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

Options

A. `+- ( 4 hat(i) + 3 hat(j) - hat(k))/(sqrt(26))`
B. `+- ( 2 hat(i) - 6 hat(j) - 3 hat(k))/( 7)`
C. `+- ( 2 hat(i) - 3 hat(j) +6 hat(k))/( 7)`
D. `+- ( 3 hat(i) -2 hat(j) + 6 hat(k))/(7)`

Answer

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: Expand the determinant

Using the determinant formula, 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 these 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: Calculate the magnitude of \(\vec{a} \times \vec{b}\)

The magnitude of the cross product is given by:

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

Calculating each term:

\[ 15^2 = 225, \quad (-10)^2 = 100, \quad 30^2 = 900 \]

Thus,

\[ |\vec{a} \times \vec{b}| = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35 \]

### Step 4: Find the unit vector

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

\[ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]

Substituting the values we calculated:

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

This simplifies to:

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

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

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