Gọi M là trung điểm của BC

Vì ΔABC đều

mà M là trug điểm của bC

nên MA vuông góc với BC 

BM=CM=a/2

\(AM=\sqrt{a^2-\left(\dfrac{1}{2}a\right)^2}=\dfrac{a\sqrt{3}}{2}\)

\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{BA}\right|=2\cdot AM=2\cdot\dfrac{a\sqrt{3}}{2}=a\sqrt{3}\)

\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=a\)

\(\left|\overrightarrow{AB}-\overrightarrow{CA}\right|=\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=a\sqrt{3}\)

vecto AB-vecto BC

=vecto AB+vecto CB

=>|vecto AB+vecto CB|=|vecto BA+vecto BC|=|2vecto BN|(Với N là trung điểm của AC)

=2xBN=a căn 3

a: A(1;3); B(-2;5); C(-4;0)

\(\overrightarrow{AB}=\left(-2-1;5-3\right)=\left(-3;2\right)\)

\(\overrightarrow{AC}=\left(-4-1;0-3\right)=\left(-5;-3\right)\)

\(\overrightarrow{BC}=\left(-4+2;0-5\right)=\left(-2;-5\right)\)

\(\overrightarrow{CB}=\left(-2+4;5-0\right)=\left(2;5\right)\)

b: \(\overrightarrow{AB}\cdot\overrightarrow{CB}=-3\cdot2+2\cdot5=-6+10=4\)

\(\overrightarrow{AC}\cdot\overrightarrow{BC}=\left(-5\right)\cdot\left(-2\right)+\left(-3\right)\cdot\left(-5\right)=10+15=25\)

c: \(\overrightarrow{AB}=\left(-3;2\right)\)

=>\(AB=\sqrt{\left(-3\right)^2+2^2}=\sqrt{13}\)

\(\overrightarrow{BC}=\left(-2;-5\right)\)

=>\(BC=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{4+25}=\sqrt{29}\)

e: \(\overrightarrow{AB}+2\cdot\overrightarrow{CB}\) =\(\left(-3+2\cdot2;2+2\cdot5\right)\)

=(-3+4;2+10)

=(1;12)

\(\overrightarrow{AB}.\overrightarrow{CB}+\overrightarrow{AC}.\overrightarrow{BC}=12\)

\(\Leftrightarrow\overrightarrow{BC}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=12\)

\(\Leftrightarrow\overrightarrow{BC}.\overrightarrow{BC}=12\)

\(\Rightarrow BC^2=12\Rightarrow BC=2\sqrt{3}\)

Chọn B.

Ta có 

mà 

\(4\cdot\overrightarrow{CI}+\overrightarrow{AC}=\overrightarrow{0}\)

=>\(4\cdot\overrightarrow{CI}=-\overrightarrow{AC}=\overrightarrow{CA}\)

=>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  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

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)

\(=\overrightarrow{AB}+\frac14\cdot\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\frac14\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\frac14\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

BC=BM+CM

=>4BM=BM+CM

=>CM=3BM

=>\(MC=\frac34BC\)

=>\(\overrightarrow{MC}=\frac34\cdot\overrightarrow{BC}=\frac34\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac34\cdot\overrightarrow{AB}+\frac34\cdot\overrightarrow{AC}\)