The unit vector perpendicular to the vectors `6hati+2hatj+3hatk and 3hati-6hatj-2hatk`, is

Options

A. `(2hati-3hatj+6hatk)/(7)`
B. `(2hati-3hatj-6hatk)/(7)`
C. `(2hati+3hatj-6hatk)/(7)`
D. `(2hati+3hatj+6hatk)/(7)`

Answer

To find the unit vector that is perpendicular to the vectors \( \mathbf{A} = 6\hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{B} = 3\hat{i} - 6\hat{j} - 2\hat{k} \), we will follow these steps:

### Step 1: Define the vectors
Let:
\[ \mathbf{A} = 6\hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \mathbf{B} = 3\hat{i} - 6\hat{j} - 2\hat{k} \]

### Step 2: Compute the cross product \( \mathbf{A} \times \mathbf{B} \)
The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 2 & 3 \\ 3 & -6 & -2 \end{vmatrix} \]

Calculating this determinant, we have:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 2 & 3 \\ -6 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & 3 \\ 3 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 2 \\ 3 & -6 \end{vmatrix} \]

Calculating the minors:
- For \( \hat{i} \):
\[ \begin{vmatrix} 2 & 3 \\ -6 & -2 \end{vmatrix} = (2)(-2) - (3)(-6) = -4 + 18 = 14 \]

- For \( \hat{j} \):
\[ \begin{vmatrix} 6 & 3 \\ 3 & -2 \end{vmatrix} = (6)(-2) - (3)(3) = -12 - 9 = -21 \]

- For \( \hat{k} \):
\[ \begin{vmatrix} 6 & 2 \\ 3 & -6 \end{vmatrix} = (6)(-6) - (2)(3) = -36 - 6 = -42 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 14\hat{i} + 21\hat{j} - 42\hat{k} \]

### Step 3: Calculate the magnitude of \( \mathbf{A} \times \mathbf{B} \)
The magnitude of the vector \( \mathbf{C} = \mathbf{A} \times \mathbf{B} \) is given by:
\[ |\mathbf{C}| = \sqrt{(14)^2 + (21)^2 + (-42)^2} \]
Calculating this:
\[ |\mathbf{C}| = \sqrt{196 + 441 + 1764} = \sqrt{2401} = 49 \]

### Step 4: Find the unit vector
The unit vector \( \mathbf{c} \) in the direction of \( \mathbf{C} \) is given by:
\[ \mathbf{c} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{14\hat{i} + 21\hat{j} - 42\hat{k}}{49} \]

### Step 5: Simplify the unit vector
Simplifying gives:
\[ \mathbf{c} = \frac{14}{49}\hat{i} + \frac{21}{49}\hat{j} - \frac{42}{49}\hat{k} = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \]

### Final Answer
Thus, the unit vector perpendicular to the given vectors is:
\[ \mathbf{c} = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \]

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