Review the reference question group A: P 1 18 1. ( 1) √ (2) √ (3) × (4) × 2.( 1) D (2) B (3) D (4) C (5)。 2(a b) 4。 As shown in figure DE = BA = MA-MB =-2/3a1/3ad-2/3a2/3bc =1/3a1/3ef =-1/3a-1. 3bab = 2/3a- 1/3bce =-Ab 5。 ( 1) AB = (8，-8) │AB│=8√2 (2)OC=(2，- 16) OD=(-8，8)。 It is proved that AB=CD Because AB=( 1,-1) and CD=( 1,-1), the connection between AB and CD*** is 7. d (-2，0) 8。 n = 29。 λ.CosB=0 cosC=4/5 1 1。 It is proved that: (2n-m) * m = 2n * m-m 2 = 2cos60-1= 0, so, (2n-m) ⊥ m 10. | a-B | = 1 14 . cosθ= 5/8 cosβ= 19/Group 20 B p 1 19 1。 adbccd2。 Proof: A ⊥ B→| AB | = | A-B || AB | = √ (AB) 2 = √ (| A | 2 | B |) So | ab | = √ (| a | 2 | b | 2) = | a-b | Duplicate certificate | AB.