Задача 30872 2 задача. Надо решающими векторным...

churalena

2018-11-08 20:43:37

2 задача. Надо решающими векторным способом!

sova

2018-11-09 11:40:55

Пусть vector{AD}=vector{a}; vector{AB}=vector{b}; vector{OS}=vector{c};
причем
|vector{a}|=|vector{b}|=4sqrt(2);
|vector{c}|=3.

vector{a}*vector{b}=0;
vector{a}*vector{c}=0;
vector{b}*vector{c}=0.

vector{BO}=(1/2)*vector{BD}=(1/2)*(vector{a}-vector{b})=(1/2)*vector{a}-(1/2)*vector{b};

vector{BS}=vector{BO}+vector{OS}=(1/2)*vector{a}-(1/2)*vector{b}+vector{c};

vector{AN}=vector{AB}+vector{BN}=vector{b}+(1/2)*vector{a}-(1/2)*vector{b}+vector{c}=
=[b](1/2)*vector{a}+(1/2)*vector{b}+vector{c}[/b]

vector{MC}=vector{MD}+vector{DC}=(1/2)*vector{AD}+vector{DC}=
=[b](1/2)*vector{a}+vector{b}[/b];

Применяем скалярное произведение векторов.

vector{AN}*vector{MC}=|vector{AN}|*|vector{MC}|*cos(vector{AN},vector{MC})
⇒
cos(vector{AN},vector{MC}) = vector{AN}*vector{MC}/|vector{AN}|*|vector{MC}|

vector{AN}*vector{MC}=((1/2)*vector{a}+(1/2)*vector{b}+vector{c})*((1/2)*vector{a}+vector{b})=

=(1/2)*vector{a}*(1/2)*vector{a}+(1/2)*vector{a}*vector{b}+
+(1/2)*vector{b}*(1/2)*vector{a}+(1/2)*vector{b}*vector{b}+
+vector{c}*(1/2)*vector{a}+vector{c}*vector{b}=

=(1/4)*|vector{a}|^2+0+0+(1/2)|vector{b}|^2+0+0=

=(1/4)*(4sqrt(2))^2+(1/2)*(4sqrt(2))^2=8+16=24;

[b]vector{AN}*vector{MC}=24[/b]

|vector{AN}|^2=((1/2)*vector{a}+(1/2)*vector{b}+vector{c})*((1/2)*vector{a}+(1/2)*vector{b}+vector{c})=

=(1/4)*|vector{a}|^2+0+0+(1/4)*|vector{b}|^2+0+0+|vector{с}|^2=

=8+8+9=25

[b]|vector{AN}|=5[/b]

|vector{МС}|^2=((1/2)*vector{a}+vector{b})*((1/2)*vector{a}+(vector{b})=

=(1/4)*|vector{a}|^2+|vector{b}|^2=

=8+32=40

[b]|vector{MC}|=2sqrt(10)[/b].

О т в е т. cos∠ (vector{AN},vector{MC}) =24/(5*sqrt(40))
∠ (vector{AN},vector{MC})=arccos (12/(5sqrt(10))