If `veca=2hati+6hatj+mhatk` and `vecb=nhati+18hatj+3hatk` are parallel to each other then values of m,n are

Options

A. 1,6
B. 6,1
C. `-1,6`
D. `-1,-6`

Explanation

To solve the problem, we need to find the values of \( m \) and \( n \) such that the vectors \( \vec{a} = 2\hat{i} + 6\hat{j} + m\hat{k} \) and \( \vec{b} = n\hat{i} + 18\hat{j} + 3\hat{k} \) are parallel.

### Step-by-step Solution:

1. **Understanding Parallel Vectors**:
Two vectors \( \vec{a} \) and \( \vec{b} \) are parallel if there exists a scalar \( \lambda \) such that:
\[ \vec{a} = \lambda \vec{b} \]

2. **Setting Up the Equation**:
Substitute the expressions for \( \vec{a} \) and \( \vec{b} \):
\[ 2\hat{i} + 6\hat{j} + m\hat{k} = \lambda (n\hat{i} + 18\hat{j} + 3\hat{k}) \]

3. **Equating Components**:
From the equation above, we can equate the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \):
- For \( \hat{i} \):
\[ 2 = \lambda n \quad \text{(1)} \]
- For \( \hat{j} \):
\[ 6 = \lambda \cdot 18 \quad \text{(2)} \]
- For \( \hat{k} \):
\[ m = \lambda \cdot 3 \quad \text{(3)} \]

4. **Finding \( \lambda \)**:
From equation (2), we can solve for \( \lambda \):
\[ \lambda = \frac{6}{18} = \frac{1}{3} \]

5. **Substituting \( \lambda \) Back**:
Substitute \( \lambda \) into equation (1) to find \( n \):
\[ 2 = \left(\frac{1}{3}\right) n \implies n = 2 \cdot 3 = 6 \]

6. **Finding \( m \)**:
Now substitute \( \lambda \) into equation (3) to find \( m \):
\[ m = \left(\frac{1}{3}\right) \cdot 3 = 1 \]

7. **Final Values**:
Thus, we find:
\[ m = 1, \quad n = 6 \]

### Conclusion:
The values of \( m \) and \( n \) such that the vectors \( \vec{a} \) and \( \vec{b} \) are parallel are:
\[ m = 1, \quad n = 6 \]