Question

Из условия DM-EF+ED+MK+x=PK-PC+FA(векторы) найдите вектор x

 

Answer 1

(ED-EF)+(DM+MK)+x=(PK-PC)+FA

FD+DK+x=CK+FA

FK+x=CK+FA

x=CK+(FA-FK)

x=CK+KA

x=CA 

Answer 2

$overrightarrow{DM}-overrightarrow{EF}+overrightarrow{ED}+overrightarrow{MK}+vec {x}=overrightarrow{PK}-overrightarrow{PC}+overrightarrow{FA}\\vec {x}=overrightarrow{PK}-overrightarrow{PC}+overrightarrow{FA}-overrightarrow{DM}+overrightarrow{EF}-overrightarrow{ED}-overrightarrow{MK}\\vec {x}=overrightarrow{PK}+overrightarrow{CP}+overrightarrow{FA}+overrightarrow{MD}+overrightarrow{EF}+overrightarrow{DE}+overrightarrow{KM}\\vec{x}=underbrace{overrightarrow{CP}+overrightarrow{PK}+overrightarrow{KM}+overrightarrow{MD}+overrightarrow{DE}+overrightarrow{EF}+overrightarrow{FA}}_{overrightarrow{CA}}$

$boxed{boldsymbol{vec{x}=overrightarrow{CA}}}$