人教A版高中同步训练数学选择性必修第一册课后习题 第1章空间向量与立体几何 空间向量的数量积运算

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1.1.2 空间向量的数量积运算
课后·训练提升
基础巩固
1.若a,b均为非零向量,则a·b=|a||b|是a与b共线的( )
A.充分不必要条件
B.必要不充分条件
C.充要条件
D.既不充分又不必要条件
答案:A
解析:a·b=|a||b|⇒cos<a,b>=1⇒<a,b>=0°,即a与b共线.反之不成立,当a与b反向共线时,a·b=-|a||b|.
2.已知向量a,b满足条件:|a|=2,|b|=√2,且a与2b-a互相垂直,则<a,b>等于( )
A.30°
B.45°
C.60°
D.90°
答案:B
解析:由已知得,a·(2b-a)=0,即2a·b=|a|2=4,所以a·b=2,
所以cos<a,b>=a·b
|a||b|=
2×√2
=√2
2
,
又0°≤<a,b>≤180°,所以<a,b>=45°.
3.已知四面体ABCD 的所有棱长都等于2,E 是棱AB 的中点,F 是棱CD 靠近C 的四等分点,则EF ⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ 等于( ) A.-1
2
B.1
2
C.-52
D.52
答案:D
解析:由题意知EF ⃗⃗⃗⃗ =1
2
AB ⃗⃗⃗⃗⃗ +BC ⃗⃗⃗⃗⃗ +14
CD ⃗⃗⃗⃗⃗ ,EF ⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =12
AB ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ +BC ⃗⃗⃗⃗⃗ ·AC
⃗⃗⃗⃗⃗ +14
CD ⃗⃗⃗⃗⃗ ·AC
⃗⃗⃗⃗⃗ , 因为AB ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =|AB ⃗⃗⃗⃗⃗ |·|AC ⃗⃗⃗⃗⃗ |cos<AB ⃗⃗⃗⃗⃗ ,AC
⃗⃗⃗⃗⃗ >=2,BC ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =|BC ⃗⃗⃗⃗⃗ |·|AC ⃗⃗⃗⃗⃗ |cos<BC ⃗⃗⃗⃗⃗ ,AC ⃗⃗⃗⃗⃗ >=2,CD ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =|CD ⃗⃗⃗⃗⃗ |·|AC ⃗⃗⃗⃗⃗ |cos<CD ⃗⃗⃗⃗⃗ ,AC ⃗⃗⃗⃗⃗ >=-2,所以EF ⃗⃗⃗⃗ ·AC
⃗⃗⃗⃗⃗ =12
×2+2+14
×(-2)=52
,故选D. 4.已知A,B,C,D 是空间中不共面的四点,若(DB ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ -2DA ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ −AC
⃗⃗⃗⃗⃗ )=0,则△ABC 一定是( ) A.直角三角形 B.等腰三角形 C.等腰直角三角形 D.等边三角形
答案:B
解析:∵(DB ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ -2DA ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ −AC ⃗⃗⃗⃗⃗ )=(DB ⃗⃗⃗⃗⃗ −DA ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ −DA ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ −AC ⃗⃗⃗⃗⃗ )=(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ −AC ⃗⃗⃗⃗⃗ )=|AB ⃗⃗⃗⃗⃗ |2-|AC ⃗⃗⃗⃗⃗ |2=0,∴|AB ⃗⃗⃗⃗⃗ |=|AC ⃗⃗⃗⃗⃗ |,即AB=AC.故△ABC 为等腰三角形.
5.(多选题)在正方体ABCD-A 1B 1C 1D 1中,关于下列四个结论,正确的是( )
A.(AA 1⃗⃗⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ )2=3AB ⃗⃗⃗⃗⃗ 2
B.A 1C ⃗⃗⃗⃗⃗⃗⃗ ·(A 1B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ −A 1A ⃗⃗⃗⃗⃗⃗⃗ )=0
C.AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 的夹角为60°
D.正方体的体积为|AB ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ | 答案:AB
解析:如图所示,(AA 1⃗⃗⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ )2=(AA 1⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ +D 1C 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ )2
=AC 1⃗⃗⃗⃗⃗⃗⃗ 2
=3AB ⃗⃗⃗⃗⃗ 2,
故A 中结论正确;
A 1C ⃗⃗⃗⃗⃗⃗⃗ ·(A 1
B 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ −A 1A ⃗⃗⃗⃗⃗⃗⃗ )=A 1
C ⃗⃗⃗⃗⃗⃗⃗ ·AB 1⃗⃗⃗⃗⃗⃗⃗ =0,故B 中结论正确;
AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 的夹角是D 1C ⃗⃗⃗⃗⃗⃗⃗ 与D 1A ⃗⃗⃗⃗⃗⃗⃗ 夹角的补角,而D 1C ⃗⃗⃗⃗⃗⃗⃗ 与D 1A ⃗⃗⃗⃗⃗⃗⃗ 的夹角为60°,故AD 1⃗⃗⃗⃗⃗⃗⃗ 与A 1B ⃗⃗⃗⃗⃗⃗⃗ 的夹角为120°,故C 中结论错误;正方体的体积为|AB ⃗⃗⃗⃗⃗ ||AA 1⃗⃗⃗⃗⃗⃗⃗ |·|AD ⃗⃗⃗⃗⃗ |,故D 中结论错误.
6.已知空间向量a,b,|a|=3√2,|b|=5,m=a+b,n=a+λb(λ∈R),<a,b>=135°,若m ⊥n,则λ的值为 . 答案:-3
10
解析:由题意知a·b=|a||b|cos<a,b>=3√2×5×(-√2
2
)=-15.由m ⊥n,得
m·n=(a+b)·(a+λb)=0,即
|a|2+(λ+1)a·b+λ|b|2=18-15(λ+1)+25λ=0,解得λ=-3
10
.
7.如图,四面体ABCD 的每条棱长都等于2,点E,F 分别为棱AB,AD 的中点,则|BC ⃗⃗⃗⃗⃗ −EF ⃗⃗⃗⃗ |= ,EF ⃗⃗⃗⃗ 与AC
⃗⃗⃗⃗⃗ 所成的角为 .
答案:√3
π2
解析:因为EF ⃗⃗⃗⃗ =1
2
BD ⃗⃗⃗⃗⃗ ,BD ⃗⃗⃗⃗⃗ ·BC
⃗⃗⃗⃗⃗ =2×2×cos π3
=2, 所以|BC ⃗⃗⃗⃗⃗ −EF ⃗⃗⃗⃗ |2
=|BC ⃗⃗⃗⃗⃗ -12
BD ⃗⃗⃗⃗⃗ |2
=|BC ⃗⃗⃗⃗⃗ |2-BC ⃗⃗⃗⃗⃗ ·BD
⃗⃗⃗⃗⃗ +14|BD ⃗⃗⃗⃗⃗ |2=4-2+14
×4=3. 所以|BC ⃗⃗⃗⃗⃗ −EF ⃗⃗⃗⃗ |=√3.
因为EF ⃗⃗⃗⃗ =1
2
BD ⃗⃗⃗⃗⃗ =12
(AD ⃗⃗⃗⃗⃗ −AB
⃗⃗⃗⃗⃗ ), 所以AC ⃗⃗⃗⃗⃗ ·EF ⃗⃗⃗⃗ =1
2
AC ⃗⃗⃗⃗⃗ ·(AD ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ )=12
(AC ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ −AC ⃗⃗⃗⃗⃗ ·AB
⃗⃗⃗⃗⃗ )=0. 又<EF ⃗⃗⃗⃗ ,AC ⃗⃗⃗⃗⃗ >∈[0,π],所以<EF ⃗⃗⃗⃗ ,AC
⃗⃗⃗⃗⃗ >=π2
. 8.已知空间向量a,b 满足|a|=3,|b|=2,且(a-2b)·(a+b)=5,则a+b 在a 上的投影向量为 . 答案:5
9a
解析:∵(a-2b)·(a+b)=5, ∴|a|2-a·b -2|b|2=5,∴a·b=-4.
∴a·(a+b)=|a|2+a·b=5,|a+b|=√|a |2
+2a ·b +|b |2
=√5. ∴cos<a,a+b>=
a ·(a+
b )|a ||a+b |
=
√5
3
, ∴a+b 在a 上的投影向量为|a+b|cos<a,a+b>·
a |a |
=√5×
√53
×13
a=5
9
a.
9.在长方体ABCD-A 1B 1C 1D 1中,AB=AA 1=2,AD=4,E 为侧面ABB 1A 1的中心,F 为A 1D 1的中点.试计算: (1)BC ⃗⃗⃗⃗⃗ ·ED 1⃗⃗⃗⃗⃗⃗⃗ ; (2)BF ⃗⃗⃗⃗ ·AB 1⃗⃗⃗⃗⃗⃗⃗ ; (3)EF ⃗⃗⃗⃗ ·FC 1⃗⃗⃗⃗⃗⃗ .
解:设AB ⃗⃗⃗⃗⃗ =a,AD ⃗⃗⃗⃗⃗ =b,AA 1⃗⃗⃗⃗⃗⃗⃗ =c,
则|a|=|c|=2,|b|=4,a·b=b·c=c·a=0.
(1)∵ED 1⃗⃗⃗⃗⃗⃗⃗ =EA 1⃗⃗⃗⃗⃗⃗⃗ +A 1D 1⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ =12
BA 1⃗⃗⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ =12
AA 1
⃗⃗⃗⃗⃗⃗⃗ −1
2
AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ =12
c-1
2
a+b,BC ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ =b, ∴BC ⃗⃗⃗⃗⃗ ·ED 1⃗⃗⃗⃗⃗⃗⃗ =b·(12
c -1
2
a +b)=|b|2=16.
(2)∵BF ⃗⃗⃗⃗ =BB 1⃗⃗⃗⃗⃗⃗⃗ +B 1A 1⃗⃗⃗⃗⃗⃗⃗⃗⃗ +A 1F ⃗⃗⃗⃗⃗⃗⃗ =AA 1⃗⃗⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ +12
AD ⃗⃗⃗⃗⃗ =c-a+1
2
b,AB 1⃗⃗⃗⃗⃗⃗⃗ =AB
⃗⃗⃗⃗⃗ +AA 1⃗⃗⃗⃗⃗⃗⃗ =a+c,
∴BF ⃗⃗⃗⃗ ·AB 1⃗⃗⃗⃗⃗⃗⃗ =(c -a +1
2
b)·(a+c)=|c|2-|a|2=0.
(3)∵EF ⃗⃗⃗⃗ =EA 1⃗⃗⃗⃗⃗⃗⃗ +A 1F ⃗⃗⃗⃗⃗⃗⃗ =12
AA 1⃗⃗⃗⃗⃗⃗⃗ −12
AB ⃗⃗⃗⃗⃗ +12
AD ⃗⃗⃗⃗⃗ =12
c-12
a+1
2
b,
FC 1⃗⃗⃗⃗⃗⃗ =FD 1⃗⃗⃗⃗⃗⃗⃗ +D 1C 1
⃗⃗⃗⃗⃗⃗⃗⃗⃗ =12
AD ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ =1
2
b+a, ∴EF ⃗⃗⃗⃗ ·FC 1⃗⃗⃗⃗⃗⃗ =(12
c -12
a +12
b)·(12
b +a)=14
|b|2-12
|a|2=2.
10.如图,在四面体OACB 中,OB=OC,AB=AC,求证:OA ⊥BC.
证明:因为OB=OC,AB=AC,OA=OA, 所以△OAB ≌△OAC,所以∠AOB=∠AOC.
所以OA ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ ·(OC ⃗⃗⃗⃗⃗ −OB ⃗⃗⃗⃗⃗ )=OA ⃗⃗⃗⃗⃗ ·OC ⃗⃗⃗⃗⃗ −OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =|OA ⃗⃗⃗⃗⃗ ||OC ⃗⃗⃗⃗⃗ |cos ∠AOC-|OA ⃗⃗⃗⃗⃗ ||OB ⃗⃗⃗⃗⃗ |cos ∠AOB=0,所以OA
⃗⃗⃗⃗⃗ ⊥BC ⃗⃗⃗⃗⃗ ,即OA ⊥BC. 能力提升
1.已知两条异面直线的方向向量分别为a,b,且|a|=|b|=1,a·b=-1
2,则这
两条异面直线所成的角为( ) A.30° B.60°
C.120°
D.150°
答案:B
2.如图,在平行六面体ABCD-A 1B 1C 1D 1中,底面是边长为1的正方形,若∠A 1AB=∠A 1AD=60°,且A 1A=3,则A 1C 的长为( )
A.√5
B.2√2
C.√14
D.√17
答案:A
解析:∵A 1C ⃗⃗⃗⃗⃗⃗⃗ =A 1A ⃗⃗⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ +BC ⃗⃗⃗⃗⃗ =-AA 1⃗⃗⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ,
∴|A 1C ⃗⃗⃗⃗⃗⃗⃗ |2=(-AA 1⃗⃗⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )2=|AA 1⃗⃗⃗⃗⃗⃗⃗ |2+|AB ⃗⃗⃗⃗⃗ |2+|AD ⃗⃗⃗⃗⃗ |2-2AA 1⃗⃗⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ -2AA 1⃗⃗⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ +2AB ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ =9+1+1-2×3×1×cos60°-2×3×1×cos60°=5, ∴|A 1C ⃗⃗⃗⃗⃗⃗⃗ |=√5.
3.如图,在直三棱柱ABC-A 1B 1C 1中,∠ABC=90°,AB=BC=1,AA 1=√2,则BA 1⃗⃗⃗⃗⃗⃗⃗ 在AC
⃗⃗⃗⃗⃗ 上的投影向量为( )
A.-√22
B.-
√22AC ⃗⃗⃗⃗⃗
C.-12
D.-12
AC
⃗⃗⃗⃗⃗ 答案:D
解析:∵BA 1⃗⃗⃗⃗⃗⃗⃗ =BA ⃗⃗⃗⃗⃗ +AA 1⃗⃗⃗⃗⃗⃗⃗ =BA ⃗⃗⃗⃗⃗ +BB 1⃗⃗⃗⃗⃗⃗⃗ ,AC ⃗⃗⃗⃗⃗ =BC ⃗⃗⃗⃗⃗ −BA ⃗⃗⃗⃗⃗ , 且BA ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ =BB 1⃗⃗⃗⃗⃗⃗⃗ ·BA ⃗⃗⃗⃗⃗ =BB 1⃗⃗⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ =0, ∴BA 1⃗⃗⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =-BA
⃗⃗⃗⃗⃗ 2=-1. 又|AC ⃗⃗⃗⃗⃗ |=√2,|BA 1⃗⃗⃗⃗⃗⃗⃗ |=√1+2=√3, ∴cos<BA 1⃗⃗⃗⃗⃗⃗⃗ ,AC
⃗⃗⃗⃗⃗ >=BA 1⃗⃗⃗⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ |BA 1⃗⃗⃗⃗⃗⃗⃗⃗ ||AC ⃗⃗⃗⃗⃗ |
=-√6
6, ∴BA 1⃗⃗⃗⃗⃗⃗⃗ 在AC ⃗⃗⃗⃗⃗ 上的投影向量为|BA 1⃗⃗⃗⃗⃗⃗⃗ |cos<BA 1⃗⃗⃗⃗⃗⃗⃗ ,AC
⃗⃗⃗⃗⃗ >·AC ⃗⃗⃗⃗⃗ |AC
⃗⃗⃗⃗⃗ |=-√2

√2
AC ⃗⃗⃗⃗⃗ =-12
AC
⃗⃗⃗⃗⃗ . 4.如图,两条异面直线a,b 所成的角为60°,在直线a,b 上分别取点A',E 和点A,F,使AA'⊥a 且AA'⊥b.若A'E=2,AF=3,EF=√23,则线段AA'的长为 .
答案:4或2
解析:由题意知FE ⃗⃗⃗⃗ =FA ⃗⃗⃗⃗⃗ +AA '⃗⃗⃗⃗⃗⃗⃗⃗ +A 'E ⃗⃗⃗⃗⃗⃗⃗ , 所以FE ⃗⃗⃗⃗ 2=(FA ⃗⃗⃗⃗⃗ +AA '⃗⃗⃗⃗⃗⃗⃗⃗ +A 'E ⃗⃗⃗⃗⃗⃗⃗ )2
=FA ⃗⃗⃗⃗⃗ 2
+AA '⃗⃗⃗⃗⃗⃗⃗⃗ 2
+A 'E ⃗⃗⃗⃗⃗⃗⃗ 2+2FA ⃗⃗⃗⃗⃗ ·AA '⃗⃗⃗⃗⃗⃗⃗⃗ +2FA ⃗⃗⃗⃗⃗ ·A 'E ⃗⃗⃗⃗⃗⃗⃗ +2AA '⃗⃗⃗⃗⃗⃗⃗⃗ ·A 'E ⃗⃗⃗⃗⃗⃗⃗ , ∵异面直线a,b 所成的角为60°,A'E=2,AF=3,EF=√23, ∴23=9+AA '⃗⃗⃗⃗⃗⃗⃗⃗ 2+4+0±2×2×3cos60°+0,∴|AA '⃗⃗⃗⃗⃗⃗⃗⃗ |=4或|AA '⃗⃗⃗⃗⃗⃗⃗⃗ |=2. 5.已知正三棱柱ABC-DEF 的侧棱长为2,底面边长为1,M 是BC 的中点,若CF 上有一点N,使MN ⊥AE,则CN
CF = .
答案:1
16
解析:设CN CF
=m.∵AE ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +BE ⃗⃗⃗⃗⃗ ,MN ⃗⃗⃗⃗⃗⃗ =1
2
BC ⃗⃗⃗⃗⃗ +m AD ⃗⃗⃗⃗⃗ ,
∴AE ⃗⃗⃗⃗⃗ ·MN ⃗⃗⃗⃗⃗⃗ =(AB ⃗⃗⃗⃗⃗ +BE
⃗⃗⃗⃗⃗ )·(12
BC ⃗⃗⃗⃗⃗ +mAD ⃗⃗⃗⃗⃗ )=12
×1×1×(-12
)+4m=0. ∴m=1
16
.
6.在四面体OABC 中,棱OA,OB,OC 两两垂直,且OA=1,OB=2,OC=3,G 为△ABC 的重心,则OG ⃗⃗⃗⃗⃗ ·(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OC ⃗⃗⃗⃗⃗ )= . 答案:14
3
解析:由已知得OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ ·OC ⃗⃗⃗⃗⃗ =OB ⃗⃗⃗⃗⃗ ·OC
⃗⃗⃗⃗⃗ =0.
如图,取BC 的中点D,连接OD,AD,则AD 经过点G,且AG=2
3
AD,所以OG
⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ +AG ⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ +23
AD ⃗⃗⃗⃗⃗ =OA ⃗⃗⃗⃗⃗ +23
(OD ⃗⃗⃗⃗⃗ −OA ⃗⃗⃗⃗⃗ )=13
OA
⃗⃗⃗⃗⃗ +23
×12
(OB ⃗⃗⃗⃗⃗ +OC ⃗⃗⃗⃗⃗ )=13
OA ⃗⃗⃗⃗⃗ +13
OB ⃗⃗⃗⃗⃗ +1
3
OC
⃗⃗⃗⃗⃗ . 所以OG ⃗⃗⃗⃗⃗ ·(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OC
⃗⃗⃗⃗⃗ )=13
(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OC ⃗⃗⃗⃗⃗ )2=13
(|OA ⃗⃗⃗⃗⃗ |2+|OB ⃗⃗⃗⃗⃗ |2+|OC
⃗⃗⃗⃗⃗ |2)=13
×(1+4+9)=143
. 7.如图,在四面体ABCD 中,AB=CD,AC=BD,E,F 分别是AD,BC 的中点,求证:EF ⊥AD,且EF ⊥BC.
证明:∵F 是BC 的中点,∴AF
⃗⃗⃗⃗⃗ =12
(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ ). 又E 是AD 的中点,∴AE ⃗⃗⃗⃗⃗ =12
AD ⃗⃗⃗⃗⃗ .
∴EF ⃗⃗⃗⃗⃗ =AF ⃗⃗⃗⃗⃗ −AE ⃗⃗⃗⃗⃗ =12
(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ )-12
AD ⃗⃗⃗⃗⃗ =1
2
(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ ).
∵|AC ⃗⃗⃗⃗⃗ |=|BD ⃗⃗⃗⃗⃗⃗ |=|AD ⃗⃗⃗⃗⃗ −AB ⃗⃗⃗⃗⃗ |, ∴AC ⃗⃗⃗⃗⃗ 2=AD ⃗⃗⃗⃗⃗ 2-2AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ 2. 同理AB ⃗⃗⃗⃗⃗ 2=AD ⃗⃗⃗⃗⃗ 2-2AC ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ 2. ∴2AD ⃗⃗⃗⃗⃗ 2-2AD ⃗⃗⃗⃗⃗ ·AB ⃗⃗⃗⃗⃗ -2AC ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ =0, 即(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ )·AD ⃗⃗⃗⃗⃗ =0.
∴EF ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ =1
2
(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ )·AD ⃗⃗⃗⃗⃗ =0,
∴EF ⃗⃗⃗⃗⃗ ⊥AD ⃗⃗⃗⃗⃗ .同理EF ⃗⃗⃗⃗⃗ ⊥BC ⃗⃗⃗⃗⃗ . ∴EF ⊥AD,且EF ⊥BC.
8.如图,在平行六面体ABCD-A 1B 1C 1D 1中,底面ABCD 是边长为1的正方形,∠BAA 1=∠DAA 1=π
3,AC 1=√26.
(1)求侧棱AA 1的长;
(2)若M,N 分别为D 1C 1,C 1B 1的中点,求AC 1⃗⃗⃗⃗⃗⃗⃗ ·MN ⃗⃗⃗⃗⃗⃗⃗ 及异面直线AC 1和MN 的夹角. 解:(1)设侧棱AA 1=x,
由题意知,AB ⃗⃗⃗⃗⃗ 2=AD ⃗⃗⃗⃗⃗ 2=1,AA 1⃗⃗⃗⃗⃗⃗⃗ 2
=x 2,AB ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ =0,AB ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ =x 2
,AD ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ =
x 2
.
∵AC 1⃗⃗⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AA 1⃗⃗⃗⃗⃗⃗⃗ ,∴AC 1⃗⃗⃗⃗⃗⃗⃗ 2
=(AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AA 1⃗⃗⃗⃗⃗⃗⃗ )2
=AB ⃗⃗⃗⃗⃗ 2+AD ⃗⃗⃗⃗⃗ 2+
AA 1⃗⃗⃗⃗⃗⃗⃗ 2
+2AB ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ +2AB ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ +2AD ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ =26, 即x 2+2x-24=0. ∵x>0,∴x=4. 故侧棱AA 1=4.
(2)∵AC 1⃗⃗⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AA 1
⃗⃗⃗⃗⃗⃗⃗ ,MN ⃗⃗⃗⃗⃗⃗⃗ =12
DB ⃗⃗⃗⃗⃗⃗ =1
2
(AB ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ ),
第11页 共11页 ∴AC 1⃗⃗⃗⃗⃗⃗⃗ ·MN ⃗⃗⃗⃗⃗⃗⃗ =12
(AB ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ )·(AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +AA 1⃗⃗⃗⃗⃗⃗⃗ ) =12(AB ⃗⃗⃗⃗⃗ 2−AD ⃗⃗⃗⃗⃗ 2+AB ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗ ·AA 1⃗⃗⃗⃗⃗⃗⃗ )=12×(1-1+2-2)=0, 故异面直线AC 1和MN 的夹角为90°.。

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