【答案】
分析:根據(jù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/0.png)
,可求出OB=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/1.png)
>OA,根據(jù)△OAB是直角三角形,分類討論,當(dāng)∠AOB=90°時或當(dāng)∠OAB=90°時,利用向量垂直的充要條件
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/2.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/4.png)
?x
1x
2+y
1y
2=0,即可求得結(jié)果.
解答:解:∵OB=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/5.png)
>OA
∴1°當(dāng)∠AOB=90°時,有2t+4=0,
解得t=-2,
2°當(dāng)∠OAB=90°時,有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/6.png)
=(t-2,-3)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/7.png)
=t(t-2)-3=0,
解得t=-1或3,
綜上t=-1,或t=-2或t=3;
又已知滿足
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/8.png)
,
即t
2+1≤16,(t∈Z)t共有7種情況,滿足三角形為直角的有3個,
△OAB不是直角三角形的概率是1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/10.png)
故答案為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/11.png)
.
點(diǎn)評:本題考查利用向量的數(shù)量積判斷兩向量的垂直關(guān)系,注意向量垂直的充要條件
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181018263037975/SYS201310241810182630379015_DA/14.png)
?x
1x
2+y
1y
2=0,和三角形是直角三角形要分類討論,體現(xiàn)了分類討論的思想,同時考查了運(yùn)算能力,屬中檔題.