Find the equation of a plane containing the point (0,-1,-1),(-4,4,4)and (4,5,1) alos show that (3,9,4) lies on the that plane

Answer

To find the equation of the plane containing the points \( A(0, -1, -1) \), \( B(-4, 4, 4) \), and \( C(4, 5, 1) \), we can use the determinant method. Here are the steps to derive the equation of the plane:

### Step 1: Set up the determinant
The general equation of a plane can be derived using the determinant of a matrix formed by the coordinates of the points. The equation of the plane can be expressed as:

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

For our points:
- \( A(0, -1, -1) \) gives \( (x_1, y_1, z_1) = (0, -1, -1) \)
- \( B(-4, 4, 4) \) gives \( (x_2, y_2, z_2) = (-4, 4, 4) \)
- \( C(4, 5, 1) \) gives \( (x_3, y_3, z_3) = (4, 5, 1) \)

### Step 2: Substitute the points into the determinant
Substituting the coordinates into the determinant, we have:

\[ \begin{vmatrix} x - 0 & y + 1 & z + 1 \\ -4 - 0 & 4 + 1 & 4 + 1 \\ 4 - 0 & 5 + 1 & 1 + 1 \end{vmatrix} = 0 \]

This simplifies to:

\[ \begin{vmatrix} x & y + 1 & z + 1 \\ -4 & 5 & 5 \\ 4 & 6 & 2 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Now we calculate the determinant:

\[ = x \begin{vmatrix} 5 & 5 \\ 6 & 2 \end{vmatrix} - (y + 1) \begin{vmatrix} -4 & 5 \\ 4 & 2 \end{vmatrix} + (z + 1) \begin{vmatrix} -4 & 5 \\ 4 & 6 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 5 & 5 \\ 6 & 2 \end{vmatrix} = (5 \cdot 2) - (5 \cdot 6) = 10 - 30 = -20 \)

2. \( \begin{vmatrix} -4 & 5 \\ 4 & 2 \end{vmatrix} = (-4 \cdot 2) - (5 \cdot 4) = -8 - 20 = -28 \)

3. \( \begin{vmatrix} -4 & 5 \\ 4 & 6 \end{vmatrix} = (-4 \cdot 6) - (5 \cdot 4) = -24 - 20 = -44 \)

Substituting these back into the determinant equation:

\[ x(-20) - (y + 1)(-28) + (z + 1)(-44) = 0 \]

Expanding this gives:

\[ -20x + 28y + 28 - 44z - 44 = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ -20x + 28y - 44z - 16 = 0 \]

Dividing through by -4 to simplify:

\[ 5x - 7y + 11z + 4 = 0 \]

### Step 5: Final equation of the plane
Thus, the equation of the plane is:

\[ 5x - 7y + 11z + 4 = 0 \]

### Step 6: Verify that point \( (3, 9, 4) \) lies on the plane
To check if the point \( (3, 9, 4) \) lies on the plane, substitute \( x = 3 \), \( y = 9 \), and \( z = 4 \) into the equation:

\[ 5(3) - 7(9) + 11(4) + 4 = 15 - 63 + 44 + 4 = 0 \]

Since the left-hand side equals zero, the point \( (3, 9, 4) \) does indeed lie on the plane.