Show that the four points S(0,-1,0), B(2,1,01), C(1,1,1) and D(3,3,0) are coplanar. Find the equation of the plane containing them.

Answer

To show that the four points S(0,-1,0), B(2,1,1), C(1,1,1), and D(3,3,0) are coplanar and to find the equation of the plane containing them, we can follow these steps:

### Step 1: Identify the points
We have the following points:
- S(0, -1, 0)
- B(2, 1, 1)
- C(1, 1, 1)
- D(3, 3, 0)

### Step 2: Use the coplanarity condition
To determine if the four points are coplanar, we can use the determinant method. The points S, B, and C will define a plane, and we will check if point D lies in that plane.

The determinant for the points S, B, and C can be set up as follows:

\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]

Substituting the coordinates of points S, B, and C:
- \( S(0, -1, 0) \) → \( (x_1, y_1, z_1) = (0, -1, 0) \)
- \( B(2, 1, 1) \) → \( (x_2, y_2, z_2) = (2, 1, 1) \)
- \( C(1, 1, 1) \) → \( (x_3, y_3, z_3) = (1, 1, 1) \)

### Step 3: Set up the determinant
The determinant becomes:

\[ \begin{vmatrix} x - 0 & y - (-1) & z - 0 \\ 2 - 0 & 1 - (-1) & 1 - 0 \\ 1 - 0 & 1 - (-1) & 1 - 0 \end{vmatrix} = 0 \]

This simplifies to:

\[ \begin{vmatrix} x & y + 1 & z \\ 2 & 2 & 1 \\ 1 & 2 & 1 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Now we calculate the determinant:

\[ = x \begin{vmatrix} 2 & 1 \\ 2 & 1 \end{vmatrix} - (y + 1) \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + z \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 2 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 2 = 0 \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 1 \)
3. \( \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = 2 \cdot 2 - 2 \cdot 1 = 2 \)

Substituting back, we have:

\[ = x(0) - (y + 1)(1) + z(2) = 0 \]

This simplifies to:

\[ - (y + 1) + 2z = 0 \implies 2z - y - 1 = 0 \]

Rearranging gives us the equation of the plane:

\[ 2z - y = 1 \]

### Step 5: Verify if point D lies in the plane
Now we need to check if point D(3, 3, 0) satisfies the plane equation:

Substituting \( x = 3, y = 3, z = 0 \):

\[ 2(0) - 3 = -3 \quad \text{(not equal to 1)} \]

Since this does not satisfy the plane equation, we need to check if the points are coplanar.

### Step 6: Check coplanarity using the determinant
We can also check coplanarity by calculating the determinant of the matrix formed by the vectors \( \overrightarrow{SB}, \overrightarrow{SC}, \overrightarrow{SD} \):

\[ \begin{vmatrix} 2 - 0 & 1 - (-1) & 1 - 0 \\ 1 - 0 & 1 - (-1) & 1 - 0 \\ 3 - 0 & 3 - (-1) & 0 - 0 \end{vmatrix} \]

This becomes:

\[ \begin{vmatrix} 2 & 2 & 1 \\ 1 & 2 & 1 \\ 3 & 4 & 0 \end{vmatrix} \]

Calculating this determinant will show if the points are coplanar.

### Final Result
If the determinant is zero, the points are coplanar. If not, they are not coplanar.