Find the vector equation of the plane passing through the points (1, 1, 1), (2, 4, 3) and (5,9,7).

Options

A. `vecr.(hati+hatj+2hatk)=0`
B. `vecr.(hati+hatj-2hatk)=2`
C. `vecr.(hati+hatj-2hatk)=0`
D. `vecr.(hati-hatj-2hatk)=4`

Answer

To find the vector equation of the plane passing through the points \( A(1, 1, 1) \), \( B(2, 4, 3) \), and \( C(5, 9, 7) \), we can follow these steps:

### Step 1: Define the position vectors
Let:
- \( \vec{A} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \)
- \( \vec{B} = \begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix} \)
- \( \vec{C} = \begin{pmatrix} 5 \\ 9 \\ 7 \end{pmatrix} \)

### Step 2: Find the direction vectors
Calculate the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = \vec{B} - \vec{A} = \begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} \]
\[ \vec{AC} = \vec{C} - \vec{A} = \begin{pmatrix} 5 \\ 9 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ 6 \end{pmatrix} \]

### Step 3: Calculate the normal vector using the cross product
The normal vector \( \vec{n} \) of the plane can be found by taking the cross product of \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} \times \begin{pmatrix} 4 \\ 8 \\ 6 \end{pmatrix} \]

Calculating the determinant:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 4 & 8 & 6 \end{vmatrix} \]

Expanding the determinant:
\[ \vec{n} = \hat{i}(3 \cdot 6 - 2 \cdot 8) - \hat{j}(1 \cdot 6 - 2 \cdot 4) + \hat{k}(1 \cdot 8 - 3 \cdot 4) \]
\[ = \hat{i}(18 - 16) - \hat{j}(6 - 8) + \hat{k}(8 - 12) \]
\[ = \hat{i}(2) - \hat{j}(-2) + \hat{k}(-4) \]
\[ = \begin{pmatrix} 2 \\ 2 \\ -4 \end{pmatrix} \]

### Step 4: Write the vector equation of the plane
The vector equation of the plane can be expressed as:
\[ \vec{r} - \vec{A} \cdot \vec{n} = 0 \]
Substituting \( \vec{A} \) and \( \vec{n} \):
\[ \vec{r} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ -4 \end{pmatrix} = 0 \]
This can be rewritten as:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ -4 \end{pmatrix} = 0 \]
\[ \Rightarrow 2(x - 1) + 2(y - 1) - 4(z - 1) = 0 \]
Simplifying gives:
\[ 2x + 2y - 4z = 0 \]

### Final Answer
The vector equation of the plane is:
\[ 2x + 2y - 4z = 0 \]

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