题目内容
在△ABC中,已知
=(2k+3,3k+1),
=(3,k)(k∈R),则
=______;若∠B=90°,则k=______.
| AB |
| AC |
| BC |
因为
=(2k+3,3k+1),
=(3,k),所以
=
-
=(3,k)-(2k+3,3k+1)=(-2k,-2k-1);
若∠B=90°,则
•
=(2k+3,3k+1)•(-2k,-2k-1)=0,即10k2+11k+1=0,解得:k=-1或k=-
.
故答案为(-2k,-2k-1);k=-1或-
.
| AB |
| AC |
| BC |
| AC |
| AB |
若∠B=90°,则
| AB |
| BC |
| 1 |
| 10 |
故答案为(-2k,-2k-1);k=-1或-
| 1 |
| 10 |
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