Vector equation of the plane `r = hati-hatj+ lamda(hati +hatj+hatk)+mu(hati – 2hatj+3hatk)` in the scalar dot product form is

Options

A. `rcdot(5i-2j+3k)=7`
B. `rcdot(5i2j-3k)=7`
C. `rcdot(5i-2j-3k)=7`
D. `rcdot(5i+2j+3k)=7`

Answer

To convert the given vector equation of the plane into scalar dot product form, we follow these steps:

### Step 1: Identify the point and direction vectors
The vector equation given is:
\[ \mathbf{r} = \hat{i} - \hat{j} + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k}) \]
From this equation, we can identify:
- A point \( P \) on the plane: \( \hat{i} - \hat{j} \)
- Two direction vectors:
- \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{B} = \hat{i} - 2\hat{j} + 3\hat{k} \)

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{N} \) to the plane can be found using the cross product of the two direction vectors \( \mathbf{A} \) and \( \mathbf{B} \).

\[ \mathbf{N} = \mathbf{A} \times \mathbf{B} \]

Calculating the cross product using the determinant:
\[ \mathbf{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} \]

Calculating the determinant:
1. For \( \hat{i} \):
\[ \hat{i} \cdot \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} = \hat{i}(1 \cdot 3 - 1 \cdot -2) = \hat{i}(3 + 2) = 5\hat{i} \]

2. For \( -\hat{j} \):
\[ -\hat{j} \cdot \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = -\hat{j}(1 \cdot 3 - 1 \cdot 1) = -\hat{j}(3 - 1) = -2\hat{j} \]

3. For \( \hat{k} \):
\[ \hat{k} \cdot \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = \hat{k}(1 \cdot -2 - 1 \cdot 1) = \hat{k}(-2 - 1) = -3\hat{k} \]

Thus, the normal vector is:
\[ \mathbf{N} = 5\hat{i} - 2\hat{j} - 3\hat{k} \]

### Step 3: Use the point-normal form of the plane
The point-normal form of a plane is given by:
\[ (\mathbf{r} - \mathbf{p}) \cdot \mathbf{n} = 0 \]
Where \( \mathbf{p} \) is a point on the plane and \( \mathbf{n} \) is the normal vector.

Substituting \( \mathbf{p} = \hat{i} - \hat{j} \) and \( \mathbf{n} = 5\hat{i} - 2\hat{j} - 3\hat{k} \):
\[ (\mathbf{r} - (\hat{i} - \hat{j})) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = 0 \]

### Step 4: Expand the equation
Let \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \):
\[ ((x - 1)\hat{i} + (y + 1)\hat{j} + z\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = 0 \]

Calculating the dot product:
\[ (x - 1) \cdot 5 + (y + 1) \cdot (-2) + z \cdot (-3) = 0 \]
\[ 5(x - 1) - 2(y + 1) - 3z = 0 \]

### Step 5: Simplify the equation
Expanding and simplifying:
\[ 5x - 5 - 2y - 2 - 3z = 0 \]
\[ 5x - 2y - 3z - 7 = 0 \]

### Final Scalar Dot Product Form
Thus, the scalar dot product form of the plane is:
\[ 5x - 2y - 3z = 7 \]