1，B 2，( 1) 1/2a+ 1/2b+ 1/2c(2) 1/2a+B+ 1/2c(3)a+ 1/2 B+c(4) 1/5a+ 1/5b+4/5c

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)