The vector equation of the plane passing through the points `(1 ,1 2) ,(0 , 2 ,3) and (4 , 5 ,6)` is

Options

A. `barr.(7hatj+7hatk)=-7`
B. `barr.(7hatj+7hatk)=7`
C. `barr.(7hatj-7hatk)=-7`
D. `barr.(7hatj-7hatk)=7`

Answer

To find the vector equation of the plane passing through the points \( A(1, 1, 2) \), \( B(0, 2, 3) \), and \( C(4, 5, 6) \), we will follow these steps:

### Step 1: Determine the vectors AB and BC
First, we need to find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).

\[ \overrightarrow{AB} = B - A = (0 - 1, 2 - 1, 3 - 2) = (-1, 1, 1) \]
\[ \overrightarrow{BC} = C - B = (4 - 0, 5 - 2, 6 - 3) = (4, 3, 3) \]

### Step 2: Find the normal vector to the plane
The normal vector \( \overrightarrow{N} \) to the plane can be found by taking the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).

\[ \overrightarrow{N} = \overrightarrow{AB} \times \overrightarrow{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 4 & 3 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ \overrightarrow{N} = \hat{i} \begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ 4 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 4 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} = 1 \cdot 3 - 1 \cdot 3 = 0 \)
2. \( \begin{vmatrix} -1 & 1 \\ 4 & 3 \end{vmatrix} = -1 \cdot 3 - 1 \cdot 4 = -3 - 4 = -7 \)
3. \( \begin{vmatrix} -1 & 1 \\ 4 & 3 \end{vmatrix} = -7 \) (same as above)

Thus, we have:

\[ \overrightarrow{N} = 0\hat{i} + 7\hat{j} + 7\hat{k} = (0, 7, 7) \]

### Step 3: Write the equation of the plane
The general equation of a plane in vector form is given by:

\[ \overrightarrow{N} \cdot (\overrightarrow{r} - \overrightarrow{r_0}) = 0 \]

Where \( \overrightarrow{r} = (x, y, z) \) and \( \overrightarrow{r_0} = (1, 1, 2) \).

Substituting \( \overrightarrow{N} = (0, 7, 7) \) and \( \overrightarrow{r_0} = (1, 1, 2) \):

\[ (0, 7, 7) \cdot ((x, y, z) - (1, 1, 2)) = 0 \]

This expands to:

\[ 0 \cdot (x - 1) + 7 \cdot (y - 1) + 7 \cdot (z - 2) = 0 \]

Which simplifies to:

\[ 7(y - 1) + 7(z - 2) = 0 \]

Further simplifying gives:

\[ y + z - 3 = 0 \]

### Step 4: Vector equation of the plane
The vector equation can be expressed as:

\[ y - z + 1 = 0 \]

### Final Answer
The vector equation of the plane passing through the points \( (1, 1, 2) \), \( (0, 2, 3) \), and \( (4, 5, 6) \) is:

\[ y - z + 1 = 0 \]