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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
人气:390 ℃ 时间:2019-08-22 10:01:27
解答
∣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
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