Question

已知向量a,b满足：|a|=1,|b|=2,|a-b|=√7。（1）求|a-2b| (2)若（a+2b）⊥ (ka-b),求实数k的值

Answer 1

解：∣a - b∣² = √(a - b)²
( √7 )² = a² - 2a•b + b²
7 = ∣a∣² - 2a•b +∣b∣²
7 = 1² - 2a•b + 2²
2a•b = -2
a•b = -1

(1) |a-2b| = √(a - 2b)²
= √(a² - 4a•b + 4b²)
= √(∣a∣² - 4a•b + 4∣b∣²)
= √(1² - 4*(-1) + 4*2²)
= √21

(2) 若（a+2b）⊥ (ka-b),
则（a+2b）• (ka-b) = 0
ka² + (2k - 1)a•b - 2b² = 0
k∣a∣² + (2k - 1)a•b - 2∣b∣² = 0
k*1² + (2k - 1)*(-1) - 2*(2)² = 0
k - 2k+ 1 - 8 = 0
k = -7

Answer 2

xvbxvcbxvbx

Answer 3

|a|=1,|b|=2, a^2=1 b^2=4
Ia-b|=√7 a^2-2ab+b^2=7 1-2ab+4=7 ab=-1
(1) |a-2b|=√(a^2-4ab+4b^2)=√(1+4+4*4)=√21
(2)（a+2b）⊥ (ka-b),
则(a+2b)*(ka-b)=0
ka^2+2kab-ab-2b^2=0
k+2k*(-1)-(-1)-2*4=0
-k+1-8=0
k=-7

Answer 4

（1）a2=1;b2=4;a2+b2-2ab=7;推出-2ab=2;a2+4b2-4ab=21;所以
|a-2b| =√21.
(2)ka2-ab+2kab-2b2=0；即k+1-2k-8=0推出k=-7