好緊急:問一問
回答 (2)
2007-06-12 4:15 am
✔ 最佳答案
Let θ be the angle between vector a and vector b.θ = 60°
a‧b = │a│‧│b│cosθ
a‧b = (4)(5)cos60°
a‧b = 10
Since ka - b and a + 2b are perpendicular.
So, (ka - b)‧(a + 2b) = 0
k│a│² + (2k - 1)a‧b - 2│b│² = 0
k(4)² + (2k - 1)(10) - 2(5)² = 0
16k + 20k - 10 - 50 = 0
k = 5/3
參考: Myself~~~
2007-06-12 4:14 am
a•b=│a││b│cos fita=4 (5)(cos 60)=10
because 垂直
(ka-b)(a+2b)=0
k│a│^2+2ka•b-a•b-2│b│^2=0
16k+20k-10-50=0
k=60/36=5/3//
because 垂直
(ka-b)(a+2b)=0
k│a│^2+2ka•b-a•b-2│b│^2=0
16k+20k-10-50=0
k=60/36=5/3//
收錄日期: 2021-05-03 16:04:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070611000051KK04069