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\(3IA+2\left(IC+DI\right)=0\Leftrightarrow3IA+2DC=0\)
\(\Leftrightarrow3IO+3OA+2DA+2AC=0\Leftrightarrow3IO+3OA-2AD-4OA=0\)
\(\Leftrightarrow3IO-OA-2AD=0\Rightarrow3IO=OA+2AD\) (1)
\(JA-2JB+2JC=0\Leftrightarrow JA+2\left(BJ+JC\right)=0\)
\(\Leftrightarrow JA+2BC=0\Leftrightarrow JO+OA+2BC=0\)
\(\Leftrightarrow JO+OA+2AD=0\Rightarrow OJ=OA+2AD\) (2)
(1); (2) \(\Rightarrow OJ=3IO\) hay I;J;O thẳng hàng
Phân tích dài quá, ko hay lắm :(
Cho tam giác ABC. Xác định điểm I, J, K thỏa các điều kiện sau: 3IA+2IC=0 ; 2JA+3JB=3BC ; KA+KB+KC=0
a: \(\overrightarrow{IB}=2\cdot\overrightarrow{IC}\)
=>C là trung điểm của IB
=>IC=CB=1/2IB
\(\overrightarrow{IK}=\overrightarrow{IB}+\overrightarrow{BK}\)
\(=2\cdot\overrightarrow{CB}+\frac12\cdot\overrightarrow{BA}=2\left(\overrightarrow{CA}+\overrightarrow{AB}\right)-\frac12\cdot\overrightarrow{AB}\)
\(=-2\cdot\overrightarrow{AC}+2\cdot\overrightarrow{AB}-\frac12\cdot\overrightarrow{AB}=-2\cdot\overrightarrow{AC}+\frac32\cdot\overrightarrow{AB}\)
b: JA=2JC
=>\(JA=\frac23AC;JC=\frac13AC\)
\(\overrightarrow{IJ}=\overrightarrow{IC}+\overrightarrow{CJ}=\overrightarrow{CB}+\frac13\cdot\overrightarrow{CA}\)
\(=\overrightarrow{CA}+\overrightarrow{AB}+\frac13\cdot\overrightarrow{CA}=\frac43\cdot\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{AB}-\frac43\cdot\overrightarrow{AC}\)
\(=\frac23\left(\frac32\cdot\overrightarrow{AB}-2\cdot\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{IK}\)
=>I,J,K thẳng hàng
\(3\overrightarrow{AP}-2\overrightarrow{AC}=\overrightarrow{0}\)
\(VT=3\left(\overrightarrow{AD}+\overrightarrow{DP}\right)-2\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\)
\(=3\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{AD}-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\overrightarrow{DP}-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\left(\overrightarrow{DC}+\overrightarrow{CP}\right)-2\overrightarrow{DC}\)
\(=\overrightarrow{AD}+3\overrightarrow{DC}+3\overrightarrow{CP}-2\overrightarrow{DC}\)
\(=\widehat{AD}+\overrightarrow{DC}+3.\dfrac{2}{3}\overrightarrow{CO}\)
\(=\overrightarrow{AD}+\overrightarrow{DC}+2.\dfrac{1}{2}\overrightarrow{CA}\)
\(=\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CA}\)
\(=\overrightarrow{AC}+\overrightarrow{CA}\)
\(=\overrightarrow{AA}=\overrightarrow{0}=VP\) (điều phải chứng minh)
\(4\cdot\overrightarrow{CI}+\overrightarrow{AC}=\overrightarrow{0}\)
=>\(4\cdot\overrightarrow{CI}=-\overrightarrow{AC}=\overrightarrow{CA}\)
=>CA=4CI
\(\overrightarrow{BI}=\overrightarrow{BC}+\overrightarrow{CI}=\overrightarrow{BC}+\frac14\cdot\overrightarrow{CA}\)
\(=-\overrightarrow{AB}+\overrightarrow{AC}-\frac14\cdot\overrightarrow{AC}=-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\)
\(\overrightarrow{BJ}=\frac12\cdot\overrightarrow{AC}-\frac23\cdot\overrightarrow{AB}\)
\(=\frac23\left(-\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{BI}\)
=>B,I,J thẳng hàng
Ta có:
\(\overrightarrow {GA} + \overrightarrow {GB} + \overrightarrow {GC} + \overrightarrow {GD} = \overrightarrow 0 \Leftrightarrow \left( {\overrightarrow {GI} + \overrightarrow {IA} } \right) + \left( {\overrightarrow {GI} + \overrightarrow {IB} } \right) + \left( {\overrightarrow {GJ} + \overrightarrow {JC} } \right) + \left( {\overrightarrow {GJ} + \overrightarrow {JD} } \right) = \overrightarrow 0 \)
\( \Leftrightarrow 2\overrightarrow {GI} + \left( {\overrightarrow {IA} + \overrightarrow {IB} } \right) + 2\overrightarrow {GJ} + \left( {\overrightarrow {JC} + \overrightarrow {JD} } \right) = \overrightarrow 0 \)
\( \Leftrightarrow 2\overrightarrow {GI} + 2\overrightarrow {GJ} = \overrightarrow 0 \Leftrightarrow 2\left( {\overrightarrow {GI} + \overrightarrow {GJ} } \right) = \overrightarrow 0 \)
\( \Leftrightarrow \overrightarrow {GI} + \overrightarrow {GJ} = \overrightarrow 0 \Rightarrow \)G là trung điểm của đoạn thẳng IJ
Vậy I, G, J thẳng hàng
A B C D I J O
\(Ta\text{ }có\text{ }:3\overrightarrow{IA}+2\overrightarrow{IC}-2\overrightarrow{ID}=0\\ \Rightarrow3\overrightarrow{IA}+2\left(\overrightarrow{IC}-\overrightarrow{ID}\right)=0\\ \Rightarrow3\overrightarrow{IA}=-2\left(\overrightarrow{IC}-\overrightarrow{ID}\right)\\ \Rightarrow3\overrightarrow{IA}=-2\overrightarrow{DC}=2\overrightarrow{BA}\\ \Rightarrow\overrightarrow{IA}=\frac{2}{3}\overrightarrow{BA}\\ \Rightarrow I;B;A\text{ thẳng hàng},I\text{ nằm giữa }A;B\left(\frac{2}{3}>0;IA< BA\right)\)
\(\text{Lại có }:\overrightarrow{JA}-2\overrightarrow{JB}+2\overrightarrow{JC}=0\\ \Rightarrow\overrightarrow{JA}=2\left(\overrightarrow{JB}-\overrightarrow{JC}\right)\\ \Rightarrow\overrightarrow{JA}=2\overrightarrow{CB}=2\overrightarrow{DA}\\ \Rightarrow J;D;A\text{ thẳng hàng},D\text{ nằm giữa }J;A\left(2>0;JA>DA\right)\)
\(\text{Lại có }:O\text{ là trung điểm }AC;BD\left(\text{Tính chất hình bình hành}\right)\\ \Rightarrow\overrightarrow{JO}=\overrightarrow{JA}+\overrightarrow{AO}=-2\overrightarrow{AD}+\frac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AB}\right)\\ =-2\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=-\frac{3}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\)
\(\text{Mặt khác }:\overrightarrow{JI}=\overrightarrow{JA}+\overrightarrow{AI}=-2\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AB}=\frac{4}{3}\left(-\frac{3}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\right)\\ \Rightarrow\overrightarrow{JI}=\frac{4}{3}\overrightarrow{JO}\\ \Rightarrow J;I;O\text{ thẳng hàng}\)
cảm ơn bạn nha