A B C D I K

a)

• \(\overrightarrow{BI}=\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\) (t/c trung điểm)

\(=\frac{1}{2}\left(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC}\right)\)

\(=\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\)

• \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)

\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}\)

\(=\overrightarrow{BA}+\frac{1}{3}\left(\overrightarrow{BC}-\overrightarrow{BA}\right)\)

\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}-\frac{1}{3}\overrightarrow{BA}\)

\(=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}\)

b) Ta có: \(\overrightarrow{BK}=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}=\frac{4}{3}\left(\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\right)=\frac{4}{3}\overrightarrow{BI}\)

=> B,K,I thẳng hàng

c) \(27\overrightarrow{MA}-8\overrightarrow{MB}=2015\overrightarrow{MC}\)

\(\Leftrightarrow27\left(\overrightarrow{MC}+\overrightarrow{CA}\right)-8\left(\overrightarrow{MC}+\overrightarrow{CB}\right)=2015\overrightarrow{MC}\)

\(\Leftrightarrow27\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{MC}-8\overrightarrow{CB}-2015\overrightarrow{MC}=\overrightarrow{0}\)

\(\Leftrightarrow-1996\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{CB}=\overrightarrow{0}\)

\(\Leftrightarrow1996\overrightarrow{CM}=8\overrightarrow{CB}-27\overrightarrow{CA}\)

\(\Leftrightarrow\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)

Vậy: Dựng điểm M sao cho \(\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)

a) Ta có: \(\overrightarrow {BC} ,\overrightarrow {PN} \) là hai vecto cùng hướng và \(\frac{1}{2}\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {PN} } \right|\)

\( \Rightarrow \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {PN} \)\( \Rightarrow \overrightarrow {AP}  + \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {AP}  + \overrightarrow {PN}  = \overrightarrow {AN} \)

b) Ta có: \(\overrightarrow {MP} ,\overrightarrow {CA} \) là hai vecto cùng hướng và \(2\left| {\overrightarrow {MP} } \right| = \left| {\overrightarrow {CA} } \right|\)

\( \Rightarrow 2\overrightarrow {MP}  = \overrightarrow {CA} \)\( \Rightarrow \overrightarrow {BC}  + 2\overrightarrow {MP}  = \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow {BA} \)

Tham khảo:

a)  M thuộc cạnh BC nên vectơ \(\overrightarrow {MB} \) và \(\overrightarrow {MC} \) ngược hướng với nhau.

Lại có: MB = 3 MC \( \Rightarrow \overrightarrow {MB}  =  - 3.\overrightarrow {MC} \)

b) Ta có: \(\overrightarrow {AM}  = \overrightarrow {AB}  + \overrightarrow {BM} \)

Mà \(BM = \dfrac{3}{4}BC\) nên \(\overrightarrow {BM}  = \dfrac{3}{4}\overrightarrow {BC} \)

\( \Rightarrow \overrightarrow {AM}  = \overrightarrow {AB}  + \dfrac{3}{4}\overrightarrow {BC} \)

Lại có: \(\overrightarrow {BC}  = \overrightarrow {AC}  - \overrightarrow {AB} \) (quy tắc hiệu)

\( \Rightarrow \overrightarrow {AM}  = \overrightarrow {AB}  + \dfrac{3}{4}\left( {\overrightarrow {AC}  - \overrightarrow {AB} } \right) = \dfrac{1}{4}.\overrightarrow {AB}  + \dfrac{3}{4}.\overrightarrow {AC} \)

Vậy \(\overrightarrow {AM}  = \dfrac{1}{4}.\overrightarrow {AB}  + \dfrac{3}{4}.\overrightarrow {AC} \)

a) Từ điểm I trên AB thỏa mãn IA = 1/2 IB ta vẽ đường song song với BC. Điểm N nằm trên đó.

B) tương tự câu a)

Dựa theo đề bài ta có hình vẽ:  A B C M N I

Ta có: MA = 2MB; BN = 5CN => BN = 5/6 BC 

Có \(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=-\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}\)

Áp dụng định lí menelaus cho tam giác ABN

\(\frac{MA}{MB}.\frac{CB}{CN}.\frac{IN}{IA}=1\)=> \(\frac{2}{1}.\frac{6}{1}.\frac{IN}{IA}=1\Rightarrow IA=12IN\)=> \(\overrightarrow{AI}=\frac{12}{13}\overrightarrow{AN}\)

Ta có: \(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=\overrightarrow{BA}+\frac{12}{13}\overrightarrow{AN}=\overrightarrow{BA}+\frac{12}{13}\left(-\overrightarrow{BA}+\frac{5}{6}\overrightarrow{BC}\right)\)rút gọn tính tiếp nhé

\(\overrightarrow{MA}+3\cdot\overrightarrow{MB}=\overrightarrow{0}\)

=>\(\overrightarrow{MA}=-3\cdot\overrightarrow{MB}\)

=>M nằm giữa A và B sao cho MA=3MB

AB=AM+MB

=3MB+MB=4BM

=>\(BM=\frac14BA;AM=\frac34AB\)

\(2\cdot\overrightarrow{NB}+3\cdot\overrightarrow{NC}=\overrightarrow{0}\)

=>\(2\cdot\overrightarrow{NB}=-3\cdot\overrightarrow{NC}\)

=>\(\overrightarrow{NB}=-\frac32\cdot\overrightarrow{NC}\)

=>N nằm giữa B và C sao cho \(NB=\frac32NC\)

NB+NC=BC

=>\(BC=\frac32NC+NC=\frac52NC\)

=>\(CN=\frac25CB;BN=\frac35CB\)

\(\overrightarrow{PM}+2\cdot\overrightarrow{PN}=\overrightarrow{0}\)

=>\(\overrightarrow{PM}=-2\cdot\overrightarrow{PN}\)

=>P nằm giữa M và N sao cho PM=2PN

PM+PN=MN

=>MN=2PN+PN=3PN

=>\(NP=\frac13NM;MP=\frac23NM\)

\(\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{NP}\)

\(=\overrightarrow{AB}+\frac35\cdot\overrightarrow{BC}+\frac13\cdot\overrightarrow{NM}=\overrightarrow{AB}+\frac35\cdot\overrightarrow{BC}+\frac13\cdot\left(\overrightarrow{NB}+\overrightarrow{BM}\right)\)

\(=\overrightarrow{AB}+\frac35\cdot\overrightarrow{BC}+\frac13\left(\frac35\cdot\overrightarrow{CB}+\frac14\cdot\overrightarrow{BA}\right)\)

\(=\overrightarrow{AB}+\frac35\cdot\overrightarrow{BC}+\frac15\cdot\overrightarrow{CB}+\frac{1}{12}\cdot\overrightarrow{BA}\)

\(=\frac{11}{12}\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{BC}=\frac{11}{12}\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AD}\)

\(=\frac{11}{12}\cdot\overrightarrow{AB}+\frac25\left(\overrightarrow{AB}+\overrightarrow{BD}\right)=\frac{11}{12}\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{BD}\)

\(=\frac{79}{60}\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{BD}\)