解法一:如图,取A为原点,AB、AS分别为y、z轴建立空间直角坐标系,
∵AC=2,BC=
![](http://thumb.1010pic.com/pic5/latex/3293.png)
,SB=
![](http://thumb.1010pic.com/pic5/latex/14414.png)
,∴B(0,
![](http://thumb.1010pic.com/pic5/latex/9221.png)
,0)、S(0,0,2
![](http://thumb.1010pic.com/pic5/latex/21.png)
)、C(2
![](http://thumb.1010pic.com/pic5/latex/14415.png)
,
![](http://thumb.1010pic.com/pic5/latex/14416.png)
,0),
![](http://thumb.1010pic.com/pic5/upload/201306/51d6015a61b4e.png)
![](http://thumb.1010pic.com/pic5/latex/14417.png)
=(2
![](http://thumb.1010pic.com/pic5/latex/14415.png)
,
![](http://thumb.1010pic.com/pic5/latex/14416.png)
,-2
![](http://thumb.1010pic.com/pic5/latex/21.png)
),
![](http://thumb.1010pic.com/pic5/latex/3653.png)
=(-2
![](http://thumb.1010pic.com/pic5/latex/14415.png)
,
![](http://thumb.1010pic.com/pic5/latex/14418.png)
,0).
(1)∵
![](http://thumb.1010pic.com/pic5/latex/14417.png)
•
![](http://thumb.1010pic.com/pic5/latex/3653.png)
=0,∴SC⊥BC.
(2)设SC与AB所成的角为α,
∵
![](http://thumb.1010pic.com/pic5/latex/230.png)
=(0,
![](http://thumb.1010pic.com/pic5/latex/9221.png)
,0),
![](http://thumb.1010pic.com/pic5/latex/14417.png)
•
![](http://thumb.1010pic.com/pic5/latex/230.png)
=4,|
![](http://thumb.1010pic.com/pic5/latex/14417.png)
||
![](http://thumb.1010pic.com/pic5/latex/230.png)
|=4
![](http://thumb.1010pic.com/pic5/latex/9221.png)
,
∴cosα=
![](http://thumb.1010pic.com/pic5/latex/14419.png)
,即为所求.
解法二:(1)∵SA⊥面ABC,AC⊥BC,AC是斜线SC在平面ABC内的射影,∴SC⊥BC.
![](http://thumb.1010pic.com/pic5/upload/201306/51d6015a86065.png)
(2)如图,过点C作CD∥AB,过点A作AD∥BC交CD于点D,
连接SD、SC,则∠SCD为异面直线SC与AB所成的角.
∵四边形ABCD是平行四边形,CD=
![](http://thumb.1010pic.com/pic5/latex/9221.png)
,SA=2
![](http://thumb.1010pic.com/pic5/latex/21.png)
,SD=
![](http://thumb.1010pic.com/pic5/latex/14420.png)
=
![](http://thumb.1010pic.com/pic5/latex/14421.png)
=5,
∴在△SDC中,由余弦定理得cos∠SCD=
![](http://thumb.1010pic.com/pic5/latex/14419.png)
,即为所求.
分析:解法一:建系,写出有关点的坐标,B,C,s,(1)要证SC⊥BC;只要证EF⊥面PAB,只要证)
![](http://thumb.1010pic.com/pic5/latex/14417.png)
•
![](http://thumb.1010pic.com/pic5/latex/3653.png)
=0即可;
(2)要求异面直线SC与AB所成的角的余弦值,只要求
![](http://thumb.1010pic.com/pic5/latex/14417.png)
与
![](http://thumb.1010pic.com/pic5/latex/230.png)
所成角的余弦值即可;
解法二:综合法证明,(1)要证SC⊥BC,只要证AC⊥BC即可;
(2)要求SC与AB所成角的余弦值,通过平移找到SC与AB所成角,解三角形即可.
点评:考查利用空间向量证明垂直和求夹角和距离问题,以及面面垂直的判定定理,体现 了转化的思想方法,l利用综合法求异面直线所成的角,关键是找出这个角,把空间角转化为平面角求解,体现了转化的思想,属中档题.