Задача 22482 4. Найти площадь параллелограмма,...

slava191

16.01.2018

4. Найти площадь параллелограмма, построенного на векторах
vector{a} = 3vector{p} + vector{q} и vector{b} = vector{р} - 2vector{q}, где |vector{р}| = 4, |vector{q}| = 1, угол между vector{p}, vector{q} = Pi/4

SOVA

16.01.2018

Находим скалярное произведение векторов vector{a} и vector{b}.

vector{a}*vector{b}=(3vector{p} +vector{q} )*(vector{р} – 2vector{q})=3vector{p}*vector{р}+vector{q}*vector{р} -6vector{p}*vector{q}-2vector{q} vector{q} =

=3*| vector{p}|^2-5|vector{р}|*| vector{q}|*cos(Pi/4)-2*|vector{q}|^2=

=3*16 - 5*4*1*sqrt(2)/2-2*1=46-10sqrt(2)

Находим косинус угла между векторами
cos(vector{a}^vector{b})=(vector{a}*vector{b})/|(vector{a}|*|vector{b}|

Находим длины векторов vector{a} и vector{b}

|vector{a}|^2=vector{a}*vector{a}=(3vector{p} +vector{q} )*(3vector{p} +vector{q} )*=9vector{p}*vector{р}+3vector{q}*vector{р} +3vector{p}*vector{q}+vector{q} vector{q} =
=9| vector{p}|^2+6|vector{р}|*| vector{q}|*cos(Pi/4)+|vector{q}|^2=
=9*16 +6*4*1*sqrt(2)/2+1=144+12sqrt(2)

|vector{b}|^2=vector{b}*vector{b}=(vector{p} -2vector{q} )*(vector{p} -2vector{q}=
=vector{p}*vector{р}-4vector{q}*vector{р} +4vector{q} vector{q} =
=| vector{p}|^2-4|vector{р}|*| vector{q}|*cos(Pi/4)+4|vector{q}|^2=
=16 -4*4*1*sqrt(2)/2+4=20-16sqrt(2)

cos(vector{a}^vector{b})=(46-10sqrt(2))/(144+12sqrt(2))*(20-16sqrt(2))=


sin(vector{a}^vector{b})=

S=|(vector{a}|*|vector{b}|*sin(vector{a}^vector{b})