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

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

step-by-step explanation:

se a cavasery6