高考复习方案2016届高考数学（理科浙江专用）二轮专题复习专题限时集训：（十二）空间向量与立体几何a word版含答案.doc

1、 1 l 1 s 1 (1 0 2) l 2 s 2 ( 1 2 2) l 1 l 2 ( ) A. 5 3B 5 3C. 2 3D 2 32 ( ) a (1 2 1) b ( 1 0 1) a (8 4 0) b ( 1 2 1) a (1 0 1) b 3 8 3 3 a ( 6 2 1) b (1 2 2) A 1 B 2 C 3 D 4 3 A(1 0 0) B(0 1 0) C(0 0 1) ABC ( ) A (1 1 1) B 2 2 2 2 2 2C 3 3 3 3 3 3D 3 3 3 3 3 34 A 1 B 1 C 1 ABC AB AC AB AC 1 2 AA 1

2、 D BC ADC 1 ABB 1 A 1 _ 5 Z12 1 M N ABCD DE AB E AE EB DE 2. ADE DE A DE B 90 P Q AE EB MQ PN PQ _ Z12 1 6 ABCD A 1 B 1 C 1 D 1 E C 1 D 1 DE AC ( ) A. 1 20B. 10 10C 10 10D 1 207 O A B C OP xOA yOB zOC (x y z R) x 2 y 3 z 2 P A B C ( ) A B C D 8 ABC A 1 B 1 C 1 AB 1 C 1 B ( ) A. 1 2B. 1 8C. 1 4D. 3

3、49 Z12 2 a ABCD A 1 B 1 C 1 D 1 M DC 1 M AD 1 ( ) Z12 2 A. 1 2 a B. 1 3 a C. 2 2 a D. 3 3 a 10 Z12 3 V ABCD ABCD VAD VAD ABCD A VD B ( ) Z12 3 A. 7 7B. 3 3C. 21 7D. 3 711 ABC A 1 B 1 C 1 BAC 1 2 AB AC AA 1 1 G E A 1 B 1 CC 1 D F AC AB ( ) GD EF DF _ 12 Z12 4 AB BC 4 AA 1 1 E ABCD AF AD 0 1 2 A 1 E F

4、 _ Z12 4 13 Z12 5 P ABCD ABCD BCD 120 AB PC 2 AP BP 2. (1) AB PC (2) B PC D Z12 5 14 Z12 6 PA ABCD CD AD BA AD CD AD AP 4 AB 1. (1) CD ADP (2)M CP BM AC C AB M Z12 6 15 ABCD AB CD AD AB BC 1 2 CD 2 E DC AE AED AE AED 1 D 1 AE D Z12 7 (1) BD 1 AE (2) D 1 AB C 5 5 CD 1 Z12 7 ( )A 1 A 解析 cos s 1 s 2

5、s 1 s 2 | | s 1 | | s 2 5 3 5 5 3 l 1 l 2 5 3 . 2 C 解析 0. C. 3 C 解析 ABC (1 1 1) 3 3 3 3 3 3 . 4. 5 3解析 AB 2 D(1 1 0) C 1 (0 2 4) AC 1 (0 2 4) AD (1 1 0) ADC 1 n 1 (x y z) n 1 AC 1 0 n 1 AD 0 2y 4z 0 x y 0 z 1 y 2 x 2 ADC 1 n 1 (2 2 1) ABB 1 A 1 n 2 (0 1 0) cos n 1 n 2 n 1 n 2 | | n 1 | | n 2 2 3 ADC

6、 1 ABB 1 A 1 5 3 . 5. 5 5 13 解析 ABCD DE AB DEBC A DE B 90 AE DEBC. E EB ED EA x y z EA EB DE 2 A(0 0 2) D(0 2 0) B(2 0 0) C(2 2 0) M(0 1 1) N(2 1 0) P(0 0 x) Q(y 0 0) MQ (y 1 1) PN (2 1 x) MQ PN MQ PN MQ PN 0 x 2y 1 |PQ | x 2 y 2 5y 2 5 2 1 5 y 0 2 |PQ | 5 5 13 . 6 B 解析 1 D D(0 0 0) E 0 1 2 1 A(1 0

7、 0) C(0 1 0) DE 0 1 2 1 AC ( 1 1 0) cos DE AC DE AC |DE | |AC | 1 2 1 4 1 2 10 10 DE AC 10 10 . 7 B 解析 x 2 y 3 z 2 OP 2OA 3OB 2OC AP AO 2OA 3(AB AO ) 2(AC AO ) AP 3AB 2AC P A B C P A B C AP mAB nAC (m n R) OP OA m(OB OA ) n(OC OA ) OP (1 m n)OA mOB nOC x 1 m n y m z n 2 3 2. B. 8 C 解析 ABC A 1 B 1 C

8、1 AB a AC b AA 1 c | | a 2. AB 1 a c | | AB 1 2 2 BC 1 b c a | | BC 1 2 2 AB 1 BC 1 (a c) (b c a) 2. cos AB 1 BC 1 | | AB 1 BC 1 | | AB 1 | | BC 1 2 2 2 2 2 1 4 . 9 D 解析 M(0 m m)(0 m a).AD 1 ( a 0 a) AD 1 s 0 2 2 0 2 2 . MD 1 (0 m a m) M AD 1 d MD 1 2 MD 1 s 0 2 m 2 a m 2 1 2 m a 2 3 2 m 2 am 1 2 a

9、2 m a 2 3 2 a 3 d 3 2 a 3 2 a a 3 1 2 a 2 3 3 a. 10 C 解析 VAD ABCD VAD ABCD AD AB ABCD AD AB AB VAD AB AV. AB AD AV 1 BV BD 2. VD E AE BE AE VD BE VD AEB A VD B AE 3 2 BE 7 2 AE AB cos AEB AE BE 21 7 . 11. 5 5 1 解析 A AB AC AA 1 x y z G 1 2 0 1 E0 1 1 2 F(x 0 0) D(0 y 0) 0x y1. GD EF 1 2 y 1 x 1 1 2

10、1 2 x y 1 2 0 y 1 2 x 1 2 |DF | x 2 y 2 5 4 x 1 5 2 1 5 x (0 1) |DF | 5 5 1 . 12. 12 5 5 解析 A DA AB AA 1 x y z FE BC K K(4 4 4 0) A 1 (0 0 1) F( 4 0 0) FK (8 4 4 0), FA 1 (4 0 1) FK FA 1 A 1 E F S S | | FK | | FA 1 sin S 2 | | FK 2 | | FA 1 2 (FK FA 1 ) 2 ( ) 8 4 2 16 (16 2 1) 4 8 4 2 32(10 2 2 1) 0

11、 1 2 1 10 S 2 144 5 S 12 5 5 . 13 解：(1) AB O PO CO AC. AP BP PO AB. ABCD BCD 120 ACB CO AB. CO PO O AB PCO PC PCO AB PC. (2) AB PC 2 AP BP 2 PO 1 OC 3 OP 2 OC 2 PC 2 OP OC. O OC OB OP x y z O xyz B(0 1 0) C( 3 0 0) P(0 0 1) D( 3 2 0) BC ( 3 1 0) PC ( 3 0 1) DC (0 2 0) DCP n 1 (1 y z) n 1 PC n 1 DC

12、n 1 PC 3 z 0 n 1 DC 2y 0 z 3 y 0 n 1 (1 0 3) BCP n 2 (1 b c) n 2 PC n 2 BC n 2 PC 3 c 0 n 2 BC 3 b 0 c 3 b 3 n 2 (1 3 3) cos n 1 n 2 n 1 n 2 |n 1 | |n 2 | 4 2 7 2 7 7 . B PC D B PC D 2 7 7 . 14 解：(1) PA ABCD PA ADP ADP ABCD. ADP ABCD AD CD AD CD ADP. (2) AD AP AB A AB AD AP x y z A(0 0 0) B(1 0 0) C

13、(4 4 0) P(0 0 4) AB (1 0 0) AC (4 4 0) AP (0 0 4) PC (4 4 4) M(x y z) PM PC (0 1) PM (x y z 4) (x y z 4) (4 4 4) x 4 y 4 z 4 4 M(4 4 4 4) BM (4 1 4 4 4 ) BM AC BM AC 0 (4 1 4 4 4) (4 4 0) 0 1 8 M 1 2 1 2 7 2 . n 1 (x 1 y 1 z 1 ) ABM n 1 AB 0 n 1 AM 0 AB (1 0 0) AM 1 2 1 2 7 2 x 1 0 1 2 x 1 1 2 y 1

14、7 2 z 1 0 z 1 1 y 1 7 n 1 (0 7 1) ABM AP ABC n 2 (0 0 4) ABC cos n 1 n 2 n 1 n 2 |n 1 | |n 2 | 4 4 50 2 10 C AB M 2 10 . 15 解：(1) AE H BE D 1 H BH. AD 1 AE D 1 E AB AE BE D 1 H AE BH AE D 1 H HB H AE HBD 1 AE BD 1 . (2) CH A(1 0 0) B(0 3 0) D 1 (0 3cos 3sin ) AB ( 1 3 0) BD 1 (0 3 3cos 3sin ) ABD 1 n (x 1 y 1 z 1 ) AB n x 1 3y 1 0 BD 1 n 3 3cos y 1 3sin z 1 0 y 1 sin x 1 3sin z 1 1 cos n ( 3sin sin 1 cos ) ABC m (0 0 1) cos m n 1 cos 3sin 2 sin 2 1 cos 2 5 5 cos 2 cos 0 cos 0 cos 1( ) 90 . CH 4 1 2 2 1 cos 120 7 Rt D 1 HC D 1 H 3 CD 1 10.