Equation of the plane through three points A, B and C with position vectors `-6i+3j+2k, 3i-2j+4k and 5i+7j+3k` is equal to

Options

A. `rcdot(i-j-7k)+23=0`
B. `rcdot(i+j+7k)=23`
C. `rcdot(i+j-7k)+23=0`
D. `rcdot(i-j-7k)=23`

Answer

To find the equation of the plane passing through the points A, B, and C with position vectors \(\vec{A} = -6\hat{i} + 3\hat{j} + 2\hat{k}\), \(\vec{B} = 3\hat{i} - 2\hat{j} + 4\hat{k}\), and \(\vec{C} = 5\hat{i} + 7\hat{j} + 3\hat{k}\), we will follow these steps:

### Step 1: Find the direction vectors
We need to find two direction vectors that lie in the plane. We can use the vectors \(\vec{AC}\) and \(\vec{AB}\).

1. **Calculate \(\vec{AC}\)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (5\hat{i} + 7\hat{j} + 3\hat{k}) - (-6\hat{i} + 3\hat{j} + 2\hat{k}) \]
\[ = 5\hat{i} + 7\hat{j} + 3\hat{k} + 6\hat{i} - 3\hat{j} - 2\hat{k} \]
\[ = (5 + 6)\hat{i} + (7 - 3)\hat{j} + (3 - 2)\hat{k} = 11\hat{i} + 4\hat{j} + 1\hat{k} \]

2. **Calculate \(\vec{AB}\)**:
\[ \vec{AB} = \vec{B} - \vec{A} = (3\hat{i} - 2\hat{j} + 4\hat{k}) - (-6\hat{i} + 3\hat{j} + 2\hat{k}) \]
\[ = 3\hat{i} - 2\hat{j} + 4\hat{k} + 6\hat{i} - 3\hat{j} - 2\hat{k} \]
\[ = (3 + 6)\hat{i} + (-2 - 3)\hat{j} + (4 - 2)\hat{k} = 9\hat{i} - 5\hat{j} + 2\hat{k} \]

### Step 2: Find the normal vector to the plane
The normal vector \(\vec{n}\) to the plane can be found using the cross product of \(\vec{AC}\) and \(\vec{AB}\).

\[ \vec{n} = \vec{AB} \times \vec{AC} \]
Using the determinant method for the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & -5 & 2 \\ 11 & 4 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -5 & 2 \\ 4 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 9 & 2 \\ 11 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 9 & -5 \\ 11 & 4 \end{vmatrix} \]
\[ = \hat{i}((-5)(1) - (2)(4)) - \hat{j}((9)(1) - (2)(11)) + \hat{k}((9)(4) - (-5)(11)) \]
\[ = \hat{i}(-5 - 8) - \hat{j}(9 - 22) + \hat{k}(36 + 55) \]
\[ = -13\hat{i} + 13\hat{j} + 91\hat{k} \]

### Step 3: Use the point-normal form of the plane equation
The equation of the plane can be expressed in the form:
\[ \vec{n} \cdot (\vec{r} - \vec{A}) = 0 \]
Substituting \(\vec{n} = -13\hat{i} + 13\hat{j} + 91\hat{k}\) and \(\vec{A} = -6\hat{i} + 3\hat{j} + 2\hat{k}\):
\[ (-13\hat{i} + 13\hat{j} + 91\hat{k}) \cdot (\vec{r} - (-6\hat{i} + 3\hat{j} + 2\hat{k})) = 0 \]
This simplifies to:
\[ (-13\hat{i} + 13\hat{j} + 91\hat{k}) \cdot (\vec{r} + 6\hat{i} - 3\hat{j} - 2\hat{k}) = 0 \]

### Step 4: Expand and simplify
Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\):
\[ (-13)(x + 6) + 13(y - 3) + 91(z - 2) = 0 \]
Expanding this:
\[ -13x - 78 + 13y - 39 + 91z - 182 = 0 \]
Combining like terms:
\[ -13x + 13y + 91z - 299 = 0 \]
Rearranging gives:
\[ 13x - 13y - 91z + 299 = 0 \]

### Final Equation
Thus, the equation of the plane is:
\[ 13x - 13y - 91z + 299 = 0 \]