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(a+kc)‖(2b-a)，求

(3)设d=（x，y)满足（d-c)‖(a+b),且d-c的膜=1求d
‖是平行的意思

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)

Answer 2

1)由题得-m+4n=3
2m+n=2
解得m=5/9 n=8/9
2)2b-a=(-5,2)
a+kc=(3+4k,2+k)
由(a+kc)‖(2b-a)得（3+4k)/(-5)=(2+k)/2
得k=-16/13
3)a+b=(2,4) d-c=(x-4,y-1)
由|a+b|=2√5则a+b方向上的单位向量为(√5/5,2√5/5）
(d-c)‖(a+b)得x-4=√5/5 y-1=2√5/5
得x=4+√5/5 y=1+2√5/5