Question

已知平面内三个向量a=(3,2),b=(-1,2),c=(4,1).(1） 求满足a=mb+nc的实数m.n（2） 若(

Answer 1

(1) a = mb + nc
(3,2) = m(-1,2) + n(4,1)
(3,2) = (-m + 4n,2m + n)
∴ -m + 4n = 3,2m + n = 2
联立方程得 m = 5/9,n = 8/9
(2) a + kc = (3,2) + k(4,1)
= (3 + 4k,2 + k)
2b - a = 2(-1,2) - (3,2)
= (-5,2)
∵ (a + kc) (2b - a)
∴ 2(3 + 4k) - (-5)(2 + k) = 0
解得 k = -16/13
(3) d - c = (x,y) - (4,1)
= (x - 4,y - 1)
a + b = (3,2) + (-1,2)
= (2,4)
∵ (d - c) (a + b)
∴ 4(x - 4) - 2(y - 1) = 0
2x - y - 7 = 0 (1)
且∣d - c∣ = 1
∴√ [(x - 4)² + (y - 1)² ] = 1
两边平方,(x - 4)² + (y - 1)² = 1 (2)
联立(1),(2)式,得 x = 4±√ 5/5 y = 1±2√5/5
解得 d = (4+√ 5/5,1+2√ 5/5) 或 (4-√ 5/5,1-2√ 5/5)