Question

平面内给定三个向量a=(3,2),b=(-1,2）,c=(4,1),（1）求满足a=mb+nc的实数m,n的值

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)