Question

1. Cho AK, BM là 2 trung tuyến của tam giác ABC. Hãy phân tích các vecto AB, BC, AC theo vecto u=AK, v=BM.

2. Cho tam giác ABC. Lấy M,N,P ll trên các đoạn AB,BC,CA sao cho AM=1/3AB, BN=1/3BC, CP=1/3CA.

CMR vecto AN + BP + CM = 0

Answer 1

Câu 1:

Gọi G là giao điểm AK và BM => G là trọng tâm \(\Delta ABC\)

\(\Rightarrow\) Theo tính chất trọng tâm \(\left\{{}\begin{matrix}AG=\frac{2}{3}AK\\BG=\frac{2}{3}BM\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{AB}=\overrightarrow{AG}+\overrightarrow{GB}=\frac{2}{3}\overrightarrow{AK}-\frac{2}{3}\overrightarrow{BM}\\ \Rightarrow\overrightarrow{AC}=\overrightarrow{AK}+\overrightarrow{KC}=\overrightarrow{AK}+\frac{1}{2}\overrightarrow{BC}\\ =\overrightarrow{AK}+\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AK}-\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\\ \Rightarrow\frac{1}{2}\overrightarrow{AC}=\overrightarrow{AK}-\frac{1}{2}\left(\frac{2}{3}\overrightarrow{AK}-\frac{2}{3}\overrightarrow{BM}\right)\\ =\overrightarrow{AK}-\frac{1}{3}\overrightarrow{AK}+\frac{1}{3}\overrightarrow{BM}\\ =\frac{2}{3}\overrightarrow{AK}+\frac{1}{3}\overrightarrow{BM}\\ \Rightarrow\overrightarrow{AC}=\frac{4}{3}\overrightarrow{AK}+\frac{2}{3}\overrightarrow{BM}\\ \Rightarrow\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{-AB}+\overrightarrow{AC}\\ =-\left(\frac{2}{3}\overrightarrow{AK}-\frac{2}{3}\overrightarrow{BM}\right)+\left(\frac{4}{3}\overrightarrow{AK}+\frac{2}{3}\overrightarrow{BM}\right)\\ =-\frac{2}{3}\overrightarrow{AK}+\frac{2}{3}\overrightarrow{BM}+\frac{4}{3}\overrightarrow{AK}+\frac{2}{3}\overrightarrow{BM}\\ =\frac{2}{3}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\)

Answer 2

1/ Theo quy tắc TĐ: \(\overrightarrow{AK}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2};\overrightarrow{BM}=\frac{\overrightarrow{BA}+\overrightarrow{BC}}{2}\)

Theo quy tắc 3 điểm: \(\overrightarrow{AB}=\overrightarrow{AK}+\overrightarrow{KB}\)

Vậy cần phân tích \(\overrightarrow{KB}\)

\(\overrightarrow{KB}=\frac{\overrightarrow{CB}}{2}=\frac{\overrightarrow{BA}-2\overrightarrow{BM}}{2}\)

\(\Rightarrow\overrightarrow{AB}=\overrightarrow{AK}+\frac{\overrightarrow{BA}-2\overrightarrow{BM}}{2}\Leftrightarrow2\overrightarrow{AB}=2\overrightarrow{AK}-\overrightarrow{AB}-2\overrightarrow{BM}\)

\(\Leftrightarrow\overrightarrow{AB}=\frac{2}{3}\overrightarrow{AK}-\frac{2}{3}\overrightarrow{BM}\)

Tìm \(\overrightarrow{BC};\overrightarrow{AC}\) tương tự

2/ Theo quy tắc 3 điểm có:

\(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}\)

\(\overrightarrow{BP}=\overrightarrow{BC}+\overrightarrow{CP}\)

\(\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{AM}\)

Cộng vế vs vế:

\(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}+\frac{1}{3}\left(\overrightarrow{BC}+\overrightarrow{CA}-\overrightarrow{BA}\right)=0\)

Answer 3

Answer 4

Câu 2:

\(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\left(\overrightarrow{AC}+\overrightarrow{CN}\right)+\left(\overrightarrow{BC}+\overrightarrow{CP}\right)+\left(\overrightarrow{CA}+\overrightarrow{AM}\right)\\ =\overrightarrow{AC}-\frac{2}{3}\overrightarrow{BC}+\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CA}+\overrightarrow{CA}+\frac{1}{3}\overrightarrow{AB}\\ =\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\left(\frac{1}{3}\overrightarrow{BC}+\frac{1}{3}\overrightarrow{CA}+\frac{1}{3}\overrightarrow{AB}\right)\\ =\frac{1}{3}\left(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}\right)=\frac{1}{3}\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=0\)