(20 1 1? Suzhou Second Model) In the triangular prism ABC-a1b1,the base △ABC is a right triangle, AC=BC=AA 1=2, and D is the midpoint of the side AA 1. ( 1

Solution: Solution: (1) As shown in the figure, a rectangular coordinate system is established with C as the origin and CA, CB, CC 1 as the coordinate axes.

C-xyz。

Then c (0 0,0,0), a (2 2,0,0), b (0 0,2,0), c1(0,0,2), b1(0 (0,2,2), d (2

So DC 1 = (-2,0,1), and b1c = (0,2,2).

So cos < dc 1, b 1c > = dc 1? b 1C | DC 1 | | b 1C | =？ 25×8=- 10 10.

That is to say, the cosine of the angle formed by DC 1 and B 1C is10.

(2) Because CB = (0 0,2,0), CA = (2 2,0,0), CC1= (0,0,2),

So CB? CA=0，CB？ CC 1=0，

So CB is the normal vector of plane ACC 1A 1.

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