Find the area of the parallelogram whose vertices are...

Question

Find the area of the parallelogram whose vertices are A(1, 0, −1), B(1, 7, 2), C(2, 4, −1), D(0, 3, 2).

Answer 1

Vector AB = (1 - 1, 7 - 0, 2 + 1) = (0, 7, 3) = 0i + 7j + 3k; vector AD= (0 - 1, 3 - 0, 2 + 1) =

(- 1, 3, 3)= - i 3j + 3k.

Then area A of parallelogram A = abs(AB×AD) = abs{det[(i j k) (0 7 3) (- 1, 3, 3)]} =

abs(12i -3j + 7k) = (12² + (- 3)² + 7²)^1/2= 202^1/2 ≈ 14.213