3 - 1)\hat{i} + (1 + 2)\hat{j} + (2 + 4)\hat{k} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

### Step 3: Identify the normal vector of the plane.
Since the plane is perpendicular to the vector \( \vec{AB} \), the normal vector \( \vec{N} \) of the plane is:
\[ \vec{N} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

### Step 4: Use the point-normal form of the plane equation.
The equation of a plane in vector form is given by:
\[ \vec{R} - \vec{B} \cdot \vec{N} = 0 \]
Where \( \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \) is the position vector of any point on the plane, and \( \vec{B} \) is the position vector of point B.

### Step 5: Substitute the values into the plane equation.
Substituting \( \vec{B} = \hat{i} - 2\hat{j} - 4\hat{k} \) and \( \vec{N} = 2\hat{i} + 3\hat{j} + 6\hat{k} \):
\[ (x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} - 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 0 \]

### Step 6: Calculate the dot product.
Calculating \( \vec{B} \cdot \vec{N} \):
\[ \vec{B} \cdot \vec{N} = (1)(2) + (-2)(3) + (-4)(6) = 2 - 6 - 24 = -28 \]

### Step 7: Write the final equation of the plane.
The equation of the plane can be written as:
\[ 2x + 3y + 6z + 28 = 0 \]

### Final Answer:
The equation of the plane through point B and perpendicular to AB is:
\[ 2x + 3y + 6z + 28 = 0 \]