To find two unit vectors orthogonal to the vectors (3, 2, 1) and (1, 1, 0), we can utilize the cross product of these two vectors. The cross product provides a vector that is perpendicular to both of the original vectors.
First, we represent the vectors as:
- Vector A: (3, 2, 1)
- Vector B: (1, 1, 0)
The cross product, denoted as A × B, is calculated using the determinant of the following matrix:
| i j k |
| 3 2 1 |
| 1 1 0 |
Calculating this determinant, we have:
- i component: (2 * 0 – 1 * 1) = -1
- j component: – (3 * 0 – 1 * 1) = -(-1) = 1
- k component: (3 * 1 – 2 * 1) = 1
So, the cross product A × B is the vector
(-1, 1, 1).
To find a unit vector in the direction of (-1, 1, 1), we need to calculate its magnitude:
Magnitude = √((-1)² + 1² + 1²) = √(1 + 1 + 1) = √3.
The unit vector in the direction of (-1, 1, 1) is obtained by dividing each of its components by the magnitude:
Unit Vector 1:
(-1/√3, 1/√3, 1/√3)
Since we need two unit vectors, the second one can be found by taking the negative of the first unit vector:
Unit Vector 2:
(1/√3, -1/√3, -1/√3)
Thus, the two unit vectors orthogonal to both (3, 2, 1) and (1, 1, 0) are:
- Vector 1: (-1/√3, 1/√3, 1/√3)
- Vector 2: (1/√3, -1/√3, -1/√3)