Find the unit vectors perpendicular to the plane of the vectors
`vec(a) = 2 hat(i) - 6 hat(j)- 3 hat(k)` and ` vec(b)= 4 hat(i) + 3 hat(j) - hat(k).`

Explanation

To find the unit vectors perpendicular to the plane of the vectors \(\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}\) and \(\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}\), we will follow these steps:

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

The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{vmatrix} \]

### Step 2: Evaluate the Determinant

Calculating the determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} = (-6)(-1) - (-3)(3) = 6 + 9 = 15 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} = (2)(-1) - (-3)(4) = -2 + 12 = 10 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} = (2)(3) - (-6)(4) = 6 + 24 = 30 \]

Putting it all together, we have:

\[ \vec{a} \times \vec{b} = 15\hat{i} - 10\hat{j} + 30\hat{k} \]

### Step 3: Find the Magnitude of the Cross Product

Now, we need to find the magnitude of \(\vec{a} \times \vec{b}\):

\[ |\vec{a} \times \vec{b}| = \sqrt{(15)^2 + (-10)^2 + (30)^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35 \]

### Step 4: Find the Unit Vector

The unit vector perpendicular to the plane formed by \(\vec{a}\) and \(\vec{b}\) is given by:

\[ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{15\hat{i} - 10\hat{j} + 30\hat{k}}{35} \]

This simplifies to:

\[ \hat{n} = \frac{15}{35}\hat{i} - \frac{10}{35}\hat{j} + \frac{30}{35}\hat{k} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \]

### Step 5: Consider the Negative Unit Vector

Since the question asks for unit vectors perpendicular to the plane, we also consider the negative of the unit vector:

\[ -\hat{n} = -\left(\frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k}\right) = -\frac{3}{7}\hat{i} + \frac{2}{7}\hat{j} - \frac{6}{7}\hat{k} \]

### Final Answer

Thus, the unit vectors perpendicular to the plane of the vectors \(\vec{a}\) and \(\vec{b}\) are:

\[ \hat{n} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \quad \text{and} \quad -\hat{n} = -\frac{3}{7}\hat{i} + \frac{2}{7}\hat{j} - \frac{6}{7}\hat{k} \]