Final answer:
To find the area of triangle ABC with given vertices, we calculate the cross product of vectors AB and AC, find its magnitude, and then take half of that to get the area. The area of triangle ABC is 22.5 square units.
Step-by-step explanation:
To determine the area of the triangle with vertices A (3,5,−1), B(7,4,2), and C(−3,−4,−7), we can use the cross product of vectors AB and AC and then find the magnitude of this cross product, since the area of the triangle is half of the area of the parallelogram formed by these two vectors. First, we find the vectors AB and AC:
- AB = B − A = (7 − 3, 4 − 5, 2 − (−1)) = (4, −1, 3)
- AC = C − A = (−3 − 3, −4 − 5, −7 − (−1)) = (−6, −9, −6)
Then we calculate the cross product AB × AC:
AB × AC = | i j k |
4 -1 3
−6 -9 -6
= i((-1)(-6) − (3)(-9)) − j((4)(-6) − (3)(−6)) + k((4)(-9) − (-1)(−6))
= i(6 + 27) − j(-24 + 18) + k(-36 + 6)
= i(33) − j(-6)+ k(-30).
The magnitude of the cross product, which is the area of the parallelogram, is:
|AB × AC| = √(33² + (−6)² + (−30)²) = √(1089 + 36 + 900) = √2025 = 45.
The area of the triangle is half of the area of the parallelogram, therefore:
Area of triangle ABC = ½|AB × AC| = ½ × 45 = 22.5 square units.