if equation of the plane is`(x-2)/(3)=(y-3)/(4)=(z-4)/(5)`convert this in vector equation of the plane

Answer

To convert the given equation of the plane \(\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}\) into vector form, we can follow these steps:

### Step 1: Identify the point and direction ratios
The given equation can be interpreted as follows:
- The point \(A(2, 3, 4)\) is a point on the plane.
- The direction ratios of the plane are given by the coefficients of \(x\), \(y\), and \(z\) in the equation, which are \(3\), \(4\), and \(5\) respectively.

### Step 2: Write the position vector of point A
The position vector \(\vec{A}\) of the point \(A(2, 3, 4)\) can be expressed as:
\[ \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]

### Step 3: Write the normal vector
The normal vector \(\vec{n}\) to the plane can be represented using the direction ratios:
\[ \vec{n} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

### Step 4: Write the vector equation of the plane
The vector equation of the plane can be expressed in the form:
\[ (\vec{r} - \vec{A}) \cdot \vec{n} = 0 \]
where \(\vec{r}\) is the position vector of any point on the plane. Substituting the values of \(\vec{A}\) and \(\vec{n}\):
\[ (\vec{r} - (2\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) = 0 \]

### Step 5: Expand the equation
Expanding the equation gives:
\[ \vec{r} \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) = 0 \]

### Step 6: Calculate the dot product
Now, we calculate the dot product:
\[ (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) = 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 = 6 + 12 + 20 = 38 \]

### Final Vector Equation
Thus, the vector equation of the plane can be written as:
\[ \vec{r} \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) - 38 = 0 \]
or equivalently,
\[ \vec{r} \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) = 38 \]