What is the equation of the plane passing ihrough the points (-2,6,-6),(-3,10,-9) and (-5,0,-6)?

Options

A. `2x-y-2z=2`
B. `2x+y+3z=0`
C. `x+y+z=6`
D. `x-y-z=3`

Answer

To find the equation of the plane passing through the points A(-2, 6, -6), B(-3, 10, -9), and C(-5, 0, -6), we can follow these steps:

### Step 1: Define the Points
Let the points be defined as follows:
- Point A: \( A(-2, 6, -6) \)
- Point B: \( B(-3, 10, -9) \)
- Point C: \( C(-5, 0, -6) \)

### Step 2: Find the Vectors BA and BC
To find the vectors BA and BC, we can use the following formulas:
- \( \vec{BA} = A - B \)
- \( \vec{BC} = C - B \)

Calculating \( \vec{BA} \):
\[ \vec{BA} = (-2, 6, -6) - (-3, 10, -9) = (-2 + 3, 6 - 10, -6 + 9) = (1, -4, 3) \]

Calculating \( \vec{BC} \):
\[ \vec{BC} = (-5, 0, -6) - (-3, 10, -9) = (-5 + 3, 0 - 10, -6 + 9) = (-2, -10, 3) \]

### Step 3: Find the Normal Vector \( \vec{n} \)
The normal vector \( \vec{n} \) to the plane can be found by taking the cross product of \( \vec{BA} \) and \( \vec{BC} \):
\[ \vec{n} = \vec{BA} \times \vec{BC} \]

Calculating the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & 3 \\ -2 & -10 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}((-4)(3) - (3)(-10)) - \hat{j}((1)(3) - (3)(-2)) + \hat{k}((1)(-10) - (-4)(-2)) \]
\[ = \hat{i}(-12 + 30) - \hat{j}(3 + 6) + \hat{k}(-10 - 8) \]
\[ = \hat{i}(18) - \hat{j}(9) - \hat{k}(18) \]
Thus, the normal vector is:
\[ \vec{n} = (18, -9, -18) \]

### Step 4: Write the Equation of the Plane
The equation of the plane can be expressed as:
\[ \vec{n} \cdot (\vec{r} - \vec{a}) = 0 \]
Where \( \vec{r} = (x, y, z) \) and \( \vec{a} = A(-2, 6, -6) \).

Substituting \( \vec{n} \) and \( \vec{a} \):
\[ (18, -9, -18) \cdot ((x + 2), (y - 6), (z + 6)) = 0 \]
This expands to:
\[ 18(x + 2) - 9(y - 6) - 18(z + 6) = 0 \]
Expanding this gives:
\[ 18x + 36 - 9y + 54 - 18z - 108 = 0 \]
Combining like terms:
\[ 18x - 9y - 18z - 18 = 0 \]
Dividing through by 9:
\[ 2x - y - 2z - 2 = 0 \]
Rearranging gives us the final equation of the plane:
\[ 2x - y - 2z = 2 \]

### Final Answer
The equation of the plane is:
\[ 2x - y - 2z = 2 \]