a, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{BC}^2\)

\(\Leftrightarrow AC^2+AB^2-2\overrightarrow{AB}.\overrightarrow{AC}=BC^2\)

\(\Leftrightarrow2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\)

\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{AB^2+AC^2-BC^2}{2}=\dfrac{5^2+8^2-7^2}{2}=20\)

b, \(2\overrightarrow{CA}.\overrightarrow{CB}=CA^2+CB^2-BC^2=CA^2\)

\(\Rightarrow\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2}{2}=\dfrac{8^2}{2}=32\)

Lời giải:

a) 

\(\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{BC}\)

\(\Rightarrow (\overrightarrow{AC}-\overrightarrow{AB})^2=\overrightarrow{BC}^2\Leftrightarrow AB^2+AC^2-2\overrightarrow{AC}.\overrightarrow{AB}=BC^2\)

\(\Leftrightarrow 2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2-BC^2\) (đpcm)

Ta có:

\(\overrightarrow{AB}.\overrightarrow{AC}=\frac{AB^2+AC^2-BC^2}{2}=\frac{5^2+8^2-7^2}{2}=20\)

\(\cos \angle A=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|}=\frac{20}{5.8}=\frac{1}{2}\)

\(\Rightarrow \angle A=60^0\)

b) 

Tương tự phần a, \(\overrightarrow{CA}.\overrightarrow{CB}=\frac{CA^2+CB^2-AB^2}{2}=\frac{8^2+7^2-5^2}{2}=44\)

1.

\(\overrightarrow{AB}.\overrightarrow{BC}=\overrightarrow{AB}.\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}.\left(-\overrightarrow{AB}\right)+\overrightarrow{AB}.\overrightarrow{AC}=-AB^2=-25\)

2.

\(\overrightarrow{AB}.\overrightarrow{BD}=\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AD}\right)=-\overrightarrow{AB}.\overrightarrow{AB}+\overrightarrow{AB}.\overrightarrow{AD}=-AB^2+0=-64\)

a) \(\overrightarrow {AB} .\overrightarrow {AC}  = 2.3.\cos \widehat {BAC} = 6.\cos {60^o} = 3\)

b)

Ta có: \(\overrightarrow {AB}  + \overrightarrow {AC}  = 2\overrightarrow {AM} \)(do M là trung điểm của BC)

\( \Leftrightarrow \overrightarrow {AM}  = \frac{1}{2}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} \)

+) \(\overrightarrow {BD}  = \overrightarrow {AD}  - \overrightarrow {AB}  = \frac{7}{{12}}\overrightarrow {AC}  - \overrightarrow {AB} \)

c) Ta có:

 \(\begin{array}{l}\overrightarrow {AM} .\overrightarrow {BD}  = \left( {\frac{1}{2}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} } \right)\left( {\frac{7}{{12}}\overrightarrow {AC}  - \overrightarrow {AB} } \right)\\ = \frac{7}{{24}}\overrightarrow {AB} .\overrightarrow {AC}  - \frac{1}{2}{\overrightarrow {AB} ^2} + \frac{7}{{24}}{\overrightarrow {AC} ^2} - \frac{1}{2}\overrightarrow {AC} .\overrightarrow {AB} \\ =  - \frac{1}{2}A{B^2} + \frac{7}{{24}}A{C^2} - \frac{5}{{24}}\overrightarrow {AB} .\overrightarrow {AC} \\ =  - \frac{1}{2}{.2^2} + \frac{7}{{24}}{.3^2} - \frac{5}{{24}}.3\\ = 0\end{array}\)

\( \Rightarrow AM \bot BD\)

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=\dfrac{5}{24}\)

\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cosA=10a^2\)

+) Ta có: \(AB \bot AC \Rightarrow \overrightarrow {AB}  \bot \overrightarrow {AC}  \Rightarrow \overrightarrow {AB} .\overrightarrow {AC}  = 0\)

+) \(\overrightarrow {AC} .\overrightarrow {BC}  = \left| {\overrightarrow {AC} } \right|.\left| {\overline {BC} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right)\)

Ta có: \(BC = \sqrt {A{B^2} + A{C^2}}  = \sqrt 2  \Leftrightarrow \sqrt {2A{C^2}}  = \sqrt 2 \)\( \Rightarrow AC = 1\)

\( \Rightarrow \overrightarrow {AC} .\overrightarrow {BC}  = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)

+) \(\overrightarrow {BA} .\overrightarrow {BC}  = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|.\cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)

a: ΔABC đều cạnh a

=>AB=AC=BC=a; \(\hat{ABC}=\hat{ACB}=\hat{BAC}=60^0\)

\(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cosBAC\)

\(=a\cdot a\cdot cos60=a^2\cdot\frac12=\frac{a^2}{2}\)

\(\overrightarrow{BC}\cdot\overrightarrow{AC}=\overrightarrow{CB}\cdot\overrightarrow{CA}\)

\(=CB\cdot CA\cdot cosACB\)

\(=a\cdot a\cdot cos60=a^2\cdot\frac12=\frac{a^2}{2}\)

b: \(3\cdot\overrightarrow{BM}=2\cdot\overrightarrow{BC}\)

=>\(\overrightarrow{BM}=\frac23\cdot\overrightarrow{BC}\)

=>\(BM=\frac23BC\) và M nằm giữa B và C

Ta có: BM+MC=BC

=>\(MC=BC-BM=BC-\frac23BC=\frac13BC\)

\(5\cdot\overrightarrow{AN}=4\cdot\overrightarrow{AC}\)

=>\(\overrightarrow{AN}=\frac45\cdot\overrightarrow{AC}\)

=>AN=4/5AC và N nằm giữa A và C

\(\overrightarrow{BN}\cdot\overrightarrow{AM}=\left(\overrightarrow{BA}+\overrightarrow{AN}\right)\left(\overrightarrow{AB}+\overrightarrow{BM}\right)\)

\(=\left(-\overrightarrow{AB}+\frac45\cdot\overrightarrow{AC}\right)\left(\overrightarrow{AB}+\frac23\cdot\overrightarrow{BC}\right)=-\overrightarrow{AB}\cdot\overrightarrow{AB}-\frac23\cdot\overrightarrow{AB}\cdot\overrightarrow{BC}+\frac45\cdot\overrightarrow{AC}\cdot\overrightarrow{AB}+\frac{8}{15}\cdot\overrightarrow{AC}\cdot\overrightarrow{BC}\)

\(=-AB\cdot AB\cdot cos0+\frac23\cdot\overrightarrow{BA}\cdot\overrightarrow{BC}+\frac45\cdot\overrightarrow{AB}\cdot\overrightarrow{AC}+\frac{8}{15}\overrightarrow{CA}\cdot\overrightarrow{CB}\)

\(=-AB^2+\frac23\cdot BA\cdot BC\cdot cos60+\frac45\cdot AB\cdot AC\cdot cos60+\frac{8}{15}\cdot CA\cdot CB\cdot cos60\)

\(=-a^2+\frac13a^2+\frac25a^2+\frac{4}{15}a^2=-\frac23a^2+\frac25a^2+\frac{4}{15}a^2=\frac{-10+6+4}{15}\cdot a^2=0\)

=>AM⊥BN

Tham khảo:

a)  \(\)\(\overrightarrow {BA}  + \overrightarrow {AC}  = \overrightarrow {BC}  \Rightarrow \left| {\overrightarrow {BC} } \right| = BC = a\)

b) Dựng hình bình hành ABDC, giao điểm của hai đường chéo là O ta có:

\(\overrightarrow {AB}  + \overrightarrow {AC}  = \overrightarrow {AD} \)

\(AD = 2AO = 2\sqrt {A{B^2} - B{O^2}}  = 2\sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}}  = a\sqrt 3 \)

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

c) \(\overrightarrow {BA}  - \overrightarrow {BC}  = \overrightarrow {BA}  + \overrightarrow {CB}  = \overrightarrow {CB}  + \overrightarrow {BA}  = \overrightarrow {CA} \)

\( \Rightarrow \left| {\overrightarrow {BA}  - \overrightarrow {BC} } \right| = \left| {\overrightarrow {CA} } \right| = CA = a\)