Cho hai vectơ ∣ ∣ ∣ → a ∣ ∣ ∣ = 3 , ∣ ∣ ∣ → b ∣ ∣ ∣ = 4 , ( → a , → b ) = 150 ∘ . Khi đó: a) → a ⋅ → b = − 6 √ 3

Ta có: a→⋅b→=|a→|⋅|b→|⋅cos(a→,b→)=3⋅4⋅cos150°=−63

\(\begin{array}{l}(\vec a + \vec b) \cdot (\vec a - \vec b) = {{\vec a}^2} - {{\vec b}^2} = |\vec a{|^2} - |\vec b{|^2} = {3^2} - {4^2} =  - 7.\\\begin{array}{*{20}{l}}{(3\vec a + \vec b) \cdot (\vec a - 2\vec b)}&{ = 3{{\vec a}^2} - 5\vec a \cdot \vec b - 2{{\vec b}^2} = 3|\vec a{|^2} - 5\vec a \cdot \vec b - 2|\vec b{|^2}}\\{}&{ = 3 \cdot {3^2} - 5( - 6\sqrt 3 ) - 2 \cdot {4^2} =  - 5 + 30\sqrt 3 .}\end{array}\end{array}\)