Find the values of `gamma` and `mu` for which `(2hati+6hatj+27hatk)xx(hati+gamma hatj+mu hatk)=vec0`

Answer

To find the values of \( \gamma \) and \( \mu \) for which the cross product

\[ (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \gamma \hat{j} + \mu \hat{k}) = \vec{0} \]

we can follow these steps:

### Step 1: Set up the cross product
The cross product can be expressed as a determinant:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \gamma & \mu \end{vmatrix} \]

### Step 2: Calculate the determinant
Using the determinant formula for a 3x3 matrix, we can expand this:

\[ = \hat{i} \begin{vmatrix} 6 & 27 \\ \gamma & \mu \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 27 \\ 1 & \mu \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 6 \\ 1 & \gamma \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ 6\mu - 27\gamma \]

2. For \( -\hat{j} \):
\[ 2\mu - 27 \]

3. For \( \hat{k} \):
\[ 2\gamma - 6 \]

Putting it all together, we have:

\[ (6\mu - 27\gamma)\hat{i} - (2\mu - 27)\hat{j} + (2\gamma - 6)\hat{k} \]

### Step 3: Set the cross product equal to zero
Since the cross product is equal to the zero vector, we can set each component to zero:

1. \( 6\mu - 27\gamma = 0 \) (Equation 1)
2. \( -2\mu + 27 = 0 \) (Equation 2)
3. \( 2\gamma - 6 = 0 \) (Equation 3)

### Step 4: Solve the equations
**From Equation 2:**
\[ -2\mu + 27 = 0 \implies 2\mu = 27 \implies \mu = \frac{27}{2} \]

**From Equation 3:**
\[ 2\gamma - 6 = 0 \implies 2\gamma = 6 \implies \gamma = 3 \]

### Step 5: Verify the results
Substituting \( \mu = \frac{27}{2} \) and \( \gamma = 3 \) back into Equation 1:

\[ 6\left(\frac{27}{2}\right) - 27(3) = 81 - 81 = 0 \]

This confirms that our values are correct.

### Final Values
Thus, the values are:

\[ \mu = \frac{27}{2}, \quad \gamma = 3 \]