Fundamental theorem of vectors and coordinate representation Master the fundamental theorem of plane vectors, orthogonal decomposition, and coordinate representation of vector operations, angles, collinearity, and perpendicularity.
Practice questions 1 过双曲线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(a > 0 , b > 0) a 2 x 2 − b 2 y 2 = 1 ( a > 0 , b > 0 ) F F F 1 2 \dfrac {1}{2} 2 1 l l l 2 已知两定点M ( 0 , 1 ) M(0,1) M ( 0 , 1 ) N ( 1 , 2 ) N(1,2) N ( 1 , 2 ) P P P M M M P P P N N N 2 \sqrt {2} 2 y = k x − 1 y=kx-1 y = k x − 1 P P P A A A B B B 3 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 4 已知0 < α < π 2 < β < π 0 < α < \dfrac{π}{2} < β < π 0 < α < 2 π < β < π a → = ( cos α , 3 ) b → = ( − 4 , sin α ) \overset{→}{a}=(\cos α,3) \overset{→}{b}=(-4,\sin α) a → = ( cos α , 3 ) b → = ( − 4 , sin α ) a → ⊥ b → \overset{→}{a}⊥ \overset{→}{b} a → ⊥ b → 5 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 ) 6 已知抛物线C C C x 2 = − 2 p y x^{2}=-2py x 2 = − 2 p y ( 2 , − 1 ) . ( 1 ) (2,-1).(1) ( 2 , − 1 ) . ( 1 ) C C C O O O C C C 0 0 0 l l l C C C M M M 7 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 8 已知过点P ( 1 , 3 2 ) P(1,\dfrac{3}{2}) P ( 1 , 2 3 ) C C C ( x − 1 ) 2 + y 2 + ( x + 1 ) 2 + y 2 = 2 a . ( Ⅰ ) \sqrt{{{(x-1)}^{2}}+{{y}^{2}}}+\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=2a.(Ⅰ) ( x − 1 ) 2 + y 2 + ( x + 1 ) 2 + y 2 = 2 a . ( Ⅰ ) C C C 9 设m ∈ R m∈R m ∈ R a → = ( m x , y + 1 ) \overrightarrow{a}=(mx,y+1) a = ( m x , y + 1 ) b → = ( x , y − 1 ) \overrightarrow{b}=(x,y-1) b = ( x , y − 1 ) a → ⊥ b → \overrightarrow{a}⊥ \overrightarrow{b} a ⊥ b 10 若平面向量a → \overrightarrow{a} a b → \overrightarrow{b} b 2 a → + b → = ( 1 , 6 ) 2 \overrightarrow{a}+ \overrightarrow{b}=(1,6) 2 a + b = ( 1 , 6 ) a → + 2 b → = ( − 4 , 9 ) \overrightarrow{a}+2 \overrightarrow{b}=(-4,9) a + 2 b = ( − 4 , 9 ) 11 设椭圆 C C C 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 ) F 1 F_{1} F 1 F 2 F_{2} F 2 2 2 \dfrac{\sqrt{2}}{2} 2 2 12 已知A ( 2 , − 3 ) A(2,-3) A ( 2 , − 3 ) A B → = ( 3 , − 2 ) \overrightarrow{AB}=(3,-2) A B = ( 3 , − 2 ) A B AB A B 13 已知向量a → = ( − 2 , 1 ) \overrightarrow{a}=(-2,1) a = ( − 2 , 1 ) b → = ( 3 , − 4 ) \overrightarrow{b}=(3,-4) b = ( 3 , − 4 ) ( a → + b → ) ⋅ ( 2 a → − b → ) ( \overrightarrow{a}+ \overrightarrow{b})⋅(2 \overrightarrow{a}- \overrightarrow{b}) ( a + b ) ⋅ ( 2 a − b ) 14 已知向量a → = ( x − 1 , 2 ) \overrightarrow{a}=(x-1,2) a = ( x − 1 , 2 ) b → = ( 2 , 1 ) \overrightarrow{b}=(2,1) b = ( 2 , 1 ) a → ⊥ b → \overrightarrow{a}⊥ \overrightarrow{b} a ⊥ b ( ) () ( ) 15 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 16 已知圆C 1 C_{1} C 1 x 2 + y 2 = r 2 ( r > 0 ) x^{2}+y^{2}=r^{2}(r > 0) x 2 + y 2 = r 2 ( r > 0 ) C 2 C_{2} C 2 x 2 = 2 p y ( p > 0 ) x^{2}=2py(p > 0) x 2 = 2 p y ( p > 0 ) ( 2 , − 2 ) ( \sqrt {2},-2) ( 2 , − 2 ) C 1 C_{1} C 1 C 2 C_{2} C 2 17 已知向量a → = ( 2 , 3 ) \overrightarrow {a} =(2 , 3) a = ( 2 , 3 ) b → = ( x , 5 ) \overrightarrow {b} =(x , 5) b = ( x , 5 ) a → ⊥ ( a → − b → ) \overrightarrow {a} ⊥( \overrightarrow {a} - \overrightarrow {b} ) a ⊥ ( a − b ) 18 a → = ( 3 2 , sin α ) \overrightarrow{a}=(\dfrac{3}{2},\sin \alpha ) a = ( 2 3 , sin α ) b → = ( cos α , 1 3 ) \overrightarrow{b}=(\cos \alpha ,\dfrac{1}{3}) b = ( cos α , 3 1 ) a → / / b → \overrightarrow{a}/\!/\overrightarrow{b} a / / b 19 已知向量a → = ( 3 , 4 ) \overrightarrow{a}=(3,4) a = ( 3 , 4 ) b → = ( sin α , cos α ) \overrightarrow{b}=(\sin α,\cos α) b = ( sin α , cos α ) a → / / b → \overrightarrow{a}/\!/ \overrightarrow{b} a / / b 20 已知向量a＝(1，2)，b＝(2，0)，若向量λa＋b与向量c＝(1，－2)共线，则实数λ＝___． 21 已知向量a → = ( − 2 , 1 ) \overrightarrow {a}=(-2,1) a = ( − 2 , 1 ) b → = ( 1 , t ) \overrightarrow {b}=(1,t) b = ( 1 , t ) ( ) (\quad) ( ) a → / / b → \overrightarrow {a}/\!/\overrightarrow {b} a / / b 22 已知向量a = ( 1 , 2 ) a=\left(1,2\right) a = ( 1 , 2 ) b = ( − 2 , t ) b=\left(-2,t\right) b = ( − 2 , t ) a / / b a/\!/b a / / b t t t 23 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 24 设向量a → = ( 4 cos α , sin α ) \overrightarrow{a}=(4\cos α,\sin α) a = ( 4 cos α , sin α ) b → = ( sin β , 4 cos β ) \overrightarrow{b}=(\sin β,4\cos β) b = ( sin β , 4 cos β ) c → = ( cos β , − 4 sin β ) ( 1 ) \overrightarrow{c}=(\cos β,-4\sin β)(1) c = ( cos β , − 4 sin β ) ( 1 ) 25 设复平面上点对应的复数z = x + y i ( x ∈ R , y ∈ R ) ( i z=x+yi(x∈R,y∈R)(i z = x + y i ( x ∈ R , y ∈ R ) ( i ) ) ) ∣ z + 2 ∣ + ∣ z − 2 ∣ = 6 |z+2|+|z-2|=6 ∣ z + 2∣ + ∣ z − 2∣ = 6 Z Z Z C 1 . C_{1}. C 1 . C 2 C_{2} C 2 x 2 − y 2 n = 1 x^{2}- \dfrac {y^{2}}{n}=1 x 2 − n y 2 = 1 26 已知向量的e 1 → = ( 1 , 0 ) \overrightarrow {e_{1}}=(1,0) e 1 = ( 1 , 0 ) e 2 → = ( 0 , 1 ) \overrightarrow {e_{2}}=(0,1) e 2 = ( 0 , 1 ) ∣ 2 e 1 → + e 2 → ∣ = ( ) |2\overrightarrow {e_{1}}+\overrightarrow {e_{2}}|=(\quad) ∣2 e 1 + e 2 ∣ = ( ) 27 24 In the grid, O A → = a \overrightarrow{OA} = a O A = a O B → = b \overrightarrow{OB} = b OB = b 28 已知椭圆C C C 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 ) A A A B B B ∣ A B ∣ = 4 |AB|=4 ∣ A B ∣ = 4 2 2 . ( Ⅰ ) \dfrac{\sqrt{2}}{2}.(Ⅰ) 2 2 . ( Ⅰ ) 29 已知双曲线x 2 a 2 − y 2 b 2 = 1 ( a > 0 , b > 0 ) \dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1(a{\rm\;>\;}0,b{\rm\;>\;}0) a 2 x 2 − b 2 y 2 = 1 ( a > 0 , b > 0 ) A A A F F F B 1 ( 0 , b ) B_{1}(0\text{,}b) B 1 ( 0 , b ) 30 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 ( ) ({ }) ( ) 31 设△ 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 c = 3 c=3 c = 3 sin ( C − π 6 ) ⋅ cos C = 1 4 \sin (C- \dfrac {π}{6})⋅\cos C= \dfrac {1}{4} sin ( C − 6 π ) ⋅ cos C = 4 1 32 已知椭圆C C C 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 ) F ( − 2 , 0 ) F(-2,0) F ( − 2 , 0 ) 6 3 \dfrac { \sqrt {6}}{3} 3 6 33 已知向量a → = ( 2 , − n ) \overrightarrow{a}=(2,-n) a = ( 2 , − n ) b → = ( S n , n + 1 ) \overrightarrow{b}=(S_{n},n+1) b = ( S n , n + 1 ) n ∈ N ∗ n∈N^{*} n ∈ N ∗ S n S_{n} S n { a n } \{a_{n}\} { a n } n n n 34 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(a>0,b>0) 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 M M M C C C 35 黄金比例被公认为是最具美感的比例，其值为K = 5 − 1 2 . K=\dfrac{\sqrt{5}-1}{2}. K = 2 5 − 1 . E ： x 2 a 2 + y 2 b 2 = 1 ( a > b > 0 ) E：\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1(a>b>0) E ： a 2 x 2 + b 2 y 2 = 1 ( a > b > 0 ) e = K e=K e = K 36 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