13. In $\triangle A B C$, points $M , N$ satisfy $\overrightarrow { A M } = 2 \overrightarrow { M C } , \overrightarrow { B N } = \overrightarrow { N C }$. If $\overrightarrow { M N } = x \overrightarrow { A B } + y \overrightarrow { A C }$, then $x =$ $\_\_\_\_$ $\_\_\_\_$ ; $y =$ $\_\_\_\_$.

6. Given vectors $\vec { a } = ( 2,1 ) , \vec { a } = ( 1 , - 2 )$, if $m \vec { a } + n \vec { b } = ( 9 , - 8 ) ( m n \in R )$, then the value of $\mathrm { m } - \mathrm { n }$ is $\_\_\_\_$ .

15. Given that $\vec { e } _ { 1 } , \vec { e } _ { 2 }$ are unit vectors in space with $\vec { e } _ { 1 } \cdot \vec { e } _ { 2 } = \frac { 1 } { 2 }$ . If the space vector $\vec { b }$ satisfies $\vec { b } \cdot \vec { e } _ { 1 } = 2 , \vec { b } \cdot \vec { e } _ { 2 } = \frac { 5 } { 2 }$ , and for all $x , y \in \mathbb{R}$ , $\left| \vec { b } - \left( x \vec { e } _ { 1 } + y \vec { e } _ { 2 } \right) \right| \geq \left| \vec { b } - \left( x _ { 0 } \vec { e } _ { 1 } + y _ { 0 } \vec { e } _ { 2 } \right) \right| = 1$ ( $x _ { 0 } , y _ { 0 } \in \mathbb{R}$ ), then $x _ { 0 } =$ $\_\_\_\_$ , $y _ { 0 } =$ $\_\_\_\_$ , $| \vec { b } | =$ $\_\_\_\_$ . III. Solution Questions: This section contains 5 questions, for a total of 74 points. Solutions should include explanations, proofs, or calculation steps.

In $\triangle A B C$, $AD$ is the median to side $BC$, and $E$ is the midpoint of $AD$. Then $\overrightarrow { E B } =$
A. $\frac { 3 } { 4 } \overrightarrow { A B } - \frac { 1 } { 4 } \overrightarrow { A C }$
B. $\frac { 1 } { 4 } \overrightarrow { A B } + \frac { 3 } { 4 } \overrightarrow { A C }$
C. $\frac { 3 } { 4 } \overrightarrow { A B } + \frac { 1 } { 4 } \overrightarrow { A C }$
D. $\frac { 1 } { 4 } \overrightarrow { A B } + \frac { 3 } { 4 } \overrightarrow { A C }$

In $\triangle A B C$, $A D$ is the median to side $B C$, and $E$ is the midpoint of $A D$. Then $\overrightarrow { E B } =$
A. $\frac { 3 } { 4 } \overrightarrow { A B } - \frac { 1 } { 4 } \overrightarrow { A C }$
B. $\frac { 1 } { 4 } \overrightarrow { A B } - \frac { 3 } { 4 } \overrightarrow { A C }$
C. $\frac { 3 } { 4 } \overrightarrow { A B } + \frac { 1 } { 4 } \overrightarrow { A C }$
D. $\frac { 1 } { 4 } \overrightarrow { A B } + \frac { 3 } { 4 } \overrightarrow { A C }$

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. $2 m + 3 n$

Suppose that $\vec { p } , \vec { q }$ and $\vec { r }$ are three non-coplanar vectors in $\mathbb { R } ^ { 3 }$. Let the components of a vector $\vec { s }$ along $\vec { p } , \vec { q }$ and $\vec { r }$ be 4,3 and 5 , respectively. If the components of this vector $\vec { s }$ along $( - \vec { p } + \vec { q } + \vec { r } ) , ( \vec { p } - \vec { q } + \vec { r } )$ and $( - \vec { p } - \vec { q } + \vec { r } )$ are $x , y$ and $z$, respectively, then the value of $2 x + y + z$ is

Let $\vec { a } = \hat { i } + \hat { j } + 2 \widehat { k } , \vec { b } = 2 \hat { i } - 3 \hat { j } + \widehat { k }$ and $\vec { c } = \hat { i } - \hat { j } + \widehat { k }$ be the three given vectors. Let $\vec { v }$ be a vector in the plane of $\vec { a }$ and $\vec { b }$ whose projection on $\vec { c }$ is $\frac { 2 } { \sqrt { 3 } }$. If $\vec { v } \cdot \hat { j } = 7$, then $\vec { v } \cdot ( \hat { i } + \hat { k } )$ is equal to
(1) 6
(2) 7
(3) 8
(4) 9

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the mid points of the diagonals $AC$ and $BD$ respectively and $( \overrightarrow { AB } - \overrightarrow { BC } ) + ( \overrightarrow { AD } - \overrightarrow { DC } ) = k \overrightarrow { FE }$, then $k$ is equal to
(1) 4
(2) $- 2$
(3) 2
(4) $- 4$

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The mid point of the arc $PQ$ is $R$. If $\overrightarrow{OP} = \vec{u}$, $\overrightarrow{OR} = \vec{v}$ and $\overrightarrow{OQ} = \alpha\vec{u} + \beta\vec{v}$, then $\alpha$, $\beta^2$ are the roots of the equation
(1) $x^2 + x - 2 = 0$
(2) $x^2 - x - 2 = 0$
(3) $3x^2 - 2x - 1 = 0$
(4) $3x^2 + 2x - 1 = 0$

Let $\vec { a } , \vec { b }$ and $\vec { c }$ be three non-zero vectors such that $\vec { b }$ and $\vec { c }$ are non-collinear if $\vec { a } + 5 \vec { b }$ is collinear with $\overrightarrow { c , b } + 6 \overrightarrow { c c }$ is collinear with $\vec { a }$ and $\vec { a } + \alpha \vec { b } + \beta \vec { c } = \overrightarrow { 0 }$, then $\alpha + \beta$ is equal to
(1) 35
(2) 30
(3) - 30
(4) - 25

Let the $\operatorname { arc } A C$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $A C$, divides the arc $A C$ such that $\frac { \text { length of } \operatorname { arc } A B } { \text { length of } \operatorname { arc } B C } = \frac { 1 } { 5 }$, and $\overrightarrow { O C } = \alpha \overrightarrow { O A } + \beta \overrightarrow { O B }$, then $\alpha + \sqrt { 2 } ( \sqrt { 3 } - 1 ) \beta$ is equal to
(1) $2 \sqrt { 3 }$
(2) $2 - \sqrt { 3 }$
(3) $5 \sqrt { 3 }$
(4) $2 + \sqrt { 3 }$

(Course 2) Q1 For $\mathbf { A } , \mathbf { B } , \mathbf { D } , \mathbf { E }$ and $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below, and for the other $\square$, enter the correct number.
Given a sphere of radius 2 with the center at point O, we have a tetrahedron ABCD whose four vertices are on the sphere. Let $\mathrm { AB } = \mathrm { BC } = \mathrm { CA } = 2$ and side BD be a diameter of the sphere.
Set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$.
(1) Let M and N denote the midpoints of segments DA and BC, respectively. Then we have
$$\overrightarrow { \mathrm { DA } } = \mathbf { A } , \quad \overrightarrow { \mathrm { MN } } = \frac { \mathbf { B } } { \mathbf { C } } + \mathbf { D }$$
(2) When the midpoint of segment MN is denoted by P and the center of gravity of triangle BCD is denoted by G, we see that
$$\overrightarrow { \mathrm { OP } } = \frac { \mathbf { E } } { \mathbf { F } } , \quad \overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { F } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$$
Also, since $\overrightarrow { \mathrm { AG } } = \frac { \mathbf { K } } { \mathbf { L } } \overrightarrow { \mathrm { AP } }$, we see that the three points A, P and G are on a straight line. (0) $\vec { a }$
(1) $\vec { b }$
(2) $\vec { c }$
(3) $\vec { a } - \vec { b }$
(4) $\vec { b } - \vec { c }$
(5) $\vec { c } - \vec { a }$ (6) $\vec { a } + \vec { b }$ (7) $\vec { b } + \vec { c }$ (8) $\vec { c } + \vec { a }$ (9) $\vec { a } + \vec { b } + \vec { c }$

Consider a trapezoid $ABCD$ where $\overline { AB }$ is parallel to $\overline { DC }$ . It is known that points $E$ and $F$ lie on diagonals $\overline { AC }$ and $\overline { BD }$ respectively, and $\overline { AB } = \frac { 2 } { 5 } \overline { DC }$ , $\overline { AE } = \frac { 3 } { 2 } \overline { EC }$ , $\overline { BF } = \frac { 2 } { 3 } \overline { FD }$ . If vector $\overrightarrow { FE }$ is expressed as $\alpha \overrightarrow { AC } + \beta \overrightarrow { AD }$ , then the real numbers $\alpha = \frac { \text{(16)} } { \text{(17)(18)} } , \beta = \frac { \text{(19)(20)} } { \text{(21)(22)} }$ . (Express as fractions in lowest terms)

Given that $P _ { 1 } , P _ { 2 } , Q _ { 1 } , Q _ { 2 } , R$ are five distinct points on a plane, where $P _ { 1 } , P _ { 2 } , R$ are not collinear, and satisfy $\overrightarrow { P _ { 1 } R } = 4 \overrightarrow { P _ { 1 } Q _ { 1 } }$ and $\overrightarrow { P _ { 2 } R } = 7 \overrightarrow { P _ { 2 } Q _ { 2 } }$, then $\overrightarrow { Q _ { 1 } Q _ { 2 } } =$ (15-1) $\overrightarrow { P _ { 1 } Q _ { 1 } } +$ (15-2)(15-3) $\overrightarrow { P _ { 2 } Q _ { 2 } }$.

On the coordinate plane, let $O$ be the origin and point $P$ have coordinates $( 2,2 )$ . Given that $\overrightarrow { O P } = \alpha \overrightarrow { O A } + \beta \overrightarrow { O B }$ , where real numbers $\alpha , \beta$ satisfy $0 \leq \alpha \leq 1,0 \leq \beta \leq 1$ . From the following options, select the possible coordinates of points $A$ and $B$.
(1) $A ( 2 , - 3 ) , B ( - 4,3 )$
(2) $A ( 3,2 ) , B ( 3,4 )$
(3) $A ( 3,4 ) , B ( 4 , - 1 )$
(4) $A ( 1,2 ) , B ( 2,1 )$
(5) $A ( 1 , - 1 ) , B ( 1,1 )$