Operations on plane vectors Master addition, subtraction, scalar multiplication, and dot product of plane vectors, their geometric meaning, collinearity, projection, and perpendicularity.
Practice questions 1 已知∣ a → ∣ = 4 \left| \overrightarrow{a}\right|=4 a = 4 ∣ b → ∣ = 2 \left| \overrightarrow{b}\right|= \sqrt{2} b = 2 a → \overrightarrow{a} a b → \overrightarrow{b} b 2 在四边形A B C D ABCD A BC D O A → + O C → = O B → + O D → \overset\rightarrow { OA }+\overset\rightarrow { OC }=\overset\rightarrow { OB }+\overset\rightarrow { OD } O A → + OC → = OB → + O D → 3 已知∣ a → ∣ = ∣ b → ∣ = 1 | \overrightarrow{a}|=| \overrightarrow{b}|=1 ∣ a ∣ = ∣ b ∣ = 1 a → \overrightarrow{a} a b → \overrightarrow{b} b 90 ∘ 90^{\circ} 9 0 ∘ c → = 2 a → + 3 b → \overrightarrow{c}=2 \overrightarrow{a}+3 \overrightarrow{b} c = 2 a + 3 b 4 (1)二进制数学1101101 ( 2 ) 1101101_{(2)} 110110 1 ( 2 ) x x x y y y { x ⩾ 0 y ⩽ x 2 x + y − 9 ⩽ 0 \left\{\begin{array}{l}x\geqslant 0\\ y\leqslant x\\ 2x+y-9\leqslant 0\end{array}\right. ⎩ ⎨ ⎧ x ⩾ 0 y ⩽ x 2 x + y − 9 ⩽ 0 5 已知非零向量a → \overrightarrow{a} a b → \overrightarrow{b} b ∣ a → ∣ = ∣ b → ∣ = ∣ a → + b → ∣ | \overrightarrow{a}|=| \overrightarrow{b}|=| \overrightarrow{a}+ \overrightarrow{b}| ∣ a ∣ = ∣ b ∣ = ∣ a + b ∣ 6 关于平面向量有下列四个命题，其中正确的命题为( ) (\text{ }\ \text{ }) ( ) a → ⋅ b → = a → ⋅ c → \overrightarrow{a}⋅\overrightarrow{b}=\overrightarrow{a}⋅\overrightarrow{c} a ⋅ b = a ⋅ c 7 已知圆的方程( x − 2 ) 2 + y 2 = 1 {{(x-2)}^{2}}+{{y}^{2}}=1 ( x − 2 ) 2 + y 2 = 1 P ( 3 , 4 ) P\left(3,4\right) P ( 3 , 4 ) A , B A,B A , B P A → ⋅ P B → = \overrightarrow{PA}\cdot \overrightarrow{PB}= P A ⋅ PB = 8 已知a → \overrightarrow {a} a b → \overrightarrow {b} b c → \overrightarrow {c} c 2 a → = 4 b → + 3 c → 2\overrightarrow {a}=4\overrightarrow {b}+3\overrightarrow {c} 2 a = 4 b + 3 c 9 已知向量a → = ( − 2 , 1 ) \overrightarrow {a}=(-2,1) a = ( − 2 , 1 ) b → = ( m , m ) \overrightarrow {b}=(m,m) b = ( m , m ) a → ⊥ ( a → + b → ) \overrightarrow {a}⊥(\overrightarrow {a}+\overrightarrow {b}) a ⊥ ( a + b ) 10 设两个向量a → \overrightarrow{a} a b → \overrightarrow{b} b ∣ a → ∣ = 2 |\overrightarrow{a} |=2 ∣ a ∣ = 2 ∣ b → ∣ = 1 |\overrightarrow{b} |=1 ∣ b ∣ = 1 ( a → + 2 b → ) ⋅ ( a → − b → ) = 1 (\overrightarrow{a} +2\overrightarrow{b} )·(\overrightarrow{a} -\overrightarrow{b} )=1 ( a + 2 b ) ⋅ ( a − b ) = 1 11 设G G G △ A B C \triangle ABC △ A BC 7 sin A ⋅ G A → + 3 sin B ⋅ G B → + 3 7 sin C ⋅ G C → = 0 → \sqrt{7}\sin A\cdot\overrightarrow {GA}+3\sin B\cdot\overrightarrow {GB}+3\sqrt{7}\sin C\cdot\overrightarrow {GC}=\overrightarrow {0} 7 sin A ⋅ G A + 3 sin B ⋅ GB + 3 7 sin C ⋅ GC = 0 12 已知F 1 F_{1} F 1 F 2 F_{2} F 2 C : x 2 a 2 − y 2 b 2 = 1 ( a > 0 , b > 0 ) C:\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a>0,b>0) C : a 2 x 2 − b 2 y 2 = 1 ( a > 0 , b > 0 ) P P P 13 在△ A B C \triangle ABC △ A BC a a a b b b c c c A A A B B B C C C A = π 3 A=\dfrac{π}{3} A = 3 π A B → ⋅ A C → = 6 \overrightarrow {AB}\boldsymbol{⋅}\overrightarrow {AC}=6 A B ⋅ A C = 6 14 已知向量a → = ( 1 , 0 ) \overrightarrow {a}=(1,0) a = ( 1 , 0 ) b → = ( − 1 , 3 ) . \overrightarrow {b}=(-1,\sqrt{3}). b = ( − 1 , 3 ) . < c → < \overrightarrow {c} < c a → > = < c → \overrightarrow {a}>=< \overrightarrow {c} a >=< c 15 过抛物线x 2 = 6 y x^{2}=6y x 2 = 6 y F F F A A A B B B M M M A B AB A B ( ) (\quad) ( ) A B AB A B y = − 3 2 y=-\dfrac{3}{2} y = − 2 3 16 已知a → = ( 2 λ sin x , sin x + cos x ) \overrightarrow{a}=(2λ\sin x,\sin x+\cos x) a = ( 2 λ sin x , sin x + cos x ) b → = ( 3 cos x , λ ( sin x − cos x ) ) ( λ > 0 ) \overrightarrow{b}=( \sqrt {3}\cos x,λ(\sin x-\cos x))(λ > 0) b = ( 3 cos x , λ ( sin x − cos x )) ( λ > 0 ) 17 已知两个单位向量a → \overrightarrow {a} a b → \overrightarrow {b} b 60 ° 60° 60° 2 a → − b → + c → = 0 2\overrightarrow {a}-\overrightarrow {b}+\overrightarrow {c}=0 2 a − b + c = 0 18 已知椭圆C : x 2 a 2 + y 2 b 2 = 1 ( a > b > 0 ) \text{C}:\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a>b>0) C : a 2 x 2 + b 2 y 2 = 1 ( a > b > 0 ) F ( 1 , 0 ) F(1,0) F ( 1 , 0 ) O O O C C C 19 在△ A B C \vartriangle ABC △ A BC D D D E E E F F F B C BC BC A C AC A C A B AB A B ( ) (\quad) ( ) A B → + A C → − A D → = 0 → \overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AD}=\overrightarrow{0} A B + A C − A D = 0 20 设向量a → = ( cos ( 2 x − π 3 ) , 2 ) , b → = ( 1 , cos 2 x ) \overrightarrow{a}=\left( \cos \left( 2x-\dfrac{\pi }{3} \right),2 \right),\overrightarrow{b}=\left( 1,{{\cos }^{2}}x \right) a = ( cos ( 2 x − 3 π ) , 2 ) , b = ( 1 , cos 2 x ) 21 已知向量a → \overrightarrow {a} a b → \overrightarrow {b} b ∣ a → ∣ = 1 |\overrightarrow {a}|=1 ∣ a ∣ = 1 ∣ b → ∣ = 2 |\overrightarrow {b}|=2 ∣ b ∣ = 2 ( a → + b → ) ⋅ a → = 0 (\overrightarrow {a}+\overrightarrow {b})\cdot\overrightarrow {a}=0 ( a + b ) ⋅ a = 0 22 已知向量a → \overrightarrow {a} a b → \overrightarrow {b} b π 4 \dfrac{π}{4} 4 π ∣ b → ∣ = 2 |\overrightarrow {b}|=\sqrt{2} ∣ b ∣ = 2 a → ⋅ ( a → + b → ) = 2 \overrightarrow {a}\boldsymbol{⋅}(\overrightarrow {a}+\overrightarrow {b})=2 a ⋅ ( a + b ) = 2 23 已知椭圆C 1 C_{1} C 1 x 2 a 2 + y 2 b 2 = 1 ( a > b > 0 ) \dfrac{x^{2}}{a^{2}}{+}\dfrac{y^{2}}{b^{2}}{=}1(a{>}b{>}0) a 2 x 2 + b 2 y 2 = 1 ( a > b > 0 ) C 2 C_{2} C 2 x 2 − y 2 3 = 1 x^{2}{-}\dfrac{y^{2}}{3}{=}1 x 2 − 3 y 2 = 1 24 若2 ( y − 1 3 a ) − 1 2 ( c + b − 3 y ) + b = 0 2\left(y-\dfrac{1}{3}a\right)-\dfrac{1}{2}(c+b-3y)+b=0 2 ( y − 3 1 a ) − 2 1 ( c + b − 3 y ) + b = 0 a a a b b b c c c y = y= y = 25 已知向量a → \overrightarrow{a} a b → \overrightarrow{b} b ∣ a → ∣ = 2 | \overrightarrow{a}|=2 ∣ a ∣ = 2 ∣ b → ∣ = 1 | \overrightarrow{b}|=1 ∣ b ∣ = 1 a → \overrightarrow{a} a 26 已知向量B A → = ( 1, − 3 ) \overrightarrow{{BA}}{=}(1{,}{-}3) B A = ( 1 , − 3 ) B C → = ( 4, − 2 ) \overrightarrow{{BC}}{=}(4{,}{-}2) BC = ( 4 , − 2 ) △ A B C {\triangle }{ABC} △ A BC ( ) ({ }) ( ) 27 已知△ A B C \triangle ABC △ A BC 6 \sqrt{6} 6 ∠ B = 45 ∘ ∠B=45^{\circ} ∠ B = 4 5 ∘ D D D B C BC BC A D = 1 AD=1 A D = 1 A D → ⋅ A C → = 3 \overrightarrow{AD}·\overrightarrow{AC}=\sqrt{3} A D ⋅ A C = 3 28 设a → = ( 2 cos x + 2 3 sin x , 1 ) \overrightarrow{a}=(2\cos x+2 \sqrt {3}\sin x,1) a = ( 2 cos x + 2 3 sin x , 1 ) b → = ( cos x , − y ) \overrightarrow{b}=(\cos x,-y) b = ( cos x , − y ) a → ⋅ b → = 0 \overrightarrow{a}⋅ \overrightarrow{b}=0 a ⋅ b = 0 29 设双曲线C C C x 2 a 2 − y 2 b 2 = 1 ( a > 0 , b > 0 ) \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\left(a>0,b>0\right) a 2 x 2 − b 2 y 2 = 1 ( a > 0 , b > 0 ) F 1 {{F}_{1}} F 1 F 2 {{F}_{2}} F 2 30 已知向量a → \overrightarrow {a} a b → \overrightarrow {b} b ∣ a → ∣ = 2 |\overrightarrow {a}|=2 ∣ a ∣ = 2 a → ⋅ b → = − 2 \overrightarrow {a}\boldsymbol{⋅}\overrightarrow {b}=-2 a ⋅ b = − 2 31 (1)已知向量a → , b → \overrightarrow{a},\overrightarrow{b} a , b 45 ° 45° 45° ∣ a → ∣ = 1 , ∣ 2 a → − b → ∣ = 10 \left|\overrightarrow{a}\right|=1,\left|2\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{10} a = 1 , 2 a − b = 10 32 若四边形 A B C D ABCD A BC D A B → + C D → = 0 \overrightarrow{AB}+ \overrightarrow{CD}=0 A B + C D = 0 ( A B → + D A → ) ⊥ A C → \left( \overrightarrow{AB}+ \overrightarrow{DA}\right)⊥ \overrightarrow{AC} ( A B + D A ) ⊥ A C 33 已知∣ b → ∣ = 3 |\overrightarrow{b}|=3 ∣ b ∣ = 3 a → \overrightarrow{a} a b → \overrightarrow{b} b − 3 2 -\dfrac{3}{2} − 2 3 a → ⋅ b → \overrightarrow{a}\cdot\overrightarrow{b} a ⋅ b 34 已知M M M △ A B C \triangle ABC △ A BC A B → ⋅ A C → = 2 3 \overrightarrow{AB}\cdot \overrightarrow{AC}=2 \sqrt {3} A B ⋅ A C = 2 3 ∠ B A C = 30 ∘ ∠BAC=30^{\circ} ∠ B A C = 3 0 ∘ △ M B C \triangle MBC △ MBC 35 在锐角三角形A B C {A} {B} {C} A B C tan A = 1 2 \tan {A} =\dfrac{1}{2} tan A = 2 1 D {D} D B C {B} {C} B C Δ A B D \Delta {A} {B} {D} Δ A B D Δ A C D \Delta {A} {CD} Δ A C D 36 已知A A A B B B O O O x 2 + y 2 = 4 x^{2}+y^{2}=4 x 2 + y 2 = 4 ∣ A B → ∣ = 2 | \overrightarrow{AB}|=2 ∣ A B ∣ = 2 O C → = 5 3 O A → − 2 3 O B → \overrightarrow{OC}= \dfrac {5}{3} \overrightarrow{OA}- \dfrac {2}{3} \overrightarrow{OB} OC = 3 5 O A − 3 2 OB 37 在△ A B C \triangle ABC △ A BC A = 90 ° A=90° A = 90° A B = 1 AB=1 A B = 1 A C = 2. AC=2. A C = 2. P P P Q Q Q A P → = λ A B → \overrightarrow{AP}=λ\overrightarrow{AB} A P = λ A B A Q → = ( 1 − λ ) A C → \overrightarrow{AQ}=(1-λ)\overrightarrow{AC} A Q = ( 1 − λ ) A C 38 若F 1 F_{1} F 1 F 2 F_{2} F 2 x 2 a 2 − y 2 b 2 = 1 \dfrac {x^{2}}{a^{2}}- \dfrac {y^{2}}{b^{2}}=1 a 2 x 2 − b 2 y 2 = 1 O O O P P P M M M