Find the area of a triangle with the given vertices
(5,-1,2), (7,-4-2), (2,-6,3)
Let (k, m, n), (r, s, t), and (u, v, w) = (5,-1,2), (7,-4-2), (2,-6,3)
Then....the area can be calculated thusly :
(1/2)absolute value of det [ 1 1 1
r-k s-m t-n
u-k v-m w-n ]
(1/2) abs det [ 1 1 1
2 -3 -4
-3 -5 1 ] write the first two columns again and calculate the determinant
(1/2) 1 1 1 1 1
2 -3 -4 2 -3
-3 -5 1 -3 -5
(1/2) abs ( [ 1 * -3 *1 + 1* -4* -3 + 1*2*-5] - [ -3*-3*1 + -5*-4*1 + 1*2*1] ) =
(1/2) abs ( [ -3 + 12 -10 ] - [ 9 + 20 + 2 ] ) =
(1/2) abs(-32) =
(1/2 (32) =
16 sq units