3. ∫ ABC-a1b1is a triangular prism and ∠ ABC = 90 ∴ b1b ⊥ bcb1b ⊥.
Then establish a coordinate system with BA as the X axis, BC as the Y axis and B 1B as the Z axis.
Vector → ba 1 (√ 3,0,√ 6) vector → am (-√ 3,0/,√6/2).
→B A 1*→AM=-3+3=0
Then →B A 1⊥→AM
4. (1) is the AC midpoint O A 1C 1 midpoint N A 1B 1 midpoint M.
∫ ABC-A1B1C1is the positive Mitsubishi column, and O is the communication midpoint ∴OB⊥AC.
Take the midpoint o of AB as the origin →OB as the X axis →CA as the Y axis →ON as the Z axis.
Establish coordinate system
A( 1/2a,0,0) B(0,√3 /2a,0) A 1( 1/2a,0,√2a) C 1(- 1/2a,0,√2a)
(2)∵M is the midpoint of A 1B 1, and ABC-a1b1is a positive Mitsubishi column.
∴C 1M⊥A 1ABB 1
Then C 1M is the normal vector of a1b1c1d1.
(PS is followed by column calculation, the answer is 30. )
5、( 1)→AB(-2,- 1,3) →AC( 1,-3,2)AB =√ 14AC =√ 14 cos∠BAC = 1/2
s = AB * AC * sin∠BAC =√ 14 *√ 14 *√3/2 = 7√3
(2) Let →a be (x, y, z)
→a*→AB=0
→a*→AC=0
│→a│=√3
X= 1 Y= 1 Z= 1
X=- 1 Y=- 1 Z=- 1
a( 1, 1, 1)
a(- 1,- 1,- 1)
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