例1设(a+3b)⊥(7a-5b),(a-4b)⊥(7a-2b) 求a与b的夹角 解 由题设知 (a+3b)(7a-5b)=0 (a-4b)(7a-2b)=0 →7|a2+16a.b-151b2=071a2-30a.b+8162=0 两式相减得46a.6=231→立-6=)6邴 代入前式有1aHb故cos(a,=a-6=6-1 -dlb121a12 1元 (a,b)=arccos 23例1 设 (a 3b) (7a 5b) , (a 4b) (7a 2b) + ⊥ − − ⊥ − 求 a与b的夹角 解 由题设知 (a + 3b)(7a − 5b) = 0 (a − 4b)(7a − 2b) = 0 7 | | 16 15 | | 0 2 2 a + a b − b = 7 | | 30 8 | | 0 2 2 a − a b + b = 两式相减得 2 46a b 23 | b | = 2 | | 2 1 a b b = 代入前式有 | a | | b | = 故 | || | cos( , ) a b a b a b = 2 1 2 | | | | = = a b 2 3 1 ( , ) arccos = = a b