已知向量$\vec {a} \vec {b} \quad \vec {a},\quad \vec {b} \text {满足} | \vec {a} | = 5,| \vec {b} | = 6,\quad \vec {a} \cdot \vec {b} = - 6$,则$\operatorname {cos} \langle \vec {a},\vec {a} + \vec {b} \rangle =$( )
本题考查平面向量夹角余弦值的计算,同时也考查了平面向量数量积的计算以及向量模的计算,考查计算能力,属于中等题.
计算出$\vec {a} \cdot ( \vec {a} + \vec {b} ),| \vec {a} + \vec {b} |$的值,利用平面向量数量积可计算出$\operatorname {cos} \langle \vec {a},\vec {a} + \vec {b} \rangle$的值.
$\because | \vec {a} | = 5,\quad | \vec {b} | = 6,\quad \vec {a} \cdot \vec {b} = - 6,$
$\therefore \vec {a} \cdot ( \vec {a} + \vec {b} ) = | \vec {a} | ^ {2} + \vec {a} \cdot \vec {b} = 5 ^ {2} - 6 = 19$.
$| \vec {a} + \vec {b} | = \sqrt {( \vec {a} + \vec {b} ) ^ {2}} = \sqrt {\vec {a} ^ {2} + 2 \vec {a} \cdot \vec {b} + \vec {b} ^ {2}} = \sqrt {25 - 2 \times 6 + 36} = 7$,
因此,$\operatorname {cos} < \vec {a},\vec {a} + \vec {b} > = \frac {\vec {a} \cdot ( \vec {a} + \vec {b} )} {| \vec {a} | \cdot | \vec {a} + \vec {b} |} = \frac {19} {5 \times 7} = \frac {19} {35}$.
故选:选项4-.