Find the unit vector perpendicular to each of the vectors 6i + 2j + 3k and 3i - 2k

To find the unit vector perpendicular to both the given vectors, we need to:

1. Find the cross product of the two vectors, which will give a vector perpendicular to both.
2. Normalize the cross product to get the unit vector.

### Step 1: Find the cross product of the two vectors
Given vectors:
- **A** = 6i + 2j + 3k
- **B** = 3i - 2k

The cross product **A × B** is calculated as the determinant of the following matrix:

| i j k |
|-------------|
| 6 2 3 |
| 3 0 -2 |

Now calculate the determinant:

**A × B** = i[(2)(-2) - (3)(0)] - j[(6)(-2) - (3)(3)] + k[(6)(0) - (2)(3)]

Simplifying each component:

- i: (2)(-2) - (3)(0) = -4
- j: (6)(-2) - (3)(3) = -12 - 9 = -21
- k: (6)(0) - (2)(3) = -6

Thus, the cross product is:

**A × B** = -4i + 21j - 6k

### Step 2: Find the magnitude of the cross product
The magnitude of the cross product **|A × B|** is:

|A × B| = √((-4)² + (21)² + (-6)²)

|A × B| = √(16 + 441 + 36) = √493

|A × B| = √493 ≈ 22.2

### Step 3: Find the unit vector
The unit vector **u** is the normalized vector, so we divide each component of **A × B** by its magnitude:

u = (1/|A × B|) * (A × B)

u = (1/22.2) * (-4i + 21j - 6k)

u ≈ -0.18i + 0.95j - 0.27k

Thus, the unit vector perpendicular to both vectors is approximately:

u ≈ -0.18i + 0.95j - 0.27k