设三角形ABC的内角A,B,C所对的边分别a,b,c,若(3b-c)cosA=acosC,S三角形
设三角形ABC的内角A,B,C所对的边分别a,b,c,若(3b-c)cosA=acosC,S三角形ABC=√2,则向量BA*向量AC等于多少
(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
打字不易，求采纳，谢谢
！
(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
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