Find the equation of the plane passing through the point (1,-2,1) and perpendicular to the line joining the point A(3,2,1) and B(1,4,2).

Answer

To find the equation of the plane that passes through the point (1, -2, 1) and is perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2), we can follow these steps:

### Step 1: Find the direction ratios of the line AB
The direction ratios of the line joining points A and B can be found using the coordinates of A and B.

- Coordinates of A: (3, 2, 1)
- Coordinates of B: (1, 4, 2)

The direction ratios (d) can be calculated as:
\[ d = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \]
\[ d = (1 - 3, 4 - 2, 2 - 1) \]
\[ d = (-2, 2, 1) \]

### Step 2: Use the direction ratios as the normal vector
Since the plane is perpendicular to the line AB, the direction ratios we found will serve as the normal vector (n) of the plane.

- Normal vector \( n = (-2, 2, 1) \)

### Step 3: Use the point-normal form of the equation of a plane
The equation of a plane in point-normal form is given by:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
Where \( (x_0, y_0, z_0) \) is a point on the plane and \( (n_x, n_y, n_z) \) are the components of the normal vector.

Using the point (1, -2, 1) and the normal vector (-2, 2, 1):
\[ -2(x - 1) + 2(y + 2) + 1(z - 1) = 0 \]

### Step 4: Simplify the equation
Now, we simplify the equation:
\[ -2x + 2 + 2y + 4 + z - 1 = 0 \]
Combining like terms:
\[ -2x + 2y + z + (2 + 4 - 1) = 0 \]
\[ -2x + 2y + z + 5 = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us the final equation of the plane:
\[ -2x + 2y + z + 5 = 0 \]

### Final Equation
Thus, the equation of the required plane is:
\[ -2x + 2y + z + 5 = 0 \]

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