已知
(1)若
(2)若
|
∵
=(1,1),a
=(x,1),b
∴
=n
+2a
=(1,1)+(2x,2)=(2x+1,3),b
=2v
-a
=(2,2)-(x,1)=(2-x,1).b
(1)∵
=3n
,v
∴(2x+1,3)=3(2-x,1),
解得x=1.
(2)∵
∥n
,v
∴2x+1=3 (2-x),∴x=1.
此时,
=(3,3),n
=(1,1),v
∵
=3n
,v
∴n与v方向相同.
已知
(1)若
(2)若
|
∵
=(1,1),a
=(x,1),b
∴
=n
+2a
=(1,1)+(2x,2)=(2x+1,3),b
=2v
-a
=(2,2)-(x,1)=(2-x,1).b
(1)∵
=3n
,v
∴(2x+1,3)=3(2-x,1),
解得x=1.
(2)∵
∥n
,v
∴2x+1=3 (2-x),∴x=1.
此时,
=(3,3),n
=(1,1),v
∵
=3n
,v
∴n与v方向相同.