Using vectors, find the area of the triangle ABC with vertices A(1,2,3), B(2,-1,4), and C(4,5,-1).

We form the vectors AB and AC.
AB = (2-1, -1-2, 4-3) = (1, -3, 1)
AC = (4-1, 5-2, -1-3) = (3, 3, -4)
We find their cross product
AB x AC = | i j k |
| 1 -3 1 |
| 3 3 -4 |
= i((-3)(-4) - (1)(3)) - j((1)(-4) - (1)(3)) + k((1)(3) - (-3)(3))
= i(12 - 3) - j(-4 - 3) + k(3 + 9)
= 9i + 7j + 12k
So, the magnitude is √(9² + 7² + 12²) = √(81 + 49 + 144) = √274 ≈ 16.55
The triangle's area is 1/2 the area of that parallelogram.
So, Area of triangle = 1/2 × 16.55 ≈ 8.275 Sq. units