△ABC中,∠C=60°且CA=2,CB=1,点M满足向量BM=2向量AM,则向量CM●向量CA=

∠60°,那一角?
BM=2AM
BA+AM=2AM
BA=AM
c^2 = a^2+b^2-2abcosC
= 1+4-2(1)(2)cos60°
= 3
c =√3
by sine rule
a/sinA = c/sinC
1/sinA = √3/sin60°
sinA = 1/2
A = 30°
cosA = √3/2
CM.CA
= (CA+AM).CA
=(CA+BA).CA
=|CA|^2 + |BA||CA|cosA
= b^2+bccosA
=4+2(√3)(√3/2)
=4+3
=7过双曲线x²/9-y²/16=1的右焦点且平行于经过一、三象限的渐近线的直线方香是