AB is a chord of a circle with centre O. M is the mid point of AB prove that OM perpendicular AB.
please give solution and explain. thank you very much :)
Amaths - Applications of vectors
2008-08-17 12:27 am
回答 (2)
2008-08-17 2:16 am
✔ 最佳答案
let (->OA) = a , (->OB) = b ,(->AB)= (->OB)-(->OA) = b - a
(->AM) = b/2 - a/2
(->OM) = (->OA) (->AM) = a b/2-a/2
= a/2 b/2
(->OM) * (->AB) = (a/2 b/2)*(-a/2 b/2) <<scalar product/dot product
= -|a|^2/4 |b|/4
by the radius of a circle , |a| = |b|
thus, (->OM) * (->AB) = 0
hence , OM is perpendicular to AB.
2008-08-17 2:13 am
OM = 0.5(OA + OB)
AB= OB - OA
OM dot AB = 0.5(OA + OB) dot (OB - OA)
= 0.5( OA dot OB - OA dot OA + OB dot OB - OB dot OA)
=0.5(OB dot OB - OA dot OA)
=0.5( (| OB |) ^2 - (| OA |) ^2
=0 , Since OA and OB have the same length, (radius)
Since OM dot AB =0
OM is perpendicular to AB
AB= OB - OA
OM dot AB = 0.5(OA + OB) dot (OB - OA)
= 0.5( OA dot OB - OA dot OA + OB dot OB - OB dot OA)
=0.5(OB dot OB - OA dot OA)
=0.5( (| OB |) ^2 - (| OA |) ^2
=0 , Since OA and OB have the same length, (radius)
Since OM dot AB =0
OM is perpendicular to AB
參考: brain
收錄日期: 2021-04-24 08:10:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080816000051KK01702