题目内容
已知向量
=(3,-4),
=(6,-3),
=(5-m,-3-m)若∠ABC为锐角,则实数m的取值范围是______.
| OA |
| OB |
| OC |
∵
=(3,1)
=(2-m,1-m),若
∥
,则有3(1-m)=2-m,解得 m=
.
由题设知,
=(-3,-1),
=(-1-m,-m),
∵∠ABC为锐角,∴
•
=3+3m+m>0,可得m>-
.
由题意知,当m=
时,
∥
.
故当∠ABC为锐角时,实数m的取值范围是 (-
,
)∪(
,+∞),
故答案为 (-
,
)∪(
,+∞).
| AB |
| AC |
| AB |
| AC |
| 1 |
| 2 |
由题设知,
| BA |
| BC |
∵∠ABC为锐角,∴
| BA |
| BC |
| 3 |
| 4 |
由题意知,当m=
| 1 |
| 2 |
| BA |
| BC |
故当∠ABC为锐角时,实数m的取值范围是 (-
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
故答案为 (-
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
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| OA |
| OB |
| 1 |
| 2 |
| AB |
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