1. Find a unit vector which is perpendicular to both of the vectors m=[4, 2, -3] and n=[2, -3, 1]

| i j k |

| 4 2 -3 |
| 2 -3 1 |
= [-7,-10,-16]/9√5

thanks steve!

To find a unit vector perpendicular to two given vectors, you can use the cross product.

The cross product of two vectors, m and n, is denoted as m × n.

Here's how you can calculate it:

Step 1: Determine the cross product of the two vectors:
m × n = (m₂n₃ - m₃n₂, m₃n₁ - m₁n₃, m₁n₂ - m₂n₁)

Substituting the values from m=[4, 2, -3] and n=[2, -3, 1], we have:
m × n = [(2 * -3) - (-3 * 1), (-3 * 4) - (4 * 1), (4 * -3) - (2 * 1)]
= [-6 + 3, -12 - 4, -12 - 2]
= [-3, -16, -14]

Step 2: Find the magnitude of the cross product:
The magnitude of a vector is denoted as |v| or ∥v∥.

The magnitude of the cross product m × n can be calculated using the formula:
|m × n| = √(m₁² + m₂² + m₃²)

Substituting the values from m × n = [-3, -16, -14], we have:
|m × n| = √((-3)² + (-16)² + (-14)²)
= √(9 + 256 + 196)
= √461
≈ 21.47

Step 3: Normalize the cross product:
To obtain a unit vector, divide the cross product vector by its magnitude:
u = (1/|m × n|) * (m × n)

Substituting the values, we get:
u = (1/21.47) * [-3, -16, -14]
u ≈ [-0.14, -0.75, -0.65]

Therefore, a unit vector perpendicular to both m=[4, 2, -3] and n=[2, -3, 1] is approximately u = [-0.14, -0.75, -0.65].