Fundamental theorem of spatial vectors and coordinate representation Master the fundamental theorem of spatial vectors, orthogonal decomposition, coordinate representation, linear operations, dot product, and projection.
Practice questions 1 已知向量a → = ( 3 , 2 , 3 ) \overrightarrow{a}=(\sqrt{3},2,3) a = ( 3 , 2 , 3 ) ( ) (\quad) ( ) ( − 3 4 , − 1 2 , − 3 4 ) (-\dfrac{\sqrt{3}}{4},-\dfrac{1}{2},-\dfrac{3}{4}) ( − 4 3 , − 2 1 , − 4 3 ) 2 设{ a → , b → , c → } \{\overrightarrow {a},\overrightarrow {b},\overrightarrow {c}\} { a , b , c } x → = a → + b → \overrightarrow {x}=\overrightarrow {a}+\overrightarrow {b} x = a + b 3 与空间向量a → = ( 1 , 2 , − 3 ) \overrightarrow {a}=(1,2,-3) a = ( 1 , 2 , − 3 ) ( ) (\quad) ( ) ( 2 , − 1 , 0 ) (2,-1,0) ( 2 , − 1 , 0 ) ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 ) ( − 1 2 , − 1 , 3 2 ) (-\dfrac{1}{2},-1,\dfrac{3}{2}) ( − 2 1 , − 1 , 2 3 ) 4 直线l 1 l_{1} l 1 e 1 → = ( 0 , 0 , 1 ) \overrightarrow {e_{1}}=(0,0,1) e 1 = ( 0 , 0 , 1 ) l 2 l_{2} l 2 e 2 → = ( 0 , 3 , 1 ) \overrightarrow {e_{2}}=(0,\sqrt{3},1) e 2 = ( 0 , 3 , 1 ) l 1 l_{1} l 1 5 已知向量a → = ( x , 2 , 2 ) \overrightarrow{a}=(x,2,2) a = ( x , 2 , 2 ) b → = ( 2 , y , − 2 ) \overrightarrow{b}=(2,y,-2) b = ( 2 , y , − 2 ) c → = ( 3 , 1 , z ) \overrightarrow{c}=(3,1,z) c = ( 3 , 1 , z ) a → / / b → \overrightarrow{a}/\!/\overrightarrow{b} a / / b 6 已知平面α α α ( 1 , 2 , − 2 ) (1,2,-2) ( 1 , 2 , − 2 ) β β β ( − 2 , − 4 , k ) (-2,-4,k) ( − 2 , − 4 , k ) α / / β α\://\:β α // β k k k ( ) (\quad) ( ) 2 2 2 − 4 -4 − 4 4 4 4 7 若A ( x , 5 − x , 2 x − 1 ) A(x,5-x,2x-1) A ( x , 5 − x , 2 x − 1 ) B ( 1 , x + 2 , 2 − x ) B(1,x+2,2-x) B ( 1 , x + 2 , 2 − x ) ∣ A B → ∣ | \overrightarrow{AB}| ∣ A B ∣ x x x ( ) () ( ) 19 19 19 − 8 7 - \dfrac {8}{7} − 7 8 8 已知 a = ( − 2 , 1 , 3 ) a=(-2,1,3) a = ( − 2 , 1 , 3 ) b = ( − 1 , 2 , 1 ) b=(-1,2,1) b = ( − 1 , 2 , 1 ) a ⊥ ( a − λ b ) a⊥( a-λb) a ⊥ ( a − λb ) λ λ λ − 2 -2 − 2 − 14 3 - \dfrac{14}{3} − 3 14 14 5 \dfrac{14}{5} 5 14 9 已知向量A B → = ( 2 , 4 , x ) \overrightarrow {AB}=(2,4,x) A B = ( 2 , 4 , x ) α α α n → = ( 1 , y , 3 ) \overrightarrow {n}=(1,y,3) n = ( 1 , y , 3 ) A B → ⊥ α \overrightarrow {AB}⊥α A B ⊥ α ( ) (\quad) ( ) 10 已知空间三点A ( 0 , 2 , 3 ) A(0,2,3) A ( 0 , 2 , 3 ) B ( − 2 , 1 , 6 ) B(-2,1,6) B ( − 2 , 1 , 6 ) C ( 1 , − 1 , 5 ) C(1,-1,5) C ( 1 , − 1 , 5 ) A B AB A B A C AC A C 11 已知点A ( − 1 , 1 , 0 ) A(-1,1,0) A ( − 1 , 1 , 0 ) B ( 1 , 2 , 0 ) B(1,2,0) B ( 1 , 2 , 0 ) C ( − 2 , − 1 , 0 ) C(-2,-1,0) C ( − 2 , − 1 , 0 ) D ( 3 , 4 , 0 ) D(3,4,0) D ( 3 , 4 , 0 ) AB → \overrightarrow{\text{AB}} AB CD → \overrightarrow{\text{CD}} CD