Benchmarks

(II) 解法1: 由(I)知, $\backslash angle\; A\text{'}OC=120^\backslash circ$ ---------------- 6分

如图建系 $O-xyz$ , $B(1,\; 0,\; 0)$ , 设 $OC=b$ , $OA\text{'}=a$ , 则 $C(0,\; b,\; 0)$ , $A\text{'}\backslash left(0,\; -\backslash frac\{1\}\{2\}a,\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}a\backslash right)$

$\backslash overrightarrow\{BC\}=(-1,\; b,\; 0)$ ---------------- 8分

平面 $A\text{'}DB$ 的法向量为 $\backslash vec\{n\}=(0,\; \backslash sqrt\{3\},\; 1)$ ---------------- 10分

$$\backslash sin\; 45^\backslash circ\; =\; \backslash left|\; \backslash cos\; \backslash langle\; \backslash vec\{n\},\; \backslash overrightarrow\{BC\}\; \backslash rangle\; \backslash right|\; =\; \backslash left|\; \backslash frac\{\backslash vec\{n\}\; \backslash cdot\; \backslash overrightarrow\{BC\}\}\{|\backslash vec\{n\}|\; |\backslash overrightarrow\{BC\}|\}\; \backslash right|\; \backslash quad\; \backslash text\{----------------\; 11分\}$$

解得 $b=\backslash sqrt\{2\}$ , $BC=\backslash sqrt\{3\}$ ---------------- 12分

解法2: 由(I)知, $\backslash angle\; A\text{'}OC=120^\backslash circ$ ---------------- 6分

过 $C$ 作 $CH\; \backslash perp\; A\text{'}O$ , $BD\; \backslash perp$ 平面 $A\text{'}OC$ ,

$\backslash therefore\; BD\; \backslash perp\; CH$ , $CH\; \backslash perp\; A\text{'}O$ , $CH\; \backslash perp$ 平面 $A\text{'}BD$

$\backslash angle\; CBH$ 就是 $BC$ 与平面 $\backslash triangle\; A\text{'}BD$ 所成角 ---------------- 9分

设 $CO=x$ , 则 $CH=\backslash frac\{\backslash sqrt\{3\}\}\{2\}x$ , $CB=\backslash sqrt\{2\}CH=\backslash frac\{\backslash sqrt\{6\}\}\{2\}x$ ,

$CB=\backslash sqrt\{OB^2+OC^2\}=\backslash sqrt\{1+x^2\}$ , 则 $\backslash frac\{\backslash sqrt\{6\}\}\{2\}x=\backslash sqrt\{1+x^2\}$

解得 $x=\backslash sqrt\{2\}$ , $BC=\backslash sqrt\{3\}$ ---------------- 12分

21. 解:

(I) 联立方程组 $\backslash begin\{cases\}\; y=kx+1\; \backslash \backslash \; x^2-4y^2=4\; \backslash end\{cases\}$ 消 $y$ 得: $(1-4k^2)x^2-8kx-8=0$ ---------------- 2分

$\backslash begin\{cases\}\; 1-4k^2\; \backslash neq\; 0\; \backslash \backslash \; \backslash Delta=32-64k^2>0\; \backslash end\{cases\}$ 解得 且 $k\; \backslash neq\; \backslash pm\; \backslash frac\{1\}\{2\}$ ---------------- 5分

(漏 $k\; \backslash neq\; \backslash pm\; \backslash frac\{1\}\{2\}$ 得 4分)

(II) 设 $M$ , $N$ 坐标分别为 $(x\_1,\; y\_1)$ , $(x\_2,\; y\_2)$ , $A(-2,\; 0)$ , 由(I)知

$$\backslash begin\{cases\}\; x\_1+x\_2=\backslash frac\{8k\}\{1-4k^2\}\; \backslash \backslash \; x\_1\; \backslash cdot\; x\_2=\backslash frac\{-8\}\{1-4k^2\}\; \backslash end\{cases\}\; \backslash quad\; \backslash text\{----------------\; 6分\}$$

直线 $MA$ 的方程为 $y=\backslash frac\{y\_1\}\{x\_1+2\}(x+2)$ , 令 $x=0$ 可得点 $P$ 坐标为 $\backslash left(0,\; \backslash frac\{2y\_1\}\{x\_1+2\}\backslash right)$

同理点 $Q$ 坐标为 $\backslash left(0,\; \backslash frac\{2y\_2\}\{x\_2+2\}\backslash right)$ ---------------- 8分

$$|PQ|=1\; \backslash Rightarrow\; \backslash left|\; \backslash frac\{y\_1\}\{x\_1+2\}\; -\; \backslash frac\{y\_2\}\{x\_2+2\}\; \backslash right|\; =\; \backslash frac\{1\}\{2\}\; \backslash Rightarrow\; \backslash left|\; \backslash frac\{(x\_1-x\_2)(1-2k)\}\{(x\_1+2)(x\_2+2)\}\; \backslash right|\; =\; \backslash frac\{1\}\{2\}$$

$$\backslash left|\; 4\backslash sqrt\{2\}\backslash sqrt\{1-2k^2\}(1-2k)\; \backslash right|\; =\; 2(2k-1)^2\; \backslash quad\; \backslash text\{----------------\; 10分\}$$

浙江省A9协作体暑假返校联考 高三数学参考答案 第4页 共6页