已知
求(1)
(2)当k为何实数时,k
|
(1)
+3a
=(2,0)+3(1,2)=( 5,2),b
则|
+3a
|=b
=25+4
,29
(2)k
-a
=k(2,0)-(1,2)=(2k-1,-2).b
设k
-a
=λ(b
+3a
),即(2k-1,-2)=λ(5,2),b
∴
,解可得2k-1=5λ -2=2λ
,k=-2 λ=-1
即k=-2时,有(-2
-a
)=-(b
+3a
),b
故k=-2时,它们反向平行.
已知
求(1)
(2)当k为何实数时,k
|
(1)
+3a
=(2,0)+3(1,2)=( 5,2),b
则|
+3a
|=b
=25+4
,29
(2)k
-a
=k(2,0)-(1,2)=(2k-1,-2).b
设k
-a
=λ(b
+3a
),即(2k-1,-2)=λ(5,2),b
∴
,解可得2k-1=5λ -2=2λ
,k=-2 λ=-1
即k=-2时,有(-2
-a
)=-(b
+3a
),b
故k=-2时,它们反向平行.