+) \(\left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = \widehat {ABC} = 60^\circ \)

+) Dựng hình bình hành ABCD, ta có: \(\overrightarrow {AD}  = \overrightarrow {BC} \)

\( \Rightarrow \left( {\overrightarrow {AB} ,\overrightarrow {BC} } \right) = \left( {\overrightarrow {AB} ,\overrightarrow {AD} } \right) = \widehat {BAD} = 120^\circ \)

+), Ta có: ABC là tam giác đều, H là trung điểm BC nên  \(AH \bot BC\)

\(\left( {\overrightarrow {AH} ,\overrightarrow {BC} } \right) = \left( {\overrightarrow {AH} ,\overrightarrow {AD} } \right) = \widehat {HAD} = 90^\circ \)

+) Hai vectơ \(\overrightarrow {BH} \) và \(\overrightarrow {BC} \)cùng hướng nên \(\left( {\overrightarrow {BH} ,\overrightarrow {BC} } \right) = 0^\circ \)

+) Hai vectơ \(\overrightarrow {HB} \) và \(\overrightarrow {BC} \)ngược hướng nên \(\left( {\overrightarrow {HB} ,\overrightarrow {BC} } \right) = 180^\circ \)

1.

Dựng \(\overrightarrow{DB'}=\overrightarrow{CB}\)

\(k\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{DB}\)

\(=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{AB}\)

\(=2\overrightarrow{AB}+\overrightarrow{B'D}+\overrightarrow{DA}\)

\(=2\overrightarrow{AB}+\overrightarrow{B'A}\)

\(=2\overrightarrow{AB}+2\overrightarrow{AB}=4\overrightarrow{AB}\)

\(\Rightarrow k=4\)

Gọi M là trung điểm IB

\(\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\left|2\overrightarrow{AM}\right|=2AM\)

Ta có \(\overrightarrow{AM}^2=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2=MI^2+IA^2-2MI.IA.cos90^o=\dfrac{1}{16}a^2+\dfrac{3}{4}a^2=\dfrac{13}{16}a^2\)

\(\Rightarrow AM=\dfrac{\sqrt{13}}{4}a\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\dfrac{\sqrt{13}}{2}a\)

a) Ta có: \(\overrightarrow {BC} ,\overrightarrow {PN} \) là hai vecto cùng hướng và \(\frac{1}{2}\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {PN} } \right|\)

\( \Rightarrow \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {PN} \)\( \Rightarrow \overrightarrow {AP}  + \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {AP}  + \overrightarrow {PN}  = \overrightarrow {AN} \)

b) Ta có: \(\overrightarrow {MP} ,\overrightarrow {CA} \) là hai vecto cùng hướng và \(2\left| {\overrightarrow {MP} } \right| = \left| {\overrightarrow {CA} } \right|\)

\( \Rightarrow 2\overrightarrow {MP}  = \overrightarrow {CA} \)\( \Rightarrow \overrightarrow {BC}  + 2\overrightarrow {MP}  = \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow {BA} \)

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:

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

\( \Rightarrow AM \bot BD\)

Lời giải:

Áp dụng các công thức sau: \(|\overrightarrow {a}|^2=\overrightarrow{a}.\overrightarrow{a}\)

\(\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{0}\) nếu \(\overrightarrow{a}\perp \overrightarrow{b}\)

Ta có:

\(BC^2.\overrightarrow{IA}+AC^2.\overrightarrow{IB}+AB^2.\overrightarrow{IC}\)

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

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

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

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

\(=\overrightarrow {BC}.\overrightarrow{0}+\overrightarrow{AC}.\overrightarrow{0}+\overrightarrow{AB}.\overrightarrow{0}=\overrightarrow {0}\)