LFM Pure and Mechanics > Vectors Introduction & 2D

6. If non-zero vectors $\mathbf { a } , \mathbf { b }$ satisfy $| \mathbf { a } | = \frac { 2 \sqrt { 2 } } { 3 } | \mathbf { b } |$ and $(\mathbf{a}-\mathbf{b})\perp(3\mathbf{a}+2\mathbf{b})$, then the angle between $\mathbf { a }$ and $\mathbf { b }$ is
A. $\frac { \pi } { 4 }$
B. $\frac { \pi } { 2 }$
C. $\frac { 3 \pi } { 4 }$
D. $\pi$

gaokao 2015 Q6 Expressing a Vector as a Linear Combination

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 $\_\_\_\_$ .

gaokao 2015 Q7 5 marks Angle or Cosine Between Vectors

Given non-zero vectors $\vec { a } , \vec { b }$ satisfying $| \vec { b } | = 4 | \vec { a } |$ and $\vec { a } \perp ( 2 \vec { a } + \vec { b } )$, the angle between $\vec { a }$ and $\vec { b }$ is
(A) $\frac { \pi } { 3 }$
(B) $\frac { \pi } { 2 }$
(C) $\frac { 2 \pi } { 3 }$
(D) $\frac { 5 \pi } { 6 }$

gaokao 2015 Q7 Inequality or Proof Involving Vectors

7. For any vectors $\vec { a } , \vec { b }$, which of the following relations always holds?
A. $| \vec { a } \bullet \vec { b } | \leq | \vec { a } | | \vec { b } |$
B. $| \vec { a } - \vec { b } | \leq | | \vec { a } | - | \vec { b } | |$
C. $( \vec { a } + \vec { b } ) ^ { 2 } = | \vec { a } + \vec { b } | ^ { 2 }$
D. $( \vec { a } + \vec { b } ) ( \vec { a } - \vec { b } ) = \vec { a } ^ { 2 } - \vec { b } ^ { 2 }$

gaokao 2015 Q7 Dot Product Computation

7. Let quadrilateral $A B C D$ be a parallelogram, $| \overrightarrow { A B } | = 6$, $| \overrightarrow { A D } | = 4$. If points $\mathrm { M }$ and $\mathrm { N }$ satisfy $\overrightarrow { B M } = 3 \overrightarrow { M C }$, $\overrightarrow { D N } = 2 \overrightarrow { N C }$, then $\overrightarrow { A M } \cdot \overrightarrow { N M } =$
(A) $20$
(B) $15$
(C) $9$
(D) $6$

gaokao 2015 Q9 Optimization of a Vector Expression

9. Given $\overrightarrow { A B } \perp \overrightarrow { A C }$, $| \overrightarrow { A B } | = \frac { 1 } { t }$, $| \overrightarrow { A C } | = t$, if point $P$ is a point in the plane of $\triangle A B C$, and $\overrightarrow { A P } = \frac { \overrightarrow { A B } } { | \overrightarrow { A B } | } + \frac { \overrightarrow { A C } } { | \overrightarrow { A C } | }$, then the maximum value of $\overrightarrow { P B } \cdot \overrightarrow { P C }$ equals
A. 13
B. 15
C. 19
D. 21

gaokao 2015 Q9 Optimization of a Vector Expression

9. Points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ move on the circle $\chi ^ { 2 } + y ^ { 2 } = 1$, and $\mathrm { AB } \perp \mathrm { BC }$. If point P has coordinates $( 2,0 )$, then the maximum value of $| \overrightarrow { P A } + \overrightarrow { P B } + \overrightarrow { P C } |$ is
A. 6
B. 7
C. 8
D. 9

gaokao 2015 Q11 Dot Product Computation

11. Given that vectors $\overrightarrow { O A } \perp \overrightarrow { O B } , | \overrightarrow { O A } | = 3$, then $\overrightarrow { O A } \bullet \overrightarrow { O B } =$ $\_\_\_\_$.

gaokao 2015 Q11 Dot Product Computation

11. Given vectors $\overrightarrow{OA} \perp \overrightarrow{AB}$ and $|\overrightarrow{OA}| = 3$, then $\overrightarrow{OA} \cdot \overrightarrow{OB} = $ $\_\_\_\_$ .

gaokao 2015 Q13 Expressing a Vector as a Linear Combination

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 $x =$ $\_\_\_\_$ $\_\_\_\_$ ; $y =$ $\_\_\_\_$.

gaokao 2015 Q13 Perpendicularity or Parallel Condition

Let vectors $\mathrm { a }$ and $\mathrm { b }$ be non-parallel. If vector $\lambda a + b$ is parallel to $\boldsymbol { a } + 2 b$, then the real number $\lambda = $ $\_\_\_\_$.

gaokao 2015 Q13 Dot Product Computation

13. In isosceles trapezoid $ABCD$, $AB \parallel DC$, $AB = 2$, $BC = 1$, $\angle ABC = 60 ^ { \circ }$. Points $E$ and $F$ are on segments $BC$ and $CD$ respectively, with $\overrightarrow { BE } = \frac { 2 } { 3 } \overrightarrow { BC }$, $\overrightarrow { DF } = \frac { 1 } { 6 } \overrightarrow { DC }$. Then the value of $\overrightarrow { AE } \cdot \overrightarrow { AF }$ is $\_\_\_\_$.

gaokao 2015 Q13 Magnitude of Vector Expression

13. Given that $\vec { e } _ { 1 } , \vec { e } _ { 2 }$ are unit vectors in the plane, and $\vec { e } _ { 1 } \cdot \vec { e } _ { 2 } = \frac { 1 } { 2 }$ . If the plane vector $\vec { b }$ satisfies $\vec { b } \cdot \vec { e } _ { 1 } = \vec { b } \cdot \vec { e } _ { 2 } = 1$ , then $| \vec { b } | =$ $\_\_\_\_$.

gaokao 2015 Q14 Dot Product Computation

14. Let vectors $a _ { k } = \left( \cos \frac { k \pi } { 6 } , \sin \frac { k \pi } { 6 } + \cos \frac { k \pi } { 6 } \right) ( k = 0,1,2 , \cdots , 12 )$, then the value of $\sum _ { k = 0 } ^ { 12 } \left( a _ { k } \cdot a _ { k + 1 } \right)$ is $\_\_\_\_$.

gaokao 2015 Q14 5 marks Optimization of a Vector Expression

In isosceles trapezoid ABCD, $\mathrm{AB} \parallel \mathrm{DC}$, $\mathrm{AB} = 2$, $\mathrm{BC} = 1$, $\angle \mathrm{ABC} = 60°$. Moving points E and F are on segments BC and DC respectively, with $\overrightarrow{\mathrm{BE}} = \lambda\overrightarrow{\mathrm{BC}}$, $\overrightarrow{\mathrm{DF}} = \frac{1}{9\lambda}\overrightarrow{\mathrm{DC}}$. Then the minimum value of $\overrightarrow{\mathrm{AE}} \cdot \overrightarrow{\mathrm{AF}}$ is .

gaokao 2015 Q15 Expressing a Vector as a Linear Combination

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.

gaokao 2017 Q4 Vector Properties and Identities (Conceptual)

Let non-zero vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}| = |\boldsymbol{a}-\boldsymbol{b}|$, then
A. $\boldsymbol{a} \perp \boldsymbol{b}$
B. $|\boldsymbol{a}| = |\boldsymbol{b}|$
C. $\boldsymbol{a} \parallel \boldsymbol{b}$
D. $|\boldsymbol{a}| > |\boldsymbol{b}|$

gaokao 2017 Q12 Optimization of a Vector Expression

12. Given that $\triangle A B C$ is an equilateral triangle with side length 2, and $P$ is a point in the plane $A B C$, the minimum value of $\overrightarrow { P A } \cdot ( \overrightarrow { P B } + \overrightarrow { P C } )$ is
A. $- 2$
B. $- \frac { 3 } { 2 }$
C. $- \frac { 4 } { 3 }$
D. $- 1$
II. Fill in the Blank: This section has 4 questions, each worth 5 points.

gaokao 2017 Q13 5 marks Magnitude of Vector Expression

Given vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ with an angle of $60 ^ { \circ }$ between them, $| \boldsymbol{a} | = 2$, and $| \boldsymbol{b} | = 1$, then $| \boldsymbol{a} + 2 \boldsymbol{b} | = $ \_\_\_\_

gaokao 2018 Q4 5 marks Dot Product Computation

Vectors $\boldsymbol { a } , \boldsymbol { b }$ satisfy $| \boldsymbol { a } | = 1 , \boldsymbol { a } \cdot \boldsymbol { b } = - 1$, then $\boldsymbol { a } \cdot ( 2 \boldsymbol { a } - \boldsymbol { b } ) =$
A. 4
B. 3
C. 2
D. 0

gaokao 2018 Q4 5 marks Dot Product Computation

Given vectors $\boldsymbol { a } , \boldsymbol { b }$ satisfying $| \boldsymbol { a } | = 1 , \boldsymbol { a } \cdot \boldsymbol { b } = - 1$, then $\boldsymbol { a } \cdot ( 2 \boldsymbol { a } - \boldsymbol { b } ) =$
A. 4
B. 3
C. 2
D. 0

gaokao 2018 Q6 5 marks Expressing a Vector as a Linear Combination

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. $\frac { 1 } { 4 } \overrightarrow { A B } + \frac { 3 } { 4 } \overrightarrow { A C }$

gaokao 2018 Q7 5 marks Expressing a Vector as a Linear Combination

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. $\frac { 1 } { 4 } \overrightarrow { A B } + \frac { 3 } { 4 } \overrightarrow { A C }$

gaokao 2018 Q13 5 marks Perpendicularity or Parallel Condition

Given vectors $a = ( 1,2 ) , b = ( 2 , - 2 ) , c = ( 1 , \lambda )$. If $c \parallel ( 2a + b )$, then $\lambda = $ $\_\_\_\_$.