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Lời giải:
Theo đề ta có: $\overrightarrow{BM}=2\overrightarrow{MC}=-2\overrightarrow{CM}$
$\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}(1)$
$=\overrightarrow{AB}-2\overrightarrow{CM}$
$\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}$
$\Rightarrow 2\overrightarrow{AM}=2\overrightarrow{AC}+2\overrightarrow{CM}(2)$
Lấy $(1)+(2)\Rightarrow 3\overrightarrow{AM}=\overrightarrow{AB}+2\overrightarrow{AC}$
$\Rightarrow \overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$
Ta có M B → = 1 3 M C → ⇔ 3 M B → = M C → ⇔ 3 B M → = C M →
A M → = A B → + B M → ⇒ 3 A M → = 3 A B → + 3 B M → ( 1 ) A M → = A C → + C M → ( 2 )
Lấy (1) trừ (2) ta được :
2 A M → = 3 A B → + 3 B M → − A C → + C M → = 3 A B → − A C → + ( 3 B M → − C M → ) = 3 A B → − A C → + 0 → = 3 A B → − A C → ⇒ A M → = 3 2 A B → − 1 2 A C → = 3 2 u → − 1 2 v →
Đáp án A
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 \(\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)
Ta có: \(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)
\(\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)
\(\Rightarrow-\overrightarrow{MB}=3\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{BM}=\dfrac{2}{3}\overrightarrow{BC}\). Mà \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}\) nên \(\overrightarrow{BM}=\dfrac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
Theo quy tắc 3 điểm, ta có
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\Rightarrow\overrightarrow{AM}=\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}\)
\(\Rightarrow\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\) hay \(\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{u}+\dfrac{3}{2}\overrightarrow{v}\)
= 3
=> 
)
= 
- 
nên 
= 
+ 
Bài 2:
vecto AM=vecto AB+vecto BM
=vecto AB+2/3vecto BC
=vecto AB+2/3*(vecto BA+vecto AC)
=1/3*vecto AB+2/3*vecto AC
Chọn B.
Ta có
mà