Задача 66730 ...

63667cd152ea863b5c7404b1

07.11.2022 16:55:53

Для треугольника ABC, построенного на векторах CA = p - q, CB = 2p+q, где |p| = 1, |q| = 2, ∠ ∝ =π/4. Найти длины сторон CA, CB и ∠ между ними, площадь треугольника, высоту h из точки B, медиану m из точки C, биссектрису n угла C

5f3ea7e3faf909182968ddd9

08.11.2022 14:14:06

★

|CA|^2=|vector{CA}|^2

|vector{CA}|^2=(vector{p}-vector{q})^2=(vector{p}-vector{q})*(vector{p}-vector{q})=vector{p}*vector{p}-vector{q}*vector{p}-vector{p}*vector{q}+vector{q}*vector{q}=

=|vector{p}|*|vector{p}|*cos0-2|vector{p}|*|vector{q}|*cos(π/4)+|vector{q}|*|vector{q}|*cos0=1*1*1-2*1*2*sqrt(2)/2+2*2*1=5-2sqrt(2)

|CA|=sqrt(5-2sqrt(2))



|CB|^2=|vector{CB}|^2

|vector{CB}|^2=(2vector{p}+vector{q})^2=(2vector{p}+vector{q})*(2vector{p}+vector{q})=4(vector{p}*(vector{p}+2(vector{q}*(vector{p}+2(vector{p}*(vector{q}+(vector{q}*(vector{q}=

=4|vector{p}|*|vector{p}|*cos0+4|vector{p}|*|vector{q}|*cos(π/4)+|vector{q}|*|vector{q}|*cos0=4*1*1*1+4*1*2*sqrt(2)/2+2*2*1=8+4sqrt(2)

|CB|=sqrt(8+4sqrt(2))



cos ∠ (vector{CA}^ vector{ CB})=(vector{CA}*vector{CB})/ (|vector{CA}|*|vector{CB}|)=


vector{CA}*vector{CB}=(vector{p}-vector{q})*(2*vector{p}+vector{q})=

=2*vector{p}*vector{p}-2*vector{q}*vector{p}+vector{p}*vector{q}-2*vector{q}*vector{q}=

=2*|vector{p}|*|vector{p}|*cos0+|vector{p}|*|vector{q}|*cos(π/4)-2|vector{q}|*|vector{q}|*cos0=4*1*1*1+1*2*sqrt(2)/2-2*2*1=sqrt(2)



cos ∠ (vector{CA}^ vector{ CB})=sqrt(2)/(sqrt(5-2sqrt(2))*sqrt(8+4sqrt(2)))


S_( Δ АВС)=(1/2)| (vector{CA} × vector{CB})|


vector{CA} × vector{CB}=(vector{p}-vector{q}) × (2vector{p}+vector{q})=

=2*vector{p} × vector{p}-2*vector{q} × vector{p}+vector{p} × vector{q}-2*vector{q} × vector{q}=

так как

vector{q} × vector{p}=-vector{p} × vector{q} ⇒

vector{CA} × vector{CB}=2*vector{p} × vector{p}+3*vector{q} × vector{p}+vector{p} × vector{q}-2*vector{q} × vector{q}

=3*vector{q} × vector{p}


|vector{CA} × vector{CB}|=

=3*|vector{p}}* |vector{q}|*sin(π/4)=3*1*2*sqrt(2)/2=3sqrt(2)


S_( Δ АВС)=(1/2)*3sqrt(2)=[b]3sqrt(2)/2[/b]


S_( Δ АВС)=(1/2)*AC* h_(b)


h_(b)=2S_( Δ АВС)/AC=3sqrt(2)/sqrt(5-2sqrt(2))