Find the vector equation of a plane passing through the point having position vector ` 2 hat(i) + hat(j) + hat(k)` and perpendicular to the vector :
`4 hat(i) - 2 hat(j) + 3hat(k)`.

Answer

To find the vector equation of a plane that passes through a given point and is perpendicular to a specified vector, we can follow these steps:

### Step 1: Identify the given information
We are given:
- A point \( A \) with position vector \( \mathbf{r_0} = 2\hat{i} + \hat{j} + \hat{k} \).
- A normal vector \( \mathbf{n} = 4\hat{i} - 2\hat{j} + 3\hat{k} \).

### Step 2: Write the general equation of a plane
The vector equation of a plane can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{r_0} \cdot \mathbf{n} \]
where \( \mathbf{r} \) is the position vector of any point on the plane, \( \mathbf{n} \) is the normal vector, and \( \mathbf{r_0} \) is the position vector of a specific point on the plane.

### Step 3: Calculate \( \mathbf{r_0} \cdot \mathbf{n} \)
We need to compute \( \mathbf{r_0} \cdot \mathbf{n} \):
\[ \mathbf{r_0} \cdot \mathbf{n} = (2\hat{i} + \hat{j} + \hat{k}) \cdot (4\hat{i} - 2\hat{j} + 3\hat{k}) \]
Calculating this dot product:
\[ = 2 \cdot 4 + 1 \cdot (-2) + 1 \cdot 3 \]
\[ = 8 - 2 + 3 = 9 \]

### Step 4: Write the equation of the plane
Now that we have \( \mathbf{r_0} \cdot \mathbf{n} = 9 \), we can write the vector equation of the plane:
\[ \mathbf{r} \cdot (4\hat{i} - 2\hat{j} + 3\hat{k}) = 9 \]

### Final Answer
The vector equation of the plane is:
\[ \mathbf{r} \cdot (4\hat{i} - 2\hat{j} + 3\hat{k}) = 9 \]

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