已知非零向量a，b满足|a|=4，|b|=3，（2a-3b）=61 （1）求a与b的夹角 （2）求|a+b|与|a-b|

如题所述

是（2a-3b）*（2a+b）=61

(1)、设a,b夹角为α, 由已知有：（2a-3b）*（2a+b）= 4a^2 - 4ab -3b^2 =4|a|^2 -4ab -3|b|^2 = 62-4ab+27 = 61, => ab=-6; 又：ab = |a||b|cosα ,从而有：cosα = -1/2, => α = 2π/3.
(2)、由(a+b)^2 = a^2 + b^2 +2ab = 16 +9 -12 = 13, => |a+b| = √13;
(a-b)^2 = a^2 + b^2 - 2ab = 16 +9 +12 = 37, => |a-b| = √37.