Find the directions ratios of the normal to the plane 2x+3y+z=7

Answer

To find the direction ratios of the normal to the plane given by the equation \(2x + 3y + z = 7\), we can follow these steps:

### Step 1: Identify the coefficients from the plane equation
The general form of a plane equation is given by:
\[ Ax + By + Cz = K \]
where \(A\), \(B\), and \(C\) are the coefficients of \(x\), \(y\), and \(z\) respectively, and \(K\) is a constant.

In our case, the equation of the plane is:
\[ 2x + 3y + z = 7 \]
From this equation, we can identify:
- \(A = 2\)
- \(B = 3\)
- \(C = 1\)

### Step 2: Write the normal vector
The normal vector to the plane can be represented as:
\[ \vec{n} = Ai + Bj + Ck \]
Substituting the values we found:
\[ \vec{n} = 2i + 3j + 1k \]

### Step 3: Determine the direction ratios
The direction ratios of the normal vector are simply the coefficients \(A\), \(B\), and \(C\). Thus, the direction ratios can be expressed as:
\[ \text{Direction Ratios} = A : B : C = 2 : 3 : 1 \]

### Conclusion
Therefore, the direction ratios of the normal to the plane \(2x + 3y + z = 7\) are:
\[ \boxed{2 : 3 : 1} \]

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