Given $\overrightarrow { A B } = ( 2,3 ) , \overrightarrow { A C } = ( 3 , t ) , | \overrightarrow { B C } | = 1$, then $\overrightarrow { A B } \cdot \overrightarrow { B C } =$
A． - 3
B． - 2
C． 2
D． 3

Let points $A , B , C$ be non-collinear. Then ``the angle between $\overrightarrow { A B }$ and $\overrightarrow { A C }$ is acute'' is ``$| \overrightarrow { A B } + \overrightarrow { A C } | > | \overrightarrow { B C } |$'' a (A) sufficient but not necessary condition (B) necessary but not sufficient condition (C) necessary and sufficient condition (D) neither sufficient nor necessary condition

7. Given non-zero vectors $a , b$ satisfying $| a | = 2 | b |$ and $( a - b ) \perp b$, the angle between $a$ and $b$ is
A. $\frac { \pi } { 6 }$
B. $\frac { \pi } { 3 }$
C. $\frac { 2 \pi } { 3 }$
D. $\frac { 5 \pi } { 6 }$

7. Given non-zero vectors $\boldsymbol { a } , \boldsymbol { b }$ satisfying $| \boldsymbol { a } | = 2 | \boldsymbol { b } |$ and $( \boldsymbol { a } - \boldsymbol { b } ) \perp \boldsymbol { b }$ , the angle between $\boldsymbol { a }$ and $\boldsymbol { b }$ is
A. $\frac { \pi } { 6 }$
B. $\frac { \pi } { 3 }$
C. $\frac { 2 \pi } { 3 }$
D. $\frac { 5 \pi } { 6 }$

Given that $\boldsymbol { a } , \boldsymbol { b }$ are unit vectors and $\boldsymbol { a } \cdot \boldsymbol { b } = 0$ , if $\boldsymbol { c } = 2 \boldsymbol { a } - \sqrt { 5 } \boldsymbol { b }$ , then $\cos \langle \boldsymbol { a } , \boldsymbol { c } \rangle =$ \_\_\_\_\_\_.

13. Given vectors $\boldsymbol { a } = ( 2,2 ) , \boldsymbol { b } = ( - 8,6 )$ , then $\cos \langle \boldsymbol { a } , \boldsymbol { b } \rangle =$

13. Given that $\boldsymbol { a } , \boldsymbol { b }$ are unit vectors and $\boldsymbol { a } \cdot \boldsymbol { b } = 0$ , if $\boldsymbol { c } = 2 \boldsymbol { a } - \sqrt { 5 } \boldsymbol { b }$ , then $\cos \langle \boldsymbol { a } , \boldsymbol { c } \rangle =$ $\_\_\_\_$ .

Given vectors $\boldsymbol { a } , \boldsymbol { b }$ satisfying $| \boldsymbol { a } | = 5 , | \boldsymbol { b } | = 6 , \boldsymbol { a } \cdot \boldsymbol { b } = - 6$ , then $\cos \langle \boldsymbol { a } , \boldsymbol { a } + \boldsymbol { b } \rangle =$
A. $- \frac { 31 } { 35 }$
B. $- \frac { 19 } { 35 }$
C. $\frac { 17 } { 35 }$
D. $\frac { 19 } { 35 }$

Given vectors $\overrightarrow { a _ { 1 } } , \overrightarrow { a _ { 2 } } , \overrightarrow { b _ { 1 } } , \overrightarrow { b _ { 2 } } , \ldots , \overrightarrow { b _ { k } } \left( k \in \mathbb{N} ^ { * } \right)$ that are pairwise non-parallel in the plane, satisfying $\left| \overrightarrow { a _ { 1 } } - \overrightarrow { a _ { 2 } } \right| = 1$ and $\left| \overrightarrow { a _ { i } } - \overrightarrow { b _ { j } } \right| \in \{ 1,2 \}$ (where $i = 1,2$ and $j = 1,2 , \ldots , k$), find the maximum value of $k$ as $\_\_\_\_$

Given unit vectors $\boldsymbol { a } , \boldsymbol { b }$ with an angle of $45 ^ { \circ }$ between them, and $k \boldsymbol { a } - \boldsymbol { b}$ is perpendicular to $\boldsymbol { a }$ , then $k = $ $\_\_\_\_$.

Let vectors $\boldsymbol { a } = ( 1 , - 1 ) , \boldsymbol { b } = ( m + 1,2 m - 4 )$ . If $\boldsymbol { a } \perp \boldsymbol { b }$ , then $m =$ $\_\_\_\_$.

10. AC

Solution: $\left| \overrightarrow { O P _ { 1 } } \right| = \left| \overrightarrow { O P _ { 2 } } \right| = 1$, so A is correct; $\left| \overrightarrow { A P _ { 1 } } \right| ^ { 2 } = 2 - 2 \cos \alpha , \left| \overrightarrow { A P _ { 2 } } \right| ^ { 2 } = 2 - 2 \cos \beta$, so B is incorrect; $\overrightarrow { O P _ { 1 } } \cdot \overrightarrow { O P _ { 2 } } \sin \alpha \sin \beta = \cos ( \alpha + \beta ) = \overrightarrow { O A } \cdot \overrightarrow { O P _ { 3 } }$, so C is correct; $\overrightarrow { O P _ { 2 } } \cdot \overrightarrow { O P _ { 3 } } = \cos ( \alpha + \beta ) \cos \beta - \sin ( \alpha + \beta ) \sin \beta \neq \overrightarrow { O P _ { 1 } } = \cos \alpha$, so D is incorrect. The answer is $AC$.

13. If vectors $\vec { a } , \vec { b }$ satisfy $| \vec { a } | = 3 , | \vec { a } - \vec { b } | = 5 , \vec { a } \cdot \vec { b } = 1$, then $| \vec { b } | =$ $\_\_\_\_$ .

14. Given vectors $\vec{a} = (3,1)$, $\vec{b} = (1,0)$, $\vec{c} = \vec{a} + k\vec{b}$. If $\vec{a} \perp \vec{c}$, then $k = \_\_\_\_$.

Given vectors $a = ( 2,1 ) , b = ( - 2,4 )$ , then $| a - b | =$
A. 2
B. 3
C. 4
D. 5

Given vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy $|\boldsymbol{a}| = 1, |\boldsymbol{b}| = \sqrt{3}, |\boldsymbol{a} - 2\boldsymbol{b}| = 3$, then $\boldsymbol{a} \cdot \boldsymbol{b} =$
A. $-2$
B. $-1$
C. $1$
D. $2$

3. In $\triangle A B C$, point $D$ is on side $A B$ with $B D = 2 D A$. Let $\overrightarrow { C A } = m$ and $\overrightarrow { C D } = n$. Then $\overrightarrow { C B } =$
A. $3 m - 2 n$
B. $- 2 m + 3 n$
C. $3 \boldsymbol { m } + 2 \boldsymbol { n }$
D. $2 m + 3 n$

9. Given a cube $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$, then
A. The angle between lines $B C _ { 1 }$ and $D A _ { 1 }$ is $90 ^ { \circ }$
B. The angle between lines $B C _ { 1 }$ and $C A _ { 1 }$ is $90 ^ { \circ }$
C. The angle between line $B C _ { 1 }$ and plane $B B _ { 1 } D _ { 1 } D$ is $45 ^ { \circ }$
D. The angle between line $B C _ { 1 }$ and plane $A B C D$ is $45 ^ { \circ }$

Given vectors $\boldsymbol { a } = ( m , 3 ) , \boldsymbol { b } = ( 1 , m + 1 )$ . If $\boldsymbol { a } \perp \boldsymbol { b }$ , then $m =$ $\_\_\_\_$ .

Let vectors $\boldsymbol { a }$ and $\boldsymbol { b }$ have an angle whose cosine is $\frac { 1 } { 3 }$, and $| \boldsymbol { a } | = 1$, $| \boldsymbol { b } | = 3$. Then $( 2 \boldsymbol { a } + \boldsymbol { b } ) \cdot \boldsymbol { b } =$ $\_\_\_\_$

Vectors $|\boldsymbol{a}| = |\boldsymbol{b}| = 1 , |\boldsymbol{c}| = \sqrt{2}$ and $\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c} = 0$ , then $\cos \langle \boldsymbol{a} - \boldsymbol{b} , \boldsymbol{b} - \boldsymbol{c} \rangle =$
A. $-\frac{1}{5}$
B. $-\frac{2}{5}$
C. $\frac{2}{5}$
D. $\frac{4}{5}$

Given vectors $\vec { a } , \vec { b }$ satisfying $| \vec { a } | = 1 , | \vec { a } + 2 \vec { b } | = 2$, and $( \vec { b } - 2 \vec { a } ) \perp \vec { b }$, then $| \vec { b } | =$
A. $\frac { 1 } { 2 }$
B. $\frac { \sqrt { 2 } } { 2 }$
C. $\frac { \sqrt { 3 } } { 2 }$
D. 1

Given vectors $\boldsymbol { a } = ( 0,1 ) , \boldsymbol { b } = ( 2 , x )$ , if $\boldsymbol { b } \perp ( \boldsymbol { b } - 4 \boldsymbol { a } )$ , then $x =$
A. $- 2$
B. $- 1$
C. $1$
D. $2$

Given vectors $\boldsymbol { a } , \boldsymbol { b }$, then ``$( \boldsymbol { a } + \boldsymbol { b } ) ( \boldsymbol { a } - \boldsymbol { b } ) = 0$'' is ``$\boldsymbol { a } = \boldsymbol { b }$ or $\boldsymbol { a } = - \boldsymbol { b }$'' a \_\_\_\_ condition.

In sailing competitions, athletes can use an anemometer to measure wind speed and direction. The measured result is called apparent wind speed in nautical science. The vector corresponding to apparent wind speed is the sum of the vector corresponding to true wind speed and the vector corresponding to ship's wind speed, where the vector corresponding to ship's wind speed has the same magnitude as the vector corresponding to ship's velocity but opposite direction. Figure 1 shows the correspondence between part of the wind force levels, names, and wind speeds. An athlete measured the vector corresponding to apparent wind speed and the vector corresponding to ship's velocity as shown in Figure 2 (the magnitude of wind speed and the magnitude of the vector are the same, unit: $\mathrm{m/s}$). Then the true wind is

Level	Wind Speed $\mathrm{m/s}$	Name
2	$1.1 \sim 3.3$	Light Breeze
3	$3.4 \sim 5.4$	Gentle Breeze
4	$5.5 \sim 7.9$	Moderate Wind
5	$8.0 \sim 10.1$	Fresh Wind

A. Light Breeze
B. Gentle Breeze
C. Moderate Wind
D. Fresh Wind