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a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)
Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}\)(D là trung điểm của BC) (1)
\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AK}\)(K là trung điểm của MN) (2)
Lấy (1) trừ (2) có: \(\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=2\left(\overrightarrow{AD}-\overrightarrow{AK}\right)\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\overrightarrow{AB}+\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)=\(\overrightarrow{KD}\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BC}\)
bài 1
a CO-OB=BA
<=.> CO = BA +OB
<=> CO=OA ( LUÔN ĐÚNG )=>ĐPCM
b AB-BC=DB
<=> AB=DB+BC
<=> AB=DC(LUÔN ĐÚNG )=> ĐPCM
Cc DA-DB=OD-OC
<=> DA+BD= OD+CO
<=> BA= CD (LUÔN ĐÚNG )=> ĐPCM
d DA-DB+DC=0
VT= DA +BD+DC
= BA+DC
Mà BA=CD(CMT)
=> VT= CD+DC=O
\(BM=2AM\Rightarrow BM=\dfrac{2}{3}AB\Rightarrow\overrightarrow{MB}=\dfrac{2}{3}\overrightarrow{AB}\)
\(AN=3CN\Rightarrow CN=\dfrac{1}{4}CA\Rightarrow\overrightarrow{CN}=\dfrac{1}{4}\overrightarrow{CA}\)
Ta có:
\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{BC}+\dfrac{1}{4}\overrightarrow{CA}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{BC}+\dfrac{1}{4}\left(\overrightarrow{CB}+\overrightarrow{BA}\right)\)
\(=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{BC}+\dfrac{1}{4}\overrightarrow{CB}+\dfrac{1}{4}\overrightarrow{BA}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{BC}-\dfrac{1}{4}\overrightarrow{BC}-\dfrac{1}{4}\overrightarrow{AB}\)
\(=\dfrac{5}{12}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{BC}\)
Lời giải:
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\frac{1}{3}\overrightarrow{BA}+\frac{3}{4}\overrightarrow{AC}\)
\(=\frac{-1}{3}\overrightarrow{AB}+\frac{3}{4}(\overrightarrow{AB}+\overrightarrow{BC})=\frac{5}{12}\overrightarrow{AB}+\frac{3}{4}\overrightarrow{BC}\)
a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
Bài 1:
Gọi M là trung điểm của AD
\(BM=\sqrt{AB^2+AM^2}=\sqrt{4a^2+\dfrac{1}{4}a^2}=\dfrac{\sqrt{17}}{2}a\)
\(\left|\overrightarrow{AB}+\overrightarrow{DB}\right|=2\cdot BM=\sqrt{17}a\)