Задача 62241 Найти диагонали параллелограмма...

61bc23f72529d46a395c741a

17.12.2021 18:01:07

Найти диагонали параллелограмма построенную на векторах 5a+b и a-3b, если |а|=2sqrt(2) , |b|=3 , а угол векторов а и b 45°.

5f3ea7e3faf909182968ddd9

17.12.2021 18:13:25

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vector{AC}=vector{AB}+vector{BC}=5 vector{a}+vector{b}+(vector{a}–3 vector{b})=6 vector{a}–2 vector{b}

vector{BD}=vector{AD}-vector{AB}=5 vector{a}+vector{b}-(vector{a}–3 vector{b})=4 vector{a}+4 vector{b}

|vector{AC}|^2=vector{AC}*vector{AC}=(6 vector{a}–2 vector{b})*(6 vector{a}–2 vector{b})=36 vector{a}*vector{a}-24 vector{a}*vector{b}+4 vector{b}*vector{b}=36*| vector{a}|*|vector{a}|cos0 ° -24*| vector{a}|*|vector{b}|cos45 °+4 |vector{b}|*|vector{b}|=

=36*(2√2)*(2√2)*1-24*(2√2)*3*(√2/2)+4*3*3*1=36*8-144+36=180

|vector{AC}|=sqrt(180}=sqrt{36*5}=6sqrt(5)

|vector{BD}|^2=vector{BD}*vector{BD}=(4 vector{a}+4 vector{b})*(4 vector{a}+4 vector{b})=16 vector{a}*vector{a}+32 vector{a}*vector{b}+16 vector{b}*vector{b}=16*| vector{a}|*|vector{a}|cos0 ° +32*| vector{a}|*|vector{b}|cos45 °+16 |vector{b}|*|vector{b}|=

=16*(2√2)*(2√2)*1+32*(2√2)*3*(√2/2)+16*3*3*1=128+192+144=464

|vector{BD}|=sqrt(464}=sqrt{16*29}=4sqrt(29)