Find equation of the plane through (2, 1, 4), (1, -1, 2) and (4, -1, 1).

Answer

To find the equation of the plane that passes through the points \( A(2, 1, 4) \), \( B(1, -1, 2) \), and \( C(4, -1, 1) \), we will follow these steps:

### Step 1: Find the direction vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

The direction vector \( \overrightarrow{AB} \) can be calculated as:
\[ \overrightarrow{AB} = B - A = (1 - 2, -1 - 1, 2 - 4) = (-1, -2, -2) \]

The direction vector \( \overrightarrow{AC} \) can be calculated as:
\[ \overrightarrow{AC} = C - A = (4 - 2, -1 - 1, 1 - 4) = (2, -2, -3) \]

### Step 2: Find the normal vector \( \overrightarrow{n} \) to the plane using the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \)

The cross product can be calculated using the determinant:
\[ \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & -2 \\ 2 & -2 & -3 \end{vmatrix} \]

Calculating the determinant:
\[ \overrightarrow{n} = \hat{i}((-2)(-3) - (-2)(-2)) - \hat{j}((-1)(-3) - (-2)(2)) + \hat{k}((-1)(-2) - (-2)(2)) \]
\[ = \hat{i}(6 - 4) - \hat{j}(3 - 4) + \hat{k}(2 - (-4)) \]
\[ = 2\hat{i} + 1\hat{j} + 6\hat{k} \]
Thus, the normal vector \( \overrightarrow{n} = (2, 1, 6) \).

### Step 3: Use the point-normal form of the equation of the plane

The equation of the plane can be expressed in the form:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \( (n_1, n_2, n_3) \) are the components of the normal vector and \( (x_0, y_0, z_0) \) is a point on the plane (we can use point \( A(2, 1, 4) \)).

Substituting the values:
\[ 2(x - 2) + 1(y - 1) + 6(z - 4) = 0 \]
Expanding this:
\[ 2x - 4 + y - 1 + 6z - 24 = 0 \]
Combining like terms:
\[ 2x + y + 6z - 29 = 0 \]

### Final Equation of the Plane
The equation of the plane is:
\[ 2x + y + 6z = 29 \]

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