Find the area of the parallelogram with vertices A(−1,2,3), B(0,4,6), C(1,1,2), and D(2,3,5).

5*sqrt3

Step-by-step explanation:

The vector AB= (0-(-1), 4-2,6-3) AB= (1,2,3)

The modul of AB is sqrt(1^2+2^2+3^2)= sqrt14

The vector AC is (1-(-1), 1-2, 2-3)= (2,-1,-1)

The modul of B is sqrt (2^2+(-1)^2+(-1)^2)= sqrt6

AB*AC= modul AB*modul AC*cosA

cosA=( 1*2+2*(-1)+3*(-1))/ sqrt14*sqrt6= -3/sqrt84=

sinB= sqrt (1- (-3/sqrt84)^2)= sqrt75/84= sqrt 25/28= 5/sqrt28

s= modul AB*modul AC*sinA= sqrt14*sqrt6* 5/ sqrt28= 5*sqrt3

no i do not : ( what is it?

answer: its the third option.

step-by-step explanation:

i just took the test.