好吧问答库 > 设△ABC的内角A,B,C所对的边分别为a,b,c,若（3b-c)cosA=acosC,S△ABC=根号2，则向量BA.向量AC=

设△ABC的内角A,B,C所对的边分别为a,b,c,若（3b-c)cosA=acosC,S△ABC=根号2，则向量BA.向量AC=

2025年04月30日 19:07

有2个网友回答

网友（1）：

(3b-c)cosA = acosC
(3b-c)[(b^2+c^2-a^2)/(2bc)] =(a^2+b^2-c^2)/(2b)
(3b-c)(b^2+c^2-a^2)=c(a^2+b^2-c^2)
3b(b^2+c^2-a^2) = 2b^2c
3(b^2+c^2-a^2) = 2bc
a^2= b^2+c^2 - (2/3)bc
by cosine-rule
(2/3)bc = 2bc cosA
cosA = 1/3
tanA = 2√2/3

S△ABC=√2
=> (1/2) |AB||AC|sinA = √2
(1/2) (AB. AC) tanA = √2
AB.AC = 2√2/tanA
BA.AC = -2√2/tanA
= -2√2/(2√2/3)
= -3

网友（2）：

= -1