Find a vector of magnitude 6 which is perpendicular to both the vectors
`vec(a)= 4 hat(i)-hat(j) + 3 hat(k)` and `vec(b) = -2 hat(i) + hat(j)- 2 hat(k).`

Explanation

To find a vector of magnitude 6 that is perpendicular to both vectors \(\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}\), we will use the cross product of the two vectors. The steps are as follows:

### Step 1: Calculate the Cross Product \(\vec{a} \times \vec{b}\)

The formula for the cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of the vectors \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \]

### Step 2: Calculate the Determinant

To calculate the determinant, we expand it as follows:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -1 & 3 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 3 \\ -2 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & -1 \\ -2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ (-1)(-2) - (3)(1) = 2 - 3 = -1 \]

2. For \(\hat{j}\):
\[ (4)(-2) - (3)(-2) = -8 + 6 = -2 \]

3. For \(\hat{k}\):
\[ (4)(1) - (-1)(-2) = 4 - 2 = 2 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = -1\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 3: Calculate the Magnitude of \(\vec{a} \times \vec{b}\)

Now, we need to find the magnitude of the cross product:
\[ |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Find the Unit Vector in the Direction of \(\vec{a} \times \vec{b}\)

The unit vector in the direction of \(\vec{a} \times \vec{b}\) is given by:
\[ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-1\hat{i} + 2\hat{j} + 2\hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Step 5: Scale the Unit Vector to Magnitude 6

To find a vector of magnitude 6, we multiply the unit vector by 6:
\[ \vec{v} = 6\hat{n} = 6\left(-\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right) = -2\hat{i} + 4\hat{j} + 4\hat{k} \]

### Step 6: Consider the Negative Direction

Since the problem asks for a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), we can also take the negative of this vector:
\[ \vec{v} = \pm(-2\hat{i} + 4\hat{j} + 4\hat{k}) \]

### Final Answer

Thus, the required vector of magnitude 6 that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) is:
\[ \vec{v} = -2\hat{i} + 4\hat{j} + 4\hat{k} \quad \text{or} \quad \vec{v} = 2\hat{i} - 4\hat{j} - 4\hat{k} \]