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正确率40.0%在$${{Δ}{A}{B}{C}}$$中，$$a， b， c$$所对的角为$$A， ~ B， ~ C$$，满足条件：$$\operatorname{s i n} C=\sqrt{3} \operatorname{s i n} B，$$$$a ( \operatorname{s i n} B-2 \operatorname{c o s} C )=( 2 c-\sqrt{3} b ) \operatorname{c o s} A，$$$$\overrightarrow{A C} \cdot\overrightarrow{A B}=6$$，则$${{B}{C}}$$边长等于（）
A
A.$${{2}}$$
B.$${\sqrt {5}}$$
C.$${{3}}$$
D.$${{4}}$$
3、['数量积的性质'， '向量的数量积的定义'， '向量的夹角']正确率60.0%已知$$\overrightarrow{a}=\ ( 1， \ 0 ) \， \ | \overrightarrow{b} |=\sqrt{2}， \ | \overrightarrow{a}-\overrightarrow{b} |=| \overrightarrow{a} |，$$则$${{a}^{→}{，}{{b}^{→}}}$$的夹角是（）
C
A.$$\frac{\pi} {2}$$
B.$$\frac{\pi} {3}$$
C.$$\frac{\pi} {4}$$
D.$$\begin{array} {l l} {\frac{\pi} {6}} \\ \end{array}$$
4、['向量的模'， '向量的数量积的定义'， '向量的线性运算']正确率40.0%设$${{M}}$$是线段$${{B}{C}}$$的中点，点$${{A}}$$在直线$${{B}{C}}$$外，$$| \overrightarrow{B C} |=6$$，且$$| \overrightarrow{A B}+\overrightarrow{A C} |=| \overrightarrow{A B}-\overrightarrow{A C} |$$，则$$| \overrightarrow{A M} |=( \textsubscript{-} )$$
C
A.$${{1}{2}}$$
B.$${{6}}$$
C.$${{3}}$$
D.$${{1}}$$
5、['向量的数量积的定义'， '投影的数量']正确率60.0%已知向量$${{b}^{→}}$$在向量$${{a}^{→}}$$方向上的投影为$${{2}}$$，且$$\left\vert\overrightarrow{a} \right\vert=1，$$则$$\overrightarrow{a} \cdot\overrightarrow{b}=( \textit{} )$$
D
A.$${{−}{2}}$$
B.$${{−}{1}}$$
C.$${{1}}$$
D.$${{2}}$$
7、['数量积的运算律'， '向量的数量积的定义'， '投影的数量']正确率40.0%已知向量$${{a}^{→}{，}{{b}^{→}}}$$满足$$\overrightarrow{a} \cdot( \overrightarrow{a}+\overrightarrow{b} )=5$$且$$\vert\overrightarrow{a} \vert=2， \vert\overrightarrow{b} \vert=1，$$则向量$${{a}^{→}}$$在向量$${{b}^{→}}$$方向的投影为$${{(}{)}}$$
B
A.$$\frac{1} {2}$$
B.$${{1}}$$
C.$$\begin{array} {l l} {\frac{3} {2}} \\ \end{array}$$
D.$${{2}}$$
8、['投影向量（投影）'， '向量的数量积的定义']正确率60.0%已知$$\left\vert\vec{a} \right\vert=6， \left\vert\vec{b} \right\vert=3， \vec{a} \cdot\vec{b}=-1 2，$$则向量$${{a}{⃗}}$$在$${{b}^{⃗}}$$方向上的投影为（）
A
A.$${{−}{4}}$$
B.$${{4}}$$
C.$${{−}{2}}$$
D.$${{2}}$$
9、['数量积的性质'， '数量积的运算律'， '向量垂直'， '向量的数量积的定义']正确率60.0%若平面向量$${{a}^{→}{，}{{b}^{→}}}$$满足$$( 2 \overrightarrow{a}-\overrightarrow{b} ) / \perp\overrightarrow{b}$$，则下列各式恒成立的是（）
C
A.$$| \overrightarrow{a}+\overrightarrow{b} |=| \overrightarrow{a} |$$
B.$$| \overrightarrow{a}+\overrightarrow{b} |=| \overrightarrow{b} |$$
C.$$| \overrightarrow{a}-\overrightarrow{b} |=| \overrightarrow{a} |$$
D.$$| \overrightarrow{a}-\overrightarrow{b} |=| \overrightarrow{b} |$$
10、['共线向量基本定理'， '数量积的运算律'， '向量的数量积的定义']正确率40.0%平行四边形$${{A}{B}{C}{D}}$$中，$$A B=2，$$$$A D=1， \， \， \， \overrightarrow{A B} \cdot\， \overrightarrow{A D}=-\sqrt{3}$$，点$${{M}}$$在边$${{C}{D}}$$上，则$$\overrightarrow{M A} \cdot\overrightarrow{M B}$$的最小值为（）
C
A.$$- \frac{1} {2}$$
B.$${{1}{−}{2}{\sqrt {2}}}$$
C.$$- \frac{3} {4}$$
D.$${{1}{−}{\sqrt {3}}}$$
1. 已知：$$\sin C = \sqrt{3} \sin B$$，$$a (\sin B - 2 \cos C) = (2c - \sqrt{3} b) \cos A$$，$$\overrightarrow{AC} \cdot \overrightarrow{AB} = 6$$
由正弦定理：$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$
代入$$\sin C = \sqrt{3} \sin B$$得：$$c = \sqrt{3} b$$
将已知等式整理：$$a \sin B - 2a \cos C = 2c \cos A - \sqrt{3} b \cos A$$
代入$$c = \sqrt{3} b$$得：$$a \sin B - 2a \cos C = 2\sqrt{3} b \cos A - \sqrt{3} b \cos A = \sqrt{3} b \cos A$$
由正弦定理：$$a = 2R \sin A$$，$$b = 2R \sin B$$，代入得：
$$2R \sin A \sin B - 4R \sin A \cos C = \sqrt{3} \cdot 2R \sin B \cos A$$
两边除以$$2R$$：$$\sin A \sin B - 2 \sin A \cos C = \sqrt{3} \sin B \cos A$$
整理：$$\sin A \sin B - \sqrt{3} \sin B \cos A = 2 \sin A \cos C$$
左边提取$$\sin B$$：$$\sin B (\sin A - \sqrt{3} \cos A) = 2 \sin A \cos C$$
由三角形内角和：$$C = \pi - A - B$$，$$\cos C = -\cos(A+B) = -\cos A \cos B + \sin A \sin B$$
代入得：$$\sin B (\sin A - \sqrt{3} \cos A) = 2 \sin A (-\cos A \cos B + \sin A \sin B)$$
展开右边：$$\sin B \sin A - \sqrt{3} \sin B \cos A = -2 \sin A \cos A \cos B + 2 \sin^2 A \sin B$$
移项：$$\sin A \sin B - 2 \sin^2 A \sin B + \sqrt{3} \sin B \cos A - 2 \sin A \cos A \cos B = 0$$
提取公因式：$$\sin B (\sin A - 2 \sin^2 A) + \cos A (\sqrt{3} \sin B - 2 \sin A \cos B) = 0$$
观察得可能$$\sin B = 0$$或特殊角，尝试$$A = \frac{\pi}{3}$$，$$B = \frac{\pi}{6}$$，则$$C = \frac{\pi}{2}$$
验证：$$\sin C = 1$$，$$\sqrt{3} \sin B = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$$，不相等，需调整。
由$$\sin C = \sqrt{3} \sin B$$，且$$C = \pi - A - B$$，设$$B = x$$，则$$C = \pi - A - x$$，$$\sin(\pi - A - x) = \sin(A + x) = \sqrt{3} \sin x$$
即$$\sin A \cos x + \cos A \sin x = \sqrt{3} \sin x$$
若$$\sin x \neq 0$$，则$$\sin A \cot x + \cos A = \sqrt{3}$$
结合另一条件，尝试$$A = \frac{\pi}{2}$$，则$$\sin A = 1$$，$$\cos A = 0$$，代入得：$$\cot x = \sqrt{3}$$，即$$x = \frac{\pi}{6}$$，则$$B = \frac{\pi}{6}$$，$$C = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$？错误，应为$$C = \pi - \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$
验证：$$\sin C = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$，$$\sqrt{3} \sin B = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$$，成立。
代入另一条件：$$a (\sin B - 2 \cos C) = (2c - \sqrt{3} b) \cos A$$
$$A = \frac{\pi}{2}$$，$$\cos A = 0$$，右边为0，左边需为0，即$$\sin B - 2 \cos C = 0$$
$$\sin \frac{\pi}{6} - 2 \cos \frac{\pi}{3} = \frac{1}{2} - 2 \cdot \frac{1}{2} = \frac{1}{2} - 1 = -\frac{1}{2} \neq 0$$，矛盾。
重新考虑，由$$\overrightarrow{AC} \cdot \overrightarrow{AB} = 6$$，即$$b c \cos A = 6$$
结合$$c = \sqrt{3} b$$，得$$\sqrt{3} b^2 \cos A = 6$$，即$$b^2 \cos A = \frac{6}{\sqrt{3}} = 2\sqrt{3}$$
由正弦定理：$$b = 2R \sin B$$，$$a = 2R \sin A$$，代入原式：
$$2R \sin A (\sin B - 2 \cos C) = (2 \cdot \sqrt{3} b - \sqrt{3} b) \cos A = \sqrt{3} b \cos A$$
即$$2R \sin A \sin B - 4R \sin A \cos C = \sqrt{3} b \cos A$$
代入$$b = 2R \sin B$$：$$2R \sin A \sin B - 4R \sin A \cos C = \sqrt{3} \cdot 2R \sin B \cos A$$
除以$$2R$$：$$\sin A \sin B - 2 \sin A \cos C = \sqrt{3} \sin B \cos A$$
即$$\sin A \sin B - \sqrt{3} \sin B \cos A = 2 \sin A \cos C$$
$$\sin B (\sin A - \sqrt{3} \cos A) = 2 \sin A \cos C$$
由$$C = \pi - A - B$$，$$\cos C = -\cos(A+B) = -\cos A \cos B + \sin A \sin B$$
代入：$$\sin B (\sin A - \sqrt{3} \cos A) = 2 \sin A (-\cos A \cos B + \sin A \sin B)$$
$$\sin B \sin A - \sqrt{3} \sin B \cos A = -2 \sin A \cos A \cos B + 2 \sin^2 A \sin B$$
移项：$$\sin A \sin B - 2 \sin^2 A \sin B + \sqrt{3} \sin B \cos A + 2 \sin A \cos A \cos B = 0$$
$$\sin B (\sin A - 2 \sin^2 A) + \cos A (\sqrt{3} \sin B + 2 \sin A \cos B) = 0$$
尝试$$A = \frac{\pi}{3}$$，则$$\sin A = \frac{\sqrt{3}}{2}$$，$$\cos A = \frac{1}{2}$$
代入：$$\sin B (\frac{\sqrt{3}}{2} - 2 \cdot \frac{3}{4}) + \frac{1}{2} (\sqrt{3} \sin B + 2 \cdot \frac{\sqrt{3}}{2} \cos B) = \sin B (\frac{\sqrt{3}}{2} - \frac{3}{2}) + \frac{1}{2} (\sqrt{3} \sin B + \sqrt{3} \cos B) = \sin B (\frac{\sqrt{3} - 3}{2}) + \frac{\sqrt{3}}{2} (\sin B + \cos B) = 0$$
即$$\frac{\sqrt{3} - 3}{2} \sin B + \frac{\sqrt{3}}{2} \sin B + \frac{\sqrt{3}}{2} \cos B = 0$$
$$\frac{2\sqrt{3} - 3}{2} \sin B + \frac{\sqrt{3}}{2} \cos B = 0$$
即$$(2\sqrt{3} - 3) \sin B + \sqrt{3} \cos B = 0$$
$$\tan B = -\frac{\sqrt{3}}{2\sqrt{3} - 3}$$，不为特殊角，可能复杂。
由$$\sin C = \sqrt{3} \sin B$$，且$$C = \pi - A - B$$，设$$A = \frac{\pi}{2}$$，则$$C = \frac{\pi}{2} - B$$，$$\sin C = \cos B = \sqrt{3} \sin B$$，即$$\cot B = \sqrt{3}$$，$$B = \frac{\pi}{6}$$，$$C = \frac{\pi}{3}$$
代入原式：$$a (\sin B - 2 \cos C) = (2c - \sqrt{3} b) \cos A$$
$$A = \frac{\pi}{2}$$，$$\cos A = 0$$，右边为0，左边需为0，即$$\sin B - 2 \cos C = 0$$
$$\sin \frac{\pi}{6} - 2 \cos \frac{\pi}{3} = \frac{1}{2} - 2 \cdot \frac{1}{2} = -\frac{1}{2} \neq 0$$，不成立。
设$$A = \frac{\pi}{6}$$，则由$$\sin C = \sqrt{3} \sin B$$，且$$C = \pi - \frac{\pi}{6} - B = \frac{5\pi}{6} - B$$，$$\sin(\frac{5\pi}{6} - B) = \sin \frac{5\pi}{6} \cos B - \cos \frac{5\pi}{6} \sin B = \frac{1}{2} \cos B - (-\frac{\sqrt{3}}{2}) \sin B = \frac{1}{2} \cos B + \frac{\sqrt{3}}{2} \sin B = \sqrt{3} \sin B$$
即$$\frac{1}{2} \cos B + \frac{\sqrt{3}}{2} \sin B = \sqrt{3} \sin B$$
$$\frac{1}{2} \cos B = \frac{\sqrt{3}}{2} \sin B$$
$$\cot B = \sqrt{3}$$，$$B = \frac{\pi}{6}$$，则$$C = \pi - \frac{\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$$
验证：$$\sin C = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$$，$$\sqrt{3} \sin B = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$$，成立。
代入另一条件：$$a (\sin B - 2 \cos C) = (2c - \sqrt{3} b) \cos A$$
$$a (\sin \frac{\pi}{6} - 2 \cos \frac{2\pi}{3}) = (2c - \sqrt{3} b) \cos \frac{\pi}{6}$$
$$a (\frac{1}{2} - 2 \cdot (-\frac{1}{2})) = (2c - \sqrt{3} b) \cdot \frac{\sqrt{3}}{2}$$
$$a (\frac{1}{2} + 1) = \frac{\sqrt{3}}{2} (2c - \sqrt{3} b)$$
$$\frac{3}{2} a = \frac{\sqrt{3}}{2} (2c - \sqrt{3} b)$$
两边乘以2：$$3a = \sqrt{3} (2c - \sqrt{3} b) = 2\sqrt{3} c - 3b$$
即$$3a + 3b = 2\sqrt{3} c$$，$$a + b = \frac{2\sqrt{3}}{3} c$$
由正弦定理：$$a = 2R \sin A = 2R \sin \frac{\pi}{6} = R$$，$$b = 2R \sin B = 2R \sin \frac{\pi}{6} = R$$，$$c = 2R \sin C = 2R \sin \frac{2\pi}{3} = 2R \cdot \frac{\sqrt{3}}{2} = \sqrt{3} R$$
代入：$$R + R = \frac{2\sqrt{3}}{3} \cdot \sqrt{3} R = \frac{2 \cdot 3}{3} R = 2R$$，成立。
现在$$\overrightarrow{AC} \cdot \overrightarrow{AB} = 6$$，即$$b c \cos A = R \cdot \sqrt{3} R \cdot \cos \frac{\pi}{6} = \sqrt{3} R^2 \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} R^2 = 6$$，得$$R^2 = 4$$，$$R = 2$$
则边长$$BC = a = R = 2$$
故选A.$$2$$
3. 已知：$$\overrightarrow{a} = (1, 0)$$，$$|\overrightarrow{b}| = \sqrt{2}$$，$$|\overrightarrow{a} - \overrightarrow{b}| = |\overrightarrow{a}|$$
设$$\overrightarrow{b} = (x, y)$$，则$$x^2 + y^2 = 2$$
$$|\overrightarrow{a} - \overrightarrow{b}| = |(1-x, -y)| = \sqrt{(1-x)^2 + y^2} = |\overrightarrow{a}| = 1$$
即$$(1-x)^2 + y^2 = 1$$
展开：$$1 - 2x + x^2 + y^2 = 1$$
代入$$x^2 + y^2 = 2$$：$$1 - 2x + 2 = 1$$，即$$3 - 2x = 1$$，$$2x = 2$$，$$x = 1$$
则$$y^2 = 2 - 1 = 1$$，$$y = \pm 1$$
$$\overrightarrow{a} \cdot \overrightarrow{b} = 1 \cdot 1 + 0 \cdot y = 1$$
夹角$$\theta$$满足：$$\cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} = \frac{1}{1 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}$$
故$$\theta = \frac{\pi}{4}$$
故选C.$$\frac{\pi}{4}$$
4. 已知：M是BC中点，|BC| = 6，且|AB + AC| = |AB - AC|
由向量加法：$$\overrightarrow{AB} + \overrightarrow{AC} = 2 \overrightarrow{AM}$$
$$\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$$
故$$|2 \overrightarrow{AM}| = |\overrightarrow{CB}|$$，即$$2 |\overrightarrow{AM}| = |\overrightarrow{CB}| = 6$$
所以$$|\overrightarrow{AM}| = 3$$
故选C.$$3$$
5. 向量$$\overrightarrow{b}$$在$$\overrightarrow{a}$$方向上的投影为2，且$$|\overrightarrow{a}| = 1$$
投影公式：$$\text{proj}_{\overrightarrow{a}} \overrightarrow{b} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}|} = 2$$
故$$\overrightarrow{a} \cdot \overrightarrow{b} = 2 \cdot 1 = 2$$
故选D.$$ 题目来源于各渠道收集，若侵权请联系下方邮箱