题目内容
已知向量
=(4,3),
=(-1,t),
=(6,8)(t∈R);
(1)若t=2,点M是线段BC上一点,且满足
=2
,求线段AM的长度;
(2)若
•
=
•
,求t的值.
| OA |
| OB |
| OC |
(1)若t=2,点M是线段BC上一点,且满足
| BM |
| MC |
(2)若
| OA |
| OB |
| OC |
| OB |
(1)由题意t=2,则
=(-1,2),
又
=
-
=
+
-
,
=2
∴
=
+
-
=
+
-
=
+
(
-
)-
=
+
-
又
=(4,3),
=(-1,2),
=(6,8)
∴
=(-
,3)
线段AM的长度是
(2)由题意
•
=
•
得-4+3t=-6+8t,解之得t=
| OB |
又
| AM |
| OM |
| OA |
| OB |
| BM |
| OA |
| BM |
| MC |
∴
| AM |
| OB |
| BM |
| OA |
| OB |
| 2 |
| 3 |
| BC |
| OA |
| OB |
| 2 |
| 3 |
| OC |
| OB |
| OA |
| 2 |
| 3 |
| OC |
| 1 |
| 3 |
| OB |
| OA |
又
| OA |
| OB |
| OC |
∴
| AM |
| 1 |
| 3 |
线段AM的长度是
| ||
| 3 |
(2)由题意
| OA |
| OB |
| OC |
| OB |
| 2 |
| 5 |
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