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Tất cả biểu thức đều là vecto, cái nào là độ dài thì nằm trong trị tuyệt đối:
\(\left|BD\right|=\sqrt{AB^2+AD^2}=a\sqrt{5}\)
\(\left|AC\right|=\sqrt{AB^2+BC^2}=a\sqrt{13}\)
a/ \(AB.BD=-BA.BD=-\left|AB\right|.\left|BD\right|.cos\widehat{ABD}\)
\(=-2a.a\sqrt{5}.\frac{2a}{a\sqrt{5}}=-4a^2\)
\(BC.BD=\left|BC\right|.\left|BD\right|.cos\widehat{DBC}=3a.a\sqrt{5}.\frac{a}{a\sqrt{5}}=3a^2\)
\(AC.BD=AC\left(BA+AD\right)=AC.BA+AC.AD\)
\(=AC.AD-AC.AB=\left|AC\right|.\left|AD\right|.cos\widehat{DAC}-\left|AB\right|.\left|AC\right|.cos\widehat{BAC}\)
\(=a.a\sqrt{13}.\frac{3a}{a\sqrt{13}}-2a.a\sqrt{13}.\frac{2a}{a\sqrt{13}}=-a^2\)
\(AC.IJ=\frac{1}{2}AC\left(AD+BC\right)=\frac{1}{2}AC.AD+\frac{1}{2}AC.BC\)
Ta có \(AC.AD=3a^2\) (ngay bên trên)
\(AC.BC=CA.CB=\left|CA\right|.\left|CB\right|.cos\widehat{BCA}=a\sqrt{13}.3a.\frac{3a}{a\sqrt{13}}=9a^2\)
\(\Rightarrow AC.IJ=6a^2\)
A B C D B' O
\(cos\left(\overrightarrow{AC};\overrightarrow{BA}\right)=cos\left(\overrightarrow{AC};\overrightarrow{AB'}\right)=cos\widehat{CAB'}=cos135^o\)\(=\dfrac{\sqrt{2}}{2}\).
\(sin\left(\overrightarrow{AC};\overrightarrow{BD}\right)=sin90^o=1\) do \(AC\perp BD\).
\(cos\left(\overrightarrow{AB};\overrightarrow{CD}\right)=cos180^o=-1\) do hai véc tơ \(\overrightarrow{AB};\overrightarrow{CD}\) ngược hướng.
\(\overrightarrow{u}=\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{BA}+\overrightarrow{AD}=2\overrightarrow{AD}+\left(\overrightarrow{DC}+\overrightarrow{BA}\right)=2\overrightarrow{AD}\)\(\Rightarrow\overrightarrow{u}\) cùng hướng \(\overrightarrow{AD}\)
\(AC=\sqrt{AB^2+BC^2}=a\sqrt{5}\)
\(BD=\sqrt{AD^2+AB^2}=a\sqrt{2}\)
\(\overrightarrow{AC}.\overrightarrow{BD}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\left(\overrightarrow{BA}+\overrightarrow{AD}\right)\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AB}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BA}+\overrightarrow{BC}.\overrightarrow{AD}\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AD}.2\overrightarrow{AD}=-\overrightarrow{AB}^2+2\overrightarrow{AD}^2\)
\(=-a^2+2a^2=a^2\)
\(cos\left(\overrightarrow{AC};\overrightarrow{BD}\right)=\dfrac{\overrightarrow{AC}.\overrightarrow{BD}}{AC.BD}=\dfrac{a^2}{a\sqrt{2}.a\sqrt{5}}=\dfrac{1}{\sqrt{10}}\)