tarea 1 continuacion

Hallar el área del triangulo que tiene vértices , P, Q, R.

P1= P (2, 0, -3)

P2= Q (1, 4, 5)

P3= R (7, 2, 9)

P1---P2= X2- X1= (1, 4, 5) – (2, 0, -3) = (-1, 4, 8)

P1---P3= X2-X1 = (7, 2, 9) – (2, 0 , -3) = ( 5, 2, 12)

A= ½ //P1—P2 x P1—P3//= (4)(12)- (2)(8)P + (-1)(12) – (5)(8)Q + (-1)(2)- (5)(4)R

= (48-16)P + (-12-40)Q + (-2-20)R

= 32P – 52Q – 22R

A

A= 56.95/2

A= 32.44 u²