解:
a = (1, √3)
|a|= 2
a·b = |a|*|b|*cos60 = 4
所以|b|= 4 b = (4, 0)或(-2√3, 2)
m·n = (a + kb)(3ka – 2b) = 0
所以3k^2 – 5k – 2 = 0
k = -1/3 或k = 2
所以不存在m + n 与a共线
解:
a = (1, √3)
|a|= 2
a·b = |a|*|b|*cos60 = 4
所以|b|= 4 b = (4, 0)或(-2√3, 2)
m·n = (a + kb)(3ka – 2b) = 0
所以3k^2 – 5k – 2 = 0
k = -1/3 或k = 2
所以不存在m + n 与a共线