LFM Pure and Mechanics

gaokao 2011 Q12 Dot Product Computation View

12. In equilateral triangle $ABC$, $D$ is a point on side $BC$. If $AB = 3, BD = 1$, then $\overrightarrow{AB} \cdot \overrightarrow{AD} =$ $\_\_\_\_$

gaokao 2011 Q18 Section Ratios and Intersection via Vectors View

18. Let $A_1, A_2, A_3, A_4$ be 4 distinct points in the plane. Then the number of points $M$ such that $\overrightarrow{MA_1} + \overrightarrow{MA_2} + \overrightarrow{MA_3} + \overrightarrow{MA_4} = \overrightarrow{0}$ is ( )
(A) 0
(B) 1
(C) 2
(D) 4
III. Solution Questions (Total: 74 points) This section contains 5 questions. When solving each question, necessary steps must be shown in the designated area on the answer sheet.

gaokao 2015 Q2 Perpendicularity or Parallel Condition View

2. Vector ${ } ^ { \mathbf { 1 } } = ( 2,4 )$ is collinear with vector ${ } ^ { \mathbf { 1 } } = ( x , 6 )$. Then the real number $x =$
(A) 2
(B) 3
(C) 4
(D) 6

gaokao 2015 Q4 Dot Product Computation View

4. Given $\vec { a } = ( 0 , - 1 ) , \vec { b } = ( - 1,2 )$, then $( 2 \vec { a } + \vec { b } ) \cdot \vec { a } =$
A. $- 1$
B. $0$
C. $1$
D. $2$

gaokao 2015 Q6 5 marks Perpendicularity or Parallel Condition View

Let $a, b$ be non-zero vectors. ``$a \cdot b = | a | | b |$'' is ``$a \parallel b$'' a

gaokao 2015 Q6 Angle or Cosine Between Vectors View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 View

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 Q19 Perpendicularity or Parallel Condition View

19. As shown in the figure, in the triangular pyramid P–ABC, $\mathrm { PA } \perp$ plane $\mathrm { ABC } , PA = 1 , AB = 1 , AC = 2 , \angle B A C = 60 ^ { \circ }$.
(1) Find the volume of the triangular pyramid P–ABC;
(2) Prove: There exists a point M on the line segment PC such that $\mathrm { AC } \perp \mathrm { BM }$, and find the value of $\frac { P M } { M C }$. [Figure]

gaokao 2017 Q4 Vector Properties and Identities (Conceptual) View

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 View

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 View

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 View

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

Load questions...

LFM Pure and Mechanics

Indices and Surds 96

Exponential Functions 197

Laws of Logarithms 242

Exponential Equations & Modelling 28

Arithmetic Sequences and Series 363

Geometric Sequences and Series 159

Differentiation from First Principles 45

Tangents, normals and gradients 140

Stationary points and optimisation 505

Applied differentiation 76

Indefinite & Definite Integrals 502

Areas Between Curves 71

Areas by integration 221

Volumes of Revolution 68

Numerical integration 49

Vectors Introduction & 2D 211

Vectors 3D & Lines 524

Travel graphs 14

SUVAT in 2D & Gravity 31

Constant acceleration (SUVAT) 78

Variable acceleration (1D) 55

Variable acceleration (vectors) 32

Forces, equilibrium and resultants 24

Newton's laws and connected particles 21

Pulley systems 6

Motion on a slope 6

Friction 8

Momentum and Collisions 34

Moments 28

Parametric curves and Cartesian conversion 9

Parametric differentiation 13

Parametric integration 6

Projectiles 49