题目内容
已知向量
=(0,1),
=(k,k),
=(1,3),若
∥
,则实数k=______.
OA |
OB |
OC |
AB |
AC |
由题意可得
=
-
=(k,k-1),
=
-
=(1,2),∵
∥
,
∴2k-(k-1)=0,解得k=-1
故答案为:-1
AB |
OB |
OA |
AC |
OC |
OA |
AB |
AC |
∴2k-(k-1)=0,解得k=-1
故答案为:-1

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