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Đặt \(\overrightarrow{u}=\overrightarrow{CD}+2\overrightarrow{CB}=\overrightarrow{CD}+2\left(\overrightarrow{CD}+\overrightarrow{DA}+\overrightarrow{AB}\right)\)
\(=3\overrightarrow{CD}+2\overrightarrow{DA}-\overrightarrow{CD}\) (do \(2\overrightarrow{AB}=\overrightarrow{DC}=-\overrightarrow{CD}\))
\(=2\overrightarrow{CD}+2\overrightarrow{DA}=2\overrightarrow{CA}\)
\(\Rightarrow\left|\overrightarrow{u}\right|=2AC=2\sqrt{AD^2+CD^2}=4\sqrt{5}\)
\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|=AC=5\)
\(\left|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CA}\right|=\left|\overrightarrow{BC}+\overrightarrow{AD}\right|=\left|2\overrightarrow{AD}\right|=2AD=8\)
Kẻ hbh ABFC
Dễ tính được ACD=530
nên ACB=37=CBF
Theo định lý cos ta tính được AF
bạn tự tính nhá mk ko có mt
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.
Ta có: (vectơ AB + vectơ AD) + vectơ AC
= vectơ AC + vectơ AC
= 2 vectơAC
=> | vectơ AB + vectơ AC + vectơ AD| = 2 vectơAC = 2a căn 2
Bài 2:
\(\left|\overrightarrow{BC}+\overrightarrow{BA}\right|=\left|\overrightarrow{AC}\right|=AC=a\sqrt{2}\)
\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=a\)
\(BC=AD=\sqrt{AC^2-AB^2}=2a\)
a/ \(T=\left|3\overrightarrow{AB}-4\overrightarrow{BC}\right|\Rightarrow T^2=9AB^2+16BC^2-24\overrightarrow{AB}.\overrightarrow{BC}\)
\(=9a^2+64a^2=73a^2\Rightarrow T=a\sqrt{73}\)
b/ \(T^2=4AB^2+9BC^2+12.\overrightarrow{BA}.\overrightarrow{BC}=4AB^2+9BC^2=40a^2\)
\(\Rightarrow T=2a\sqrt{10}\)
c/ \(T=\left|\overrightarrow{AD}+3\overrightarrow{BC}\right|=\left|\overrightarrow{AD}+3\overrightarrow{AD}\right|=\left|4\overrightarrow{AD}\right|=4AD=8a\)
d/ \(T=\left|2\overrightarrow{DC}-3\overrightarrow{DC}\right|=\left|-\overrightarrow{DC}\right|=CD=AB=a\)
Đặt \(\overrightarrow{u}=\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{DC}=3.\overrightarrow{AB}+\overrightarrow{AD}\) (do \(\overrightarrow{DC}=2\overrightarrow{AB}\))
\(\Rightarrow\left|\overrightarrow{u}\right|^2=\left(3\overrightarrow{AB}+\overrightarrow{AD}\right)^2=9AB^2+AD^2+6\overrightarrow{AB}.\overrightarrow{AD}=9AB^2+AD^2=10AB^2\)
\(\Rightarrow\left|\overrightarrow{u}\right|=AB\sqrt{10}=2\sqrt{10}\)