Cho hai vectơ $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4,(\overrightarrow{a},\overrightarrow{b})={150}^{°}$. Khi đó:

a) \(\vec a \cdot \vec b =  - 6\sqrt 3 \)

b) \((\vec a + \vec b) \cdot (\vec a - \vec b) = 7.\)

c) \((3\vec a + \vec b) \cdot (\vec a - 2\vec b) =  - 5 + 30\sqrt 3 \)

d) \((3\vec a + \vec b) \cdot (\vec a - 2\vec b) = 5 + 30\sqrt 3 \)

Ta có: \(\overrightarrow {CM}  = \overrightarrow {BM}  - \overrightarrow {BC}  = \frac{1}{2}\overrightarrow {BA}  - \overrightarrow {BC} \).

Vì \(G\) là trọng tâm của tam giác \(ACM\) nên

\(3\overrightarrow {BG}  = \overrightarrow {BA}  + \overrightarrow {BM}  + \overrightarrow {BC}  = \overrightarrow {BA}  + \frac{1}{2}\overrightarrow {BA}  + \overrightarrow {BC}  = \frac{3}{2}\overrightarrow {BA}  + \overrightarrow {BC}  \Rightarrow \overrightarrow {BG}  = \frac{1}{2}\overrightarrow {BA}  + \frac{1}{3}\overrightarrow {BC} .\)

Vì \(ABCD\) là hình chữ nhật nên \(BC = AD = 3a,\overrightarrow {BC}  \cdot \overrightarrow {BA}  = 0\).

Ta có: \(\overrightarrow {BG}  \cdot \overrightarrow {CM}  = \left( {\frac{1}{2}\overrightarrow {BA}  + \frac{1}{3}\overrightarrow {BC} } \right) \cdot \left( {\frac{1}{2}\overrightarrow {BA}  - \overrightarrow {BC} } \right) = \frac{1}{4}{\overrightarrow {BA} ^2} - \frac{1}{3}\overrightarrow {BA}  \cdot \overrightarrow {BC}  - \frac{1}{3}{\overrightarrow {BC} ^2}\)

\( = \frac{1}{4}{(4a)^2} - \frac{1}{3} \cdot 4a \cdot 3a - \frac{1}{3}{(3a)^2} =  - 3{a^2}.\)

a) Ta có: \(\vec a - \vec b = ( - 6;2) \Rightarrow \vec a(\vec a - \vec b) =  - 2( - 6) + 3.2 = 18\);

\(\vec a + \vec b = (2;4),2\vec a - \vec b = ( - 8;5) \Rightarrow (\vec a + \vec b)(2\vec a - \vec b) = 2( - 8) + 4.5 = 4\).

b) Ta có: \(\vec c = (m;1)\). Vì \(\vec c \bot \vec a\) nên \(\vec a \cdot \vec c = 0 \Rightarrow  - 2m + 3 \cdot 1 = 0 \Rightarrow m = \frac{3}{2}\).

c) Gọi \(\vec d = (x;y)\). Ta có: \(\left\{ {\begin{array}{*{20}{l}}{\vec a \cdot \vec d = 4}\\{\vec b \cdot \vec d =  - 2}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{ - 2x + 3y = 4}\\{4x + y =  - 2}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x =  - \frac{5}{7}}\\{y = \frac{6}{7}}\end{array}} \right.} \right.} \right.\) Vậy \(\vec d = \left( { - \frac{5}{7};\frac{6}{7}} \right)\).