Question

求这题第二第三问，要有过程，谢谢！

Answer 1

a=i-2j
b=-2i+kj
(2)
a+2b = -3i + (2k-2)j

|a+2b| =3
9 +(2k-2)^2 =9
4k^2-8k +4 =0
k^2-2k+1=0
k=1
x: a,b 的夹角
a+2b = -3i + (2k-2)j = -3i
|a|= √5, |b|=√5
(a+2b).(a+2b) =|a+2b|^2
|a|^2+4|b|^2 +4|a||b|cosx =9
5+20 + 20cosx =9
cosx = -4/5
x=arccos(-4/5)
(3)
a.b <0
(i-2j).(-2i+kj) <0
-2 -2k <0
k> 1

Answer 2

(2) (-3)^2+(2K-2)^2=9 得k=1
cosX=ab/|a||b|=-4/5

(3)钝角就是cosX<0
cosX=ab/|a||b|<0
就是ab<0
ab=-2-2k<0
得k>-1