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

Given vertices of ΔABC are A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1)
So, AB = OB - OA = (2i - j + 4k) - (i + 2j + 3k) = i - 3j + k
and AC = OC - OA = (4i + 5j - k) - (i + 2j + 3k) = 3i + 3j - 4k
Now AB × AC = | i j k |
| 1 -3 1 |
| 3 3 -4 |
= 9i + 7j + 12k
⇒ |AB × AC| = √9² + 7² + 12² = √274
ar(ABC) = 1/2 |AB × AC| = 1/2 √274 Sq.units.