Задача 30509 Найдите координаты вектора MN, где точки...

vk442921024

2018-10-25 00:22:00

Найдите координаты вектора MN, где точки M и N лежат на отрезках AC и BD, причем AM/MC = 1/2 и BN/BD = 2/3.

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sova

2018-10-25 13:28:09

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1.
АМ:МC=1/2 ⇒ λ =1/2
x_(M)=(x_(A)+ λ x_(C))/(1+ λ )=(7+(1/2)*(-3))/(3/2)=11/3
y_(M)=(y_(A)+ λ y_(C))/(1+ λ )=(0+(1/2)*7)/(3/2)=7/3
z_(M)=(z_(A)+ λ z_(C))/(1+ λ )=(-1+(1/2)*3)/(3/2)=1/3
BN:ND=2/3 ⇒ λ =2/3
x_(N)=(x_(B)+ λ x_(D))/(1+ λ )=(-6+(2/3)*2)/(5/3)=-8/5;
y_(N)=(y_(B)+ λ y_(D))/(1+ λ )=(-8+(2/3)*(-8))/(5/3)=-8
z_(N)=(z_(B)+ λ z_(D))/(1+ λ )=(-5+(2/3)*3)/(5/3)=-1/5

vector{MN}=((-8/5)-(11/3); -8-(7/3); (-1/5)-(1/3))=(-79/15; -31/3; -8/15)
2.
vector{AC}=(-3-7;7-0;3-(-1))=(-10;7;4)
|vector{AC}|=sqrt((-10)^2+7^2+4^2)=sqrt(165)
3.
vector{BD}=(2-(-6);-8-(-8);3-(-5))=(8;0;8)
|vector{BD}|=8sqrt(2)
сos α =8/8sqrt(2)=1/sqrt(2)
cos β=0
cos γ 8/8sqrt(2)=1/sqrt(2)

4.
vector{AB}=(-13;-7;-4)
vector{DA}=(5;8;-4)

vector{AB}*vector{DA}=(-13)*5+(-7)*8+(-4)*(-4)=-65-56+16=-105

5.
пр_(vector{CA})(vector{BD}=|vector{BD}|*cos(vector{CA},vector{BD})=

=(vector{CA}*vector{BD})/|vector{CA}|=

=(10*8+(-7)*0+4*8)/sqrt(165)=128/sqrt(165)

6.
vector{BC}=(-3-(-6);7-(-8);3-(-5))=(3;15;8)
|vector{BC}|=sqrt(3^2+15^2+8^2)=sqrt(298)
vector{AD}=-vector{DA}=(-5;-8;4)
|vector{BC}|=sqrt((-5)^2+(-8)^2+4^2)=sqrt(105)

cos(vector{BC},vector{AD})=(vector{BC}*vector{AD})/(|vector{BC}|*|vector{AD}|)=

=(3*(-5)+15*(-8)+8*4)/(sqrt(298)*sqrt(105))=-103/(sqrt(298)*sqrt(105))

7.vector{СB}=(-3;-15;-8)
vector{ВA}=(13;7;4)
далее см приложение

8.
vector{AB}=(-13;-7;-4)
vector{AC}=(-10;7;4)

далее см. приложение