Find the Cartesian equations of a plane passing through A (1,2,3) and directions ratios of it's normal are 3,2,5.

Answer

To find the Cartesian equation of a plane that passes through the point \( A(1, 2, 3) \) and has direction ratios of its normal as \( 3, 2, 5 \), we can follow these steps:

### Step 1: Identify the normal vector and the point
The normal vector \( \mathbf{n} \) can be represented using its direction ratios:
\[ \mathbf{n} = 3\mathbf{i} + 2\mathbf{j} + 5\mathbf{k} \]
The point through which the plane passes is given as:
\[ A(1, 2, 3) \]

### Step 2: Write the general equation of the plane
The general equation of a plane can be expressed in the form:
\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0 \]
where \( \mathbf{r} \) is a position vector of any point \( (x, y, z) \) on the plane, and \( \mathbf{a} \) is the position vector of the point \( A \).

### Step 3: Substitute the values into the equation
The position vector \( \mathbf{r} \) can be written as:
\[ \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \]
The position vector \( \mathbf{a} \) for point \( A(1, 2, 3) \) is:
\[ \mathbf{a} = 1\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \]
Now substituting into the equation:
\[ (3\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) \cdot ((x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) - (1\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})) = 0 \]

### Step 4: Simplify the equation
This simplifies to:
\[ (3\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) \cdot ((x - 1)\mathbf{i} + (y - 2)\mathbf{j} + (z - 3)\mathbf{k}) = 0 \]
Now, performing the dot product:
\[ 3(x - 1) + 2(y - 2) + 5(z - 3) = 0 \]

### Step 5: Expand and rearrange the equation
Expanding this gives:
\[ 3x - 3 + 2y - 4 + 5z - 15 = 0 \]
Combining like terms results in:
\[ 3x + 2y + 5z - 22 = 0 \]
Thus, the Cartesian equation of the plane is:
\[ 3x + 2y + 5z = 22 \]

### Final Answer
The Cartesian equation of the plane is:
\[ 3x + 2y + 5z = 22 \]

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