Question

设向量a=(cosα,sinα)，b=(cosβ,sinβ)

（1）若a-b=(-2/3,1/3),求cos<a,b>
（2）若cos<a,b>=60°，那么t为何值│a-tb│的值最小？

Answer 1

(1) cos<a, b> = a•b /∣a∣∣b∣
= (cosα, sinα)•(cosβ, sinβ) / [√(cos²α+sin²α) * √(cos²β+sin²β)]
= (cosαcosβ + sinαsinβ) / √1 * √1
= cosαcosβ + sinαsinβ

∵a - b = (-2/3, 1/3)
∴(cosα, sinα) - (cosβ, sinβ) = (-2/3, 1/3)
(cosα-cosβ, sinα-sinβ) = (-2/3, 1/3)

比较系数, 得cosα - cosβ = -2/3 ~ (1)
sinα - sinβ = 1/3 ~ (2)

(1)² + (2)²: (cos²α - 2cosαcosβ + cos²β) + (sin²α - 2sinαsinβ + sin²β) = 4/9 + 1/9
化简: 2 - 2(cosαcosβ + sinαsinβ) = 5/9
得: cosαcosβ + sinαsinβ = 13/18

∴所求: cos<a, b> = 13/18

(2) ∣a - tb∣= √(a - tb)²
= √(a² - 2ta•b + b²)
= √(∣a∣² - 2t∣a∣∣b∣cos<a, b> + ∣b∣²)
= √[(cos²α+sin²α) - 2t√(cos²α+sin²α)*√(cos²β+sin²β)cos60° + (cos²β+sin²β)]
= √(1 - 2t * 1 * 1 * 1/2 + 1)
= √(2 - t)

∵∣a - tb∣≥ 0

∴ √(2 - t)≥ 0
得t = 2时, ∣a - tb∣取得最小值.