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).
cupoosta [38]
Answer:
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
POR N EASILY YDIPGIDILDGDIPDY
Answer:
15
Step-by-step explanation:
3a+9=54
-9 -9
3a=45
:3 :3
a=15
I do not get what the Question that you are asking is.
Answer:
the formula
Step-by-step explanation:
substitute the formula with the numbers give n-2 is number of sides -2 fyi
