Find the vector and cartesian equations of the plane passing through the points (2,2,-1), (3,4,2) and (7,06) also find the vector equation of a plane passing through (4,3,1) and parallel to the plane obtained above.

Answer

To find the vector and Cartesian equations of the plane passing through the points \( A(2, 2, -1) \), \( B(3, 4, 2) \), and \( C(7, 0, 6) \), we will follow these steps:

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

The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (3 - 2, 4 - 2, 2 - (-1)) = (1, 2, 3) \]

The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} = (7 - 2, 0 - 2, 6 - (-1)) = (5, -2, 7) \]

### Step 2: Find the normal vector \( \vec{n} \) to the plane

To find the normal vector \( \vec{n} \), we take the cross product of \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 5 & -2 & 7 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 2 & 3 \\ -2 & 7 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 5 & 7 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} \]
\[ = \hat{i} (2 \cdot 7 - 3 \cdot (-2)) - \hat{j} (1 \cdot 7 - 3 \cdot 5) + \hat{k} (1 \cdot (-2) - 2 \cdot 5) \]
\[ = \hat{i} (14 + 6) - \hat{j} (7 - 15) + \hat{k} (-2 - 10) \]
\[ = 20\hat{i} + 8\hat{j} - 12\hat{k} \]

Thus, the normal vector is:
\[ \vec{n} = (20, 8, -12) \]

### Step 3: Write the vector equation of the plane

The vector equation of the plane can be written as:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \( \vec{a} \) is a position vector of point \( A(2, 2, -1) \).

Calculating \( \vec{a} \cdot \vec{n} \):
\[ \vec{a} \cdot \vec{n} = (2, 2, -1) \cdot (20, 8, -12) = 2 \cdot 20 + 2 \cdot 8 + (-1) \cdot (-12) = 40 + 16 + 12 = 68 \]

Thus, the vector equation of the plane is:
\[ \vec{r} \cdot (20, 8, -12) = 68 \]

### Step 4: Write the Cartesian equation of the plane

The Cartesian equation can be derived from the vector equation:
\[ 20x + 8y - 12z = 68 \]

### Step 5: Find the vector equation of the plane passing through point \( (4, 3, 1) \) and parallel to the above plane

Since the new plane is parallel to the original plane, it will have the same normal vector \( \vec{n} = (20, 8, -12) \).

Using point \( (4, 3, 1) \):
\[ \vec{r} \cdot (20, 8, -12) = \vec{b} \cdot (20, 8, -12) \]
where \( \vec{b} = (4, 3, 1) \).

Calculating \( \vec{b} \cdot \vec{n} \):
\[ \vec{b} \cdot \vec{n} = (4, 3, 1) \cdot (20, 8, -12) = 4 \cdot 20 + 3 \cdot 8 + 1 \cdot (-12) = 80 + 24 - 12 = 92 \]

Thus, the vector equation of the new plane is:
\[ \vec{r} \cdot (20, 8, -12) = 92 \]

### Summary of Equations

1. **Vector Equation of the Plane through points A, B, C:**
\[ \vec{r} \cdot (20, 8, -12) = 68 \]

2. **Cartesian Equation of the Plane through points A, B, C:**
\[ 20x + 8y - 12z = 68 \]

3. **Vector Equation of the Plane through point (4, 3, 1) and parallel to the first plane:**
\[ \vec{r} \cdot (20, 8, -12) = 92 \]