Cho tam giác \(ABC\) có \(AB = a,AC = 2a,\widehat A = 60^\circ \). \(M\) là điểm thỏa mãn \(2\overrightarrow {MA}  + 3\overrightarrow {MB}  = \overrightarrow 0 \).

a) Vì \(\Delta ABC\) vuông tại \(A\), ta có:

\(BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} = 2\).

\(\cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = \cos \widehat {ABC} = \frac{{AB}}{{BC}} = \frac{1}{2} \Rightarrow \widehat {ABC} = 60^\circ \Rightarrow \widehat {ACB} = 30^\circ \).

\(\cos \left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right) = \cos \left( {\overrightarrow {CD} ,\overrightarrow {CB} } \right) = \cos \widehat {BCD} = \cos 150^\circ = - \frac{{\sqrt 3 }}{2}\).

b) \(\overrightarrow {BA} \cdot \overrightarrow {BC} \)\( = \left| {\overrightarrow {BA} } \right| \cdot \left| {\overrightarrow {BC} } \right| \cdot \cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = 1 \cdot 2 \cdot \frac{1}{2} = 1\).

\(\overrightarrow {AC} \cdot \overrightarrow {CB} = \left| {\overrightarrow {AC} } \right| \cdot \left| {\overrightarrow {CB} } \right| \cdot \cos \left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right) = \sqrt 3 \cdot 2 \cdot \left( { - \frac{{\sqrt 3 }}{2}} \right) = - 3\).

a) Có \(\overrightarrow {AB} \cdot \overrightarrow {AC} = \left| {\overrightarrow {AB} } \right| \cdot \left| {\overrightarrow {AC} } \right| \cdot \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = 2 \cdot 3 \cdot \cos 60^\circ = 3\).

 \(\overrightarrow {AB} \cdot \overrightarrow {BC} = \overrightarrow {AB} \cdot \left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) = \overrightarrow {AB} \cdot \overrightarrow {AC} - {\overrightarrow {AB} ^2} = 3 - {2^2} = - 1\).

b) Có \(\overrightarrow {MN} = \overrightarrow {MA} + \overrightarrow {AN} = - \frac{1}{2}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AC} \).

Có \(\overrightarrow {MP} = \overrightarrow {MB} + \overrightarrow {BP} = \frac{1}{2}\overrightarrow {AB} + 2\overrightarrow {BC} = \frac{1}{2}\overrightarrow {AB} + 2\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) = - \frac{3}{2}\overrightarrow {AB} + 2\overrightarrow {AC} \).

Do đó \(\overrightarrow {MN} = \frac{1}{3}\overrightarrow {MP} \). Suy ra \(M,N,P\) thẳng hàng.