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Do ABCD là hình vuông nên AC vuông góc BD
Do đó:
\(P=\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BC}+\overrightarrow{BA}+\overrightarrow{BD}\right)=\left(\overrightarrow{AB}+\overrightarrow{AC}\right).2\overrightarrow{BD}\)
\(=2\overrightarrow{AB}.\overrightarrow{BD}+2\overrightarrow{AC}.\overrightarrow{BD}=2\overrightarrow{AB}.\overrightarrow{BC}=2a.a.cos135^0=-a^2\sqrt{2}\)
Ta có: \(AC = BD = \sqrt {A{B^2} + B{C^2}} = \sqrt {{a^2} + {a^2}} = a\sqrt 2 \)
+) \(AB \bot AD \Rightarrow \overrightarrow {AB} \bot \overrightarrow {AD} \Rightarrow \overrightarrow {AB} .\overrightarrow {AD} = 0\)
+) \(\overrightarrow {AB} .\overrightarrow {AC} = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = a.a\sqrt 2.\cos 45^\circ = a^2\)
+) \(\overrightarrow {AC} .\overrightarrow {CB} = \left| {\overrightarrow {AC} } \right|.\left| {\overrightarrow {CB} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right) = a\sqrt 2 .a.\cos 135^\circ = - {a^2}\)
+) \(AC \bot BD \Rightarrow \overrightarrow {AC} \bot \overrightarrow {BD} \Rightarrow \overrightarrow {AC} .\overrightarrow {BD} = 0\)
Chú ý
\(\overrightarrow {a} \bot \overrightarrow {b} \Leftrightarrow \overrightarrow {a} .\overrightarrow {b} = 0\)
Gọi H là điểm nằm trên cạnh AB sao cho B là trung điểm AH.
Ta có: AH = 2AB = 2a.
Áp dụng định lí Pytago vào tam giác ADH ta có:
Đáp án B
a, \(AC=\dfrac{AB}{sin45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}=a.a\sqrt{2}.cos45^o=a^2\)
b, \(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)
\(=\overrightarrow{AC}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BC}\)
\(=AC.BD.cos90^o+AC.AD.cos45^o\)
\(=a\sqrt{2}.a\sqrt{2}.0+a\sqrt{2}.a.\dfrac{\sqrt{2}}{2}=a^2\)
c, \(\overrightarrow{AB}.\overrightarrow{BD}=AB.BD.cos135^o=-a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2\)
d, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\left(2\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{BC}.\left(\overrightarrow{AD}+\overrightarrow{BD}\right)\)
\(=\overrightarrow{BC}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BD}\)
\(=AD^2+BC.BD.cos45^o\)
\(=a^2+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=2a^2\)
e, \(\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right)\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\right)\)
\(=\left(\overrightarrow{AC}+\overrightarrow{AC}\right)\left(\overrightarrow{DB}+\overrightarrow{DB}\right)\)
\(=4.\overrightarrow{AC}.\overrightarrow{DB}=4.AC.DB.cos90^o=0\)
\(\overrightarrow{AC}.\overrightarrow{BD}=\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\left(\overrightarrow{BA}+\overrightarrow{AD}\right)\)
\(=\overrightarrow{AD}.\overrightarrow{BA}+\overrightarrow{AD}^2+\overrightarrow{DC}.\overrightarrow{BA}+\overrightarrow{DC}.\overrightarrow{AD}\)
\(=\overrightarrow{AD}^2-\overrightarrow{AB}.\overrightarrow{DC}=a^2-a.2a=-a^2\)
vì ABCD là hình vuông nên áp dụng pytago
AC2=Ab2+BC2 =a2+a2=2a2
=>AC=a\(\sqrt{2}\)
(AB+AC).(BC+BD+BA)=(a\(\sqrt{2}\)+a).(a\(\sqrt{2}\)+a+a)=4a2+(2\(\sqrt{2}\)+\(\sqrt{2}\))a2