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1.4.2(1) 用空间向量研究异面直线所成角和线面角

2022-09-02 20:35:07  阅读:178  来源: 互联网

标签:1.4 overrightarrow 异面 cdot dfrac sqrt 角和线 vec 平面


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基础知识

求异面直线a ,b所成的角

已知\(a\),\(b\)为两异面直线,\(A\) ,\(C\)与\(B\) ,\(D\)分别是\(a\),\(b\)上的任意两点,\(a\),\(b\)所成的角为\(θ\),
则 \(\cos \theta=|\cos <\overrightarrow{A C}, \overrightarrow{B D}>|=\dfrac{|\overrightarrow{A C} \cdot \overrightarrow{B D}|}{|\overrightarrow{A C}||\overrightarrow{B D}|}\) .
解释
①向量 \(\overrightarrow{A C}\), \(\overrightarrow{B D}\)所成角 \(<\overrightarrow{A C}, \overrightarrow{B D}>\)的范围是\((0 ,π]\),而异面直线\(AC\),\(BD\)所成的角范围是 \(\left[0, \dfrac{\pi}{2}\right]\);
② \(<\overrightarrow{A C}, \overrightarrow{B D}>\)与\(θ\)的关系相等或互补;
image.png image.png
故 \(\cos \theta=|\cos <\overrightarrow{A C}, \overrightarrow{B D}>|\),不要漏了“绝对值符号”.
 

【例】在正方体\(ABCD-A'B'C'D'\)中,直线\(A'B\)与直线\(AD'\)夹角\(θ=\)\(\underline{\quad \quad}\),向量 \(\overrightarrow{B A^{\prime}}\)与直线\(\overrightarrow{A D^{\prime}}\)所成角等于\(\underline{\quad \quad}\) .
image.png
解析 因为\(A' B//CD'\),所以\(∠AD'C\)是直线\(A'B\)与直线\(AD'\)夹角,
又因为\(∆ACD'\)是等边三角形,所以\(θ=60°\),而向量 \(\overrightarrow{B A^{\prime}}\)与直线\(\overrightarrow{A D^{\prime}}\)所成角等于\(120°\).
 

求直线l和平面α所成的角

设直线\(l\)方向向量为\(\vec{a}\),平面\(β\)法向量为\(\vec{n}\),直线与平面所成的角为\(θ\),\(\vec{a}\)与 \(\vec{n}\)的夹角为\(α\),
则 \(θ\)为\(α\) 的余角或\(α\)的补角的余角,即有 \(\sin \theta=|\cos \alpha|=\dfrac{|\vec{a} \cdot \vec{n}|}{|\vec{a}||\vec{n}|}\) .
解释 如下图,当 \(\theta=\dfrac{\pi}{2}-\alpha\)时, \(\sin \theta=\cos \alpha\);当 \(\theta=\alpha-\dfrac{\pi}{2}\)时, \(\sin \theta=-\cos \alpha\);
image.png image.png
在求直线\(l\)和平面\(β\)所成的角实际过程中,较难判断平面\(β\)的法向量的方向,但不管如何均有\(\sin θ=|\cos α|\).
 

【例1】在正方体\(ABCD-A'B'C'D'\)中,直线\(BD'\)与平面\(ABCD\)所成角为\(θ\),向量 \(\overrightarrow{D^{\prime}D}\)与向量\(\overrightarrow{BD^{\prime} }\)所成角为\(α\),判断\(\cosα\)与sinθ的关系.
image.png
解析 因为\(DD'⊥\)平面\(ABCD\),所以直线\(BD'\)与平面\(ABCD\)所成角\(θ=∠D'BD\),
而平面\(ABCD\)的法向量 \(\overrightarrow{D^{\prime}D}\)与向量\(\overrightarrow{BD^{\prime} }\)所成角\(α=∠ED'D\),由图易得 \(\theta=\alpha-\dfrac{\pi}{2}\).
所以\(\sin θ=-\cos α\),
若把平面\(ABCD\)的法向量 \(\overrightarrow{D^{\prime}D}\)改为 \(\overrightarrow{DD^{\prime}}\),则 \(\theta=\dfrac{\pi}{2}-\alpha\)时, \(\sin θ=\cos α\).
 
【例2】若直线\(l\)的方向向量\(\vec{a}=(1,0,1)\),而平面\(ABC\)的一个法向量\(\vec{n}=(-1,1,0)\),则直线\(l\)与平面\(ABC\)所成角等于\(\underline{\quad \quad}\) .
解析 设直线\(l\)与平面\(ABC\)所成角为\(θ\),则 \(\sin \theta=\mid \cos \langle\vec{a}, \vec{n}>|=\dfrac{|\vec{a} \cdot \vec{n}|}{|\vec{a}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{2} \cdot \sqrt{2}}=\dfrac{1}{2}\),
所以直线\(l\)与平面\(ABC\)所成角是\(30°\).
 

基本方法

【题型1】求异面直线所成角

【典题1】 已知空间四点\(A(0,1,0)\), \(B\left(1,0, \dfrac{1}{2}\right)\),\(C(0,0,1)\), \(D\left(1,1, \dfrac{1}{2}\right)\),则异面直线\(AB\),\(CD\)所成的角的余弦值为\(\underline{\quad \quad}\).
解析 设 \(\overrightarrow{A B}=\vec{a}=\left(1,-1, \dfrac{1}{2}\right)\), \(\overrightarrow{C D}=\vec{b}=\left(1,1,-\dfrac{1}{2}\right)\).
\(\vec{a} \cdot \vec{b}=1-1-\dfrac{1}{4}=-\dfrac{1}{4}\), \(|\vec{a}|=\sqrt{1+1+\dfrac{1}{4}}=\dfrac{3}{2}\), \(|\vec{b}|=\sqrt{1+1+\dfrac{1}{4}}=\dfrac{3}{2}\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{-\dfrac{1}{4}}{\dfrac{3}{2} \times \dfrac{3}{2}}=-\dfrac{1}{9}\),
\(∴\)异面直线\(AB\),\(CD\)所成的角的余弦值为 \(\dfrac{1}{9}\).
 

【典题2】如图,\(ABC-A_1 B_1 C_1\)是直三棱柱,\(∠BCA=90°\),点\(E\)、\(F\)分别是\(A_1 B_1\) 、\(A_1 C_1\)的中点,若\(BC=CA=AA_1\),则\(BE\)与\(AF\)所成角的余弦值为\(\underline{\quad \quad}\).
image.png
解析 \(∵ABC-A_1 B_1 C_1\)直三棱柱,\(∠BCA=90°\),
\(∴\)以\(C\)为原点,\(CA\)为\(x\)轴,\(CB\)为\(y\)轴,\(CC_1\)为\(z\)轴,建立空间直角坐标系,
image.png
点\(E\)、\(F\)分别是\(A_1 B_1\)、\(A_1 C_1\)的中点,设\(BC=CA=AA_1=2\),
则\(B(0,2,0)\),\(A(2,0,0)\),\(E(1,1,2)\),\(F(1,0,2)\),
\(\overrightarrow{B E}=(1,-1,2)\), \(\overrightarrow{A F}=(-1,0,2)\),
设\(BE\)与\(AF\)所成角为\(θ\),则 \(\cos \theta=\dfrac{|\overrightarrow{B E} \cdot \overrightarrow{A F}|}{|\overrightarrow{B E}| \cdot|\overrightarrow{A F}|}=\dfrac{3}{\sqrt{6} \cdot \sqrt{5}}=\dfrac{\sqrt{30}}{10}\),
\(∴BE\)与\(AF\)所成角的余弦值为 \(\dfrac{\sqrt{30}}{10}\).
 

巩固练习

1 在正方体\(ABCD﹣A_1 B_1 C_1 D_1\)中,点\(E\),\(F\)分别是棱\(AB\),\(BC\)的中点,则直线\(CE\)与\(D_1 F\)所成角的大小为(  )
  A.\(\dfrac{\pi}{6}\) \(\qquad \qquad\) B.\(\dfrac{\pi}{4}\) \(\qquad \qquad\) C. \(\dfrac{\pi}{3}\) \(\qquad \qquad\) D. \(\dfrac{\pi}{2}\)
image.png
 

2 如图,\(S\)是三角形\(ABC\)所在平面外的一点,\(SA=SB=SC\),且 \(\angle A S B=\angle B S C=\angle C S A=\dfrac{\pi}{2}\),\(M\)、\(N\)分别是AB和SC的中点,则异面直线\(SM\)与\(BN\)所成角的余弦值为\(\underline{\quad \quad}\).
image.png
 

3 已知正四棱锥\(V﹣ABCD\)底面中心为\(O\),\(E\),\(F\)分别为\(VA\),\(VC\)的中点,底面边长为\(2\),高为\(4\),建立适当的空间直角坐标系,求异面直线\(BE\)与\(DF\)所成角的正切值.
 
 

参考答案

  1. 答案 \(D\)
    解析 以\(D\)为原点,\(DA\)为\(x\)轴,\(DC\)为\(y\)轴,\(DD_1\)为\(z\)轴,建立空间直角坐标系,
    设正方体\(ABCD﹣A_1 B_1 C_1 D_1\)中棱长为\(2\),
    则\(C(0,2,0)\),\(E(2,1,0)\),\(D_1 (0,0,2)\),\(F(1,2,0)\),
    \(\overrightarrow{C E}=(2,-1,0)\), \(\overrightarrow{D_{1} F}=(1,2,-2)\),
    设直线\(CE\)与\(D_1 F\)所成角的大小为\(θ\),则 \(\cos \theta=\dfrac{\left|\overrightarrow{C E} \cdot \overrightarrow{D_{1} F}\right|}{|\overrightarrow{C E}| \cdot\left|\overrightarrow{D_{1} F}\right|}=0\),
    \(\therefore \theta=\dfrac{\pi}{2}\).
    \(∴\)直线\(CE\)与\(D_1 F\)所成角的大小为 \(\dfrac{\pi}{2}\).
    故选:\(D\).
    image.png

  2. 答案 \(\dfrac{\sqrt{10}}{5}\)
    解析 \(∵\angle A S B=\angle B S C=\angle C S A=\dfrac{\pi}{2}\),
    \(∴\)以\(S\)为坐标原点,分别以\(SC\),\(SB\),\(SA\)所在直线为\(x,y,z\)轴建立空间直角坐标系,
    设\(SA=SB=SC=2a\),则\(S(0,0,0)\),\(B(0,2a,0)\),\(M(0,a,a)\),\(N(a,0,0)\),
    则 \(\overrightarrow{S M}=(0, a, a)\), \(\overrightarrow{B N}=(a,-2 a, 0)\),
    \(\therefore \cos <\overrightarrow{S M}, \overrightarrow{B N}>=\dfrac{|\overrightarrow{S M} \cdot \overrightarrow{B N}|}{|\overrightarrow{S M}| \cdot|\overrightarrow{B N}|}=\dfrac{2 a^{2}}{\sqrt{2} a \cdot \sqrt{5} a}=\dfrac{\sqrt{10}}{5}\).
    \(∴\)异面直线\(SM\)与\(BN\)所成角的余弦值为 \(\dfrac{\sqrt{10}}{5}\).
    image.png

  3. 答案 \(\dfrac{4 \sqrt{10}}{3}\)
    解析 以底面正方形\(ABCD\)中心\(O\)为原点,以\(OA\)为\(x\)轴,\(OB\)为\(y\)轴,\(OV\)为\(z\)轴,
    建立空间直角坐标系,
    image.png
    则 \(A(\sqrt{2}, 0,0)\), \(B(0, \sqrt{2}, 0)\), \(C(-\sqrt{2}, 0,0)\), \(D(0,-\sqrt{2}, 0)\),\(V(0,0,4)\), \(E\left(\dfrac{\sqrt{2}}{2}, 0,2\right)\), \(F\left(-\dfrac{\sqrt{2}}{2}, 0,2\right)\),
    \(\overrightarrow{B E}=\left(\dfrac{\sqrt{2}}{2},-\sqrt{2}, 2\right)\), \(\overrightarrow{D F}=\left(-\dfrac{\sqrt{2}}{2}, \sqrt{2}, 2\right)\),
    设向量\(BE\)和\(DF\)成角为\(θ\),
    \(\cos \theta=|\cos <\overrightarrow{B E}, \overrightarrow{D F}>|=\left|\dfrac{\overrightarrow{B E} \cdot \overrightarrow{D F}}{|\overrightarrow{B E}| \cdot|\overrightarrow{D F}|}\right|\)\(=\left|\dfrac{-\dfrac{1}{2}-2+4}{\sqrt{\dfrac{1}{2}+2+4} \cdot \sqrt{\dfrac{1}{2}+2+4}}\right|=\dfrac{3}{13}\),
    \(\therefore \tan \theta=\dfrac{\sin \theta}{\cos \theta}=\dfrac{4 \sqrt{10}}{3}\).
    \(∴\)异面直线\(BE\)与\(DF\)所成角的正切值为 \(\dfrac{4 \sqrt{10}}{3}\).
     

【题型2】求线面角

【典题1】 如图,在棱长为\(1\)的正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(M\)为\(BC\)中点,则直线\(MD\)与平面\(AB_1 C\)所成角的正弦值是(  )
image.png
 A.\(\dfrac{1}{5}\) \(\qquad \qquad\) B.\(\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad\)C.\(\dfrac{\sqrt{10}}{5}\) \(\qquad \qquad\) D.\(\dfrac{\sqrt{15}}{5}\)
解析 以\(D_1\)为原点,\(D_1 A_1\),\(D_1 C_1\),\(D_1 D\)分别为\(x,y,z\)轴,建立空间直角坐标系,
image.png
显然平面\(AB_1 C\)的法向量 \(\overrightarrow{D_{1} B}=(1,1,1)\),
\(D(0,0,1)\), \(M\left(\dfrac{1}{2}, 1,1\right)\), \(\overrightarrow{D M}=\left(\dfrac{1}{2}, 1,0\right)\),
\(\cos <\overrightarrow{D_{1} B}, \overrightarrow{D M}>=\dfrac{\dfrac{1}{2}+1}{\sqrt{3} \cdot \dfrac{\sqrt{5}}{2}}=\dfrac{3}{\sqrt{15}}=\dfrac{\sqrt{15}}{5}\),
由直线\(MD\)与平面\(AB_1 C\)所成角的正弦值等于 \(\mid \cos <\overrightarrow{D_{1} B}, \overrightarrow{D M}>|=\dfrac{\sqrt{15}}{5}\),
故选:\(D\).
 

【典题2】 如图,已知\(AB\)是圆柱底面圆的一条直径,\(OP\)是圆柱的一条母线,\(C\)为底面圆上一点,且\(AC//OB\), \(O P=A B=\sqrt{2} O A\),则直线\(PC\)与平面\(PAB\)所成角的正弦值为\(\underline{\quad \quad}\) .
image.png
解析 \(∵AB\)是圆柱底面圆的一条直径,\(∴∠AOB=90^∘\),
\(\because O P=A B=\sqrt{2} O A,\),\(∴∠BAO=45^∘\),
\(∵AC//OB\),\(∴∠OAC=90^∘\);
\(∵AB\)是圆柱的底面圆的直径,\(∴∠ACB=90^∘\),
又\(∠OAB=45^∘\),\(∴\)四边形\(OACB\)为正方形,
设\(AB=2\),
如图建立空间直角坐标系\(O-xyz\),
可知 \(A(\sqrt{2}, 0,0)\), \(B(0, \sqrt{2}, 0)\),\(P(0,0,2)\), \(C(\sqrt{2}, \sqrt{2}, 0)\),
image.png
设平面\(PAB\)的法向量为 \(\vec{n}=(x, y, z)\), \(\overrightarrow{A B}=(-\sqrt{2}, \sqrt{2}, 0)\), \(\overrightarrow{A P}=(-\sqrt{2}, 0,2)\),
\(\therefore\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A B}=0 \\ \vec{n} \cdot \overrightarrow{A P}=0 \end{array}\right.\),即 \(\left\{\begin{array}{l} -\sqrt{2} x+\sqrt{2} y=0 \\ -\sqrt{2} x+2 z=0 \end{array}\right.\),
取 \(x=\sqrt{2}\),则 \(\vec{n}=(\sqrt{2}, \sqrt{2}, 1)\),
又 \(\overrightarrow{P C}=(\sqrt{2}, \sqrt{2},-2)\),
设直线\(PC\)与平面\(PAB\)所成角为\(θ\),
\(\therefore \sin \theta=\mid \cos \langle\vec{n}, \overrightarrow{P C}>|=\dfrac{|\overrightarrow{\vec{r}} \cdot \overrightarrow{P C}|}{|\vec{n}| \cdot|\overrightarrow{P C}|}=\dfrac{\sqrt{10}}{10}\),
所以直线\(PC\)与平面\(PAB\)所成角的正弦值为 \(\dfrac{\sqrt{10}}{10}\).
 

【典题3】 如图,在四棱锥\(P-ABCD\)中,\(PA⊥\)平面\(ABCD\),\(∠ABC=∠BAD=90°\),\(AD=AP=4\),\(AB=BC=2\),\(M\),\(N\)分别为线段\(PC\),\(AD\)上的点(不在端点).当\(N\)为\(AD\)中点时,是否存在\(M\),使得直线\(MN\)与平面\(PBC\)所成角的正弦值为 \(\dfrac{2 \sqrt{5}}{5}\),若存在,求出\(MC\)的长,若不存在,说明理由.
image.png
解析 假设存在存在\(M(a,b,c)\),使得直线\(MN\)与平面\(PBC\)所成角的正弦值为 \(\dfrac{2 \sqrt{5}}{5}\),\(\overrightarrow{C M}=\lambda \overrightarrow{C P}\).
则\((a-2,b-2,c)=λ(-2,-2,4)\),解得\(a=2-2λ\),\(b=2-2λ\),\(c=4λ\),
\(∴M(2-2λ,2-2λ,4λ)\),
则 \(\overrightarrow{M N}=(2 \lambda-2,2 \lambda,-4 \lambda)\), \(\overrightarrow{B C}=(0,2,0)\), \(\overrightarrow{B P}=(-2,0,4)\),
设平面\(PBC\)的法向量\(\vec{p} =(a,b,c)\),
则 \(\left\{\begin{array}{l} \vec{p} \cdot \overrightarrow{B C}=2 b=0 \\ \vec{p} \cdot \overrightarrow{B P}=-2 a+4 c=0 \end{array}\right.\),取\(a=2\),得\(\vec{p}=(2,0,1)\),
\(∵\)直线\(MN\)与平面\(PBC\)所成角的正弦值为 \(\dfrac{2 \sqrt{5}}{5}\),
\(\therefore \dfrac{|\overrightarrow{M N} \cdot \vec{p}|}{|\overrightarrow{M N}| \cdot|\vec{p}|}=\dfrac{4}{\sqrt{(2 \lambda-2)^{2}+(2 \lambda)^{2}+(-4 \lambda)^{2}} \cdot \sqrt{20}}=\dfrac{2 \sqrt{5}}{5}\),
整理,得\(24λ^2-8λ+3=0\),无解,
\(∴\)不存在\(M\),使得直线\(MN\)与平面\(PBC\)所成角的正弦值为 \(\dfrac{2 \sqrt{5}}{5}\).
image.png
 

巩固练习

1 如图,在正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(E\),\(F\)分别是上底棱的中点,\(AB_1\)与平面\(B_1 D_1 EF\)所成的角的大小是(  )
image.png
  A.\(30°\) \(\qquad \qquad\) B.\(45°\) \(\qquad \qquad\) C.\(60°\) \(\qquad \qquad\) D.\(90°\)
 

2 如图,正三棱柱\(ABC-A_1 B_1 C_1\)的所有棱长都相等,\(E,F,G\)分别为\(AB\),\(AA_1\),\(A_1 C_1\)的中点,则\(EF\)与平面\(B_1 GF\)所成角的正弦值为 \(\underline{\quad \quad}\) .
image.png
 

3如图,在四棱锥\(P-ABCD\)中,\(BC//\)平面\(PAD\),\(AD=CD=2BC=4\),\(∠BCD=60^∘\),点\(G\)为\(AD\)的中点.
(1)证明:\(CD//\)平面\(PBG\);
(2)若平面\(PBD⊥\)平面\(ABCD\),且\(PB=PD=2\),求直线\(PA\)与平面\(PCD\)所成角的正弦值.
image.png
 

4 在四棱锥\(P-ABCD\)中,\(PD⊥\)底面\(ABCD\),\(CD//AB\),\(AD=DC=CB=1\),\(AB=2\), \(D P=\sqrt{3}\).
(1)证明:\(BD⊥PA\);
(2)求\(PD\)与平面\(PAB\)所成的角的正弦值.
image.png
 
 

参考答案

  1. 答案 \(B\)
    解析 以\(D_1\)为坐标原点,\(D_1 A_1\),\(D_1 C_1\),\(D_1 D\)为\(x,y,z\)轴建立空间直角坐标系,
    image.png
    则\(A(1,0,1)\),\(B_1 (1,1,0)\),\(D_1 (0,0,0)\), \(E\left(0, \dfrac{1}{2}, 1\right)\),
    设平面\(D_1 B_1 E\)的法向量为 \(\vec{n}=(x, y, z)\),
    则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{D_{1} B_{1}}=x+y=0 \\ \vec{n} \cdot \overrightarrow{D_{1} E}=\dfrac{1}{2} y+z=0 \end{array}\right.\),
    又 \(\overrightarrow{A B_{1}}=(0,1,-1)\),
    设\(AB_1\)与平面\(B_1 D_1 EF\)所成的角为\(θ\),则 \(\sin \theta=\left|\cos <\vec{n}, \overrightarrow{A B}_{1}>\right|=\dfrac{\sqrt{2}}{2}\),
    故\(AB_1\)与平面\(B_1 D_1 EF\)所成的角为 \(\dfrac{\pi}{4}\).
    故选:\(B\).

  2. 答案 \(A\)
    解析 设正三棱柱的棱长为\(2\),取\(AC\)的中点\(D\),
    连接\(DG\),\(DB\),分别以\(DB\),\(DC\),\(DG\)所在的直线为\(x\)轴,\(y\)轴,\(z\)轴建立空间直角坐标糸,如图所示,
    image.png
    则 \(B_{1}(\sqrt{3}, 0,2)\),\(G(0,0,2)\), \(E\left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}, 0\right)\),\(F(0,-1,1)\),
    \(\overrightarrow{E F}=\left(-\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}, 1\right)\), \(\overrightarrow{G F}=(0,-1,-1)\), \(\overrightarrow{B_{1} G}=(-\sqrt{3}, 0,0)\),
    设平面\(B_1 GF\)的法向量为\(\vec{n}=(x,y,z)\),
    则 \(\left\{\begin{array}{l} \overrightarrow{B_{1} G} \cdot \vec{n}=-\sqrt{3} x=0 \\ \overrightarrow{G F} \cdot \vec{n}=-y-z=0 \end{array}\right.\),取\(y=1\),则\(z=-1\),\(x=0\),
    故\(\vec{n}=(0,1,-1)\)为平面\(B_1 GF\)的一个法向量,
    \(EF\)与平面\(B_1 GF\)所成角为\(θ\),则 \(\sin \theta=\left|\dfrac{\vec{n} \cdot \overrightarrow{E F}}{|\vec{n}| \cdot \overrightarrow{E F}}\right|=\left|\dfrac{-\dfrac{1}{2}-1}{\sqrt{2} \cdot \sqrt{2}}\right|=\dfrac{3}{4}\),
    \(EF\)与平面\(B_1 GF\)所成角的正弦值为 \(\dfrac{3}{4}\).
    故选:\(A\).

  3. 答案 (1)略 (2) \(\dfrac{\sqrt{42}}{14}\)
    解析 (1)证明:因为\(BC//\)平面\(PAD\),平面\(PAD∩\)平面\(ABCD=AD\),\(BC⊂\)平面\(ABCD\),
    所以\(BC//AD\),
    又因为点\(G\)为\(AD\)的中点,所以 \(G D=\dfrac{1}{2} A D\),
    因为\(AD=2BC\),所以\(BC=\dfrac{1}{2} A D\),所以\(BC=GD\),
    又由\(BC//GD\),
    所以四边形\(BCDG\)为平行四边形,所以\(CD//BG\),
    因为\(CD⊄\)平面\(PBG\),\(BG⊂\)平面\(PBG\),
    所以\(CD//\)平面\(PBG\);
    (2)解:连接\(GC\)交\(BD\)于\(O\),连接\(PO\),
    因为四边形\(BCDG\)为平行四边形,
    所以\(O\)为\(BD\)的中点,
    又因为\(PB=PD\),所以\(PO⊥BD\),
    因为平面\(PBD⊥\)平面\(ABCD\),\(PO⊂\)平面 \(PBD\),
    所以\(PO⊥\)平面\(ABCD\),
    又因为\(GC⊂\)平面\(ABCD\),所以\(PO⊥GC\),
    在\(△BCD\)中,由余弦定理可得\(BD^2=BC^2+CD^2-2BC⋅CD\cos⁡60^∘=12\),
    所以 \(B D=2 \sqrt{3}\),所以 \(O B=\dfrac{1}{2} B D=\sqrt{3}\),
    因为\(BC^2+BD^2=CD^2\),
    所以\(△BCD\)为直角三角形,即\(∠DBC=90^∘\),
    过点\(O\)作\(BC\)的平行线交\(GB\)于\(H\),
    以\(OH\)为\(x\)轴,\(OB\)为\(y\)轴,\(OP\)为\(z\)轴建立空间直角坐标系,如图所示,
    image.png
    可得\(P(0,0,1)\), \(A(2,-\sqrt{3}, 0)\), \(C(2, \sqrt{3}, 0)\), \(D(0,-\sqrt{3}, 0)\),
    所以 \(\overrightarrow{P A}=(2,-\sqrt{3},-1)\), \(\overrightarrow{P C}=(2, \sqrt{3},-1)\), \(\overrightarrow{P D}=(0,-\sqrt{3},-1)\),
    设平面\(PCD\)的法向量为\(\vec{m}=(x,y,z)\),则 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{P C}=2 x+\sqrt{3} y-z=0 \\ \vec{m} \cdot \overrightarrow{P D}=-\sqrt{3} y-z=0 \end{array}\right.\),
    取\(y=1\),可得 \(x=-\sqrt{3}\), \(z=-\sqrt{3}\),所以 \(\vec{m}=(-\sqrt{3}, 1,-\sqrt{3})\),
    设\(PA\)与平面\(PCD\)所成的角为\(θ\),则 \(\mid \sin \theta=\dfrac{|\overrightarrow{P A} \cdot \vec{m}|}{|\overrightarrow{P A}| \cdot|\vec{m}|}=\dfrac{2 \sqrt{3}+\sqrt{3}-\sqrt{3}}{\sqrt{7} \cdot \sqrt{8}}=\dfrac{\sqrt{42}}{14}\),
    所以\(PA\)与平面\(PCD\)所成的角的正弦值为 \(\dfrac{\sqrt{42}}{14}\).

  4. 答案 (1)略 (2) \(\dfrac{\sqrt{5}}{5}\)
    解析 (1)证明:\(∵PD⊥\)底面\(ABCD\),\(B D \subset\) 面\(ABCD\),\(∴PD⊥BD\),
    取\(AB\)中点\(E\),连接\(DE\),
    \(∵AD=DC=CB=1\),\(AB=2\),
    \(∴∠DAB=60^∘\),
    又 \(\because A E=\dfrac{1}{2} A B=A D=1\),
    \(\therefore D E=1\), \(\therefore D E=\dfrac{1}{2} A B\),
    \(∴△ABD\)为直角三角形,且\(AB\)为斜边,\(∴BD⊥AD\),
    又\(PD⋂AD=D\),\(PD⊂\)面\(PAD\),\(AD⊂\)面\(PAD\),
    \(∴BD⊥\)面\(PAD\),
    又\(PA⊂\)面\(PAD\),\(∴BD⊥PA\);
    (2)由(1)知,\(PD,AD,BD\)两两互相垂直,故建立如图所示的空间直角坐标系,
    image.png
    \(B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{3}\),
    则\(D(0,0,0)\),\(A(1,0,0)\), \(B(0, \sqrt{3}, 0)\), \(P(0,0, \sqrt{3})\),
    \(\therefore \overrightarrow{P D}=(0,0,-\sqrt{3})\), \(\overrightarrow{P A}=(1,0,-\sqrt{3})\), \(\overrightarrow{A B}=(-1, \sqrt{3}, 0)\),
    设平面\(PAB\)的一个法向量为 \(\vec{n}=(x, y, z)\),
    则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{P A}=x-\sqrt{3} z=0 \\ \vec{n} \cdot \overrightarrow{A B}=-x+\sqrt{3} y=0 \end{array}\right.\),则可取\(\vec{n}=(\sqrt{3},1,1)\),
    设\(PD\)与平面\(PAB\)所成的角为\(θ\),则 \(\sin \theta=|\cos <\overrightarrow{P D}, \vec{n}>|=\left|\dfrac{\overrightarrow{P D} \cdot \vec{n}}{|\overrightarrow{P D} \| \vec{n}|}\right|=\dfrac{\sqrt{5}}{5}\),
    \(∴PD\)与平面\(PAB\)所成的角的正弦值为 \(\dfrac{\sqrt{5}}{5}\).

分层练习

【A组---基础题】

1在正方体\(ABCD﹣A_1 B_1 C_1 D_1\)中,\(E\)为棱\(BB_1\)的中点,则异面直线\(A_1 E\)与\(AC\)所成角的余弦值为(  )
 A. \(\dfrac{1}{5}\) \(\qquad \qquad\) B. \(\dfrac{\sqrt{10}}{5}\) \(\qquad \qquad\) C. \(\dfrac{2 \sqrt{5}}{5}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{10}}{10}\)
image.png
 

2在四面体\(ABCD\)中,\(AB\),\(BC\),\(BD\)两两垂直,\(AB=BC=BD=4\),\(E\)、\(F\)分别为棱\(BC\)、\(AD\)的中点,则直线\(EF\)与平面\(ACD\)所成角的余弦值(  )
 A. \(\dfrac{1}{3}\) \(\qquad \qquad\) B. \(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\) C. \(\dfrac{2 \sqrt{2}}{3}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{6}}{3}\)
 

3 若异面直线\(l_1\),\(l_2\)的方向向量分别是 \(\vec{a}=(0,-2,-1)\), \(\vec{b}=(2,0,4)\),则异面直线\(l_1\)与\(l_2\)的夹角的余弦值等于\(\underline{\quad \quad}\) .
 

4 如图,在四棱锥\(P-ABCD\)中,\(PA⊥\)平面\(ABCD\),\(∠BAD=90^∘\), \(P A=A B=B C=\dfrac{1}{2} A D\),\(BC//AD\),已知\(Q\)是边\(PD\)的中点,则\(CQ\)与平面\(ABCD\)所成角的正弦值为\(\underline{\quad \quad}\) .
image.png
 

5 正方体\(ABCD-A_1 B_1 C_1 D_1\)棱长为\(2\),\(E\)是棱\(AB\)的中点,\(F\)是四边形\(AA_1 D_1 D\)内一点(包含边界),且 \(\overrightarrow{F E} \cdot \overrightarrow{F D}=-\dfrac{3}{4}\),当三棱锥\(F-AED\)的体积最大时,\(EF\)与平面\(ABB_1 A_1\)所成角的正弦值为\(\underline{\quad \quad}\).
 

6 如图所示,三棱柱\(ABC-A_1 B_1 C_1\)中,\(CA=CB\),\(AB=AA_1\),\(∠BAA_1=60°\).
(1)证明:\(AB⊥A_1 C\);
(2)若平面\(ABC⊥\)平面\(AA_1 B_1 B\),\(AB=CB=2\),求直线\(A_1 C\)与平面\(BB_1 C_1 C\)所成角的正弦值.
image.png
 

7 如图,已知\(ABCD\)和\(CDEF\)都是直角梯形,\(AB//DC\),\(DC//EF\),\(AB=5\),\(DC=3\),\(EF=1\),\(∠BAD=∠CDE=60^∘\),二面角\(F-DC-B\)的平面角为\(60^∘\).设\(M,N\)分别为\(AE\),\(BC\)的中点.
(Ⅰ)证明:\(FN⊥AD\);
(Ⅱ)求直线\(BM\)与平面\(ADE\)所成角的正弦值.
image.png
 
 

参考答案

  1. 答案 \(B\)
    解析 以\(D\)为原点,\(DA\)为\(x\)轴,\(DC\)为\(y\)轴,\(DD_1\)为\(z\)轴,建立空间直角坐标系,
    设正方体\(ABCD-A_1 B_1 C_1 D_1\)中棱长为\(2\),
    则\(A_1 (2,0,2)\),\(E(2,2,1)\),\(A(2,0,0)\),\(C(0,2,0)\),
    \(\overrightarrow{A_{1} E}=(0,2,-1)\), \(\overrightarrow{A C}=(-2,2,0)\),
    设异面直线\(A_1 E\)与\(AC\)所成角为\(θ\),则 \(\cos \theta=\dfrac{\left|\overrightarrow{A_{1} E} \cdot \overrightarrow{A C}\right|}{\mid \overrightarrow{\overrightarrow{A_{1} E}|\cdot| \overrightarrow{A C} \mid}}=\dfrac{4}{\sqrt{5} \cdot \sqrt{8}}=\dfrac{\sqrt{10}}{5}\).
    \(∴\)异面直线\(A_1 E\)与\(AC\)所成角的余弦值为 \(\dfrac{\sqrt{10}}{5}\).
    故选:\(B\).
    image.png

  2. 答案 \(C\)
    解析 如图,以\(BC\),\(BD\),\(BA\)为坐标轴建立空间直角坐标系,
    则\(A(0,0,4)\),\(B(0,0,0)\),\(C(4,0,0)\),\(D(0,4,0)\),\(E(2,0,0)\),\(F(0,2,2)\),
    \(\therefore \overrightarrow{A C}=(4,0,-4)\), \(\overrightarrow{A D}=(0,4,-4)\), \(\overrightarrow{E F}=(-2,2,2)\),
    设平面\(ACD\)的法向量为 \(\vec{n}=(x, y, z)\),则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A C}=0 \\ \vec{n} \cdot \overrightarrow{A D}=0 \end{array}\right.\),
    即 \(\left\{\begin{array}{l} 4 x-4 z=0 \\ 4 y-4 z=0 \end{array}\right.\),令\(x=1\)可得\(\vec{n}=(1,1,1)\),
    \(\therefore \cos <\vec{n}, \overrightarrow{E F}>=\dfrac{\vec{n} \cdot \overrightarrow{E F}}{|\vec{n}||\overrightarrow{E F}|}=\dfrac{2}{\sqrt{3} \times 2 \sqrt{3}}=\dfrac{1}{3}\),
    设直线\(EF\)与平面\(ACD\)所成角为\(θ\),则 \(\sin \theta=\dfrac{1}{3}\),
    \(\therefore \cos \theta=\dfrac{2 \sqrt{2}}{3}\).
    故选:\(C\).
    image.png

  3. 答案 \(\dfrac{2}{5}\)
    解析 \(\because \vec{a}=(0,-2,-1)\), \(\vec{b}=(2,0,4)\),
    \(\therefore \vec{a} \cdot \vec{b}=0 \times 2+(-2) \times 0+(-1) \times 4=-4\),
    \(|\vec{a}|=\sqrt{0+(-2)^{2}+(-1)^{2}}=\sqrt{5}\), \(|\vec{b}|=\sqrt{2^{2}+0+4^{2}}=2 \sqrt{5}\),
    \(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{-4}{\sqrt{5} \times 2 \sqrt{5}}=-\dfrac{2}{5}\),
    异面直线\(l_1\)与\(l_2\)的夹角的余弦值等于 \(\dfrac{2}{5}\).

  4. 答案 \(\dfrac{\sqrt{5}}{5}\)
    解析 以\(A\)为坐标原点,以\(AD\)所在直线为\(x\)轴,\(AB\)所在直线为\(y\)轴,\(AP\)所在直线为\(z\)轴,空间直角坐标系\(A-xyz\),
    则\(C=(1,1,0)\), \(Q\left(1,0, \dfrac{1}{2}\right)\), \(\overrightarrow{C Q}=\left(0,-1, \dfrac{1}{2}\right)\),
    取平面\(ABCD\)的一个法向量\(\vec{n}=(0,0,1)\),则 \(\cos <\vec{n}, \overrightarrow{C Q}>=\dfrac{\dfrac{1}{2}}{1 \times \dfrac{\sqrt{5}}{2}}=\dfrac{\sqrt{5}}{5}\),
    即\(CQ\)与平面\(ABCD\)所成角的正弦值为 \(\dfrac{\sqrt{5}}{5}\).
    image.png

  5. 答案 \(\dfrac{2}{3}\)
    解析 如图,以\(A\)为坐标原点,建立空间直角坐标系,
    image.png
    则\(A(0,0,0)\),\(E(1,0,0)\),\(D(0,2,0)\),
    设\(F(0,m,n)\),\(m∈[0,2]\),\(n∈[0,2]\),
    \(\therefore \overrightarrow{F E} \cdot \overrightarrow{F D}=m^{2}-2 m+n^{2}=-\dfrac{3}{4}\),
    \(\therefore S_{\triangle A D E}\)为定值,要想三棱锥\(F-AED\)的体积最大,则\(F\)到底面\(ADE\)的距离最大,
    其中 \(n^{2}=-\dfrac{3}{4}-m^{2}+2 m=-(m-1)^{2}+\dfrac{1}{4}\),
    \(∴\)当\(m=1\)时,\(n^2\)取得最大值为 \(\dfrac{1}{4}\),
    \(∵n∈[0,2]\),\(∴n\)的最大值为 \(\dfrac{1}{2}\),
    \(\therefore F\left(0,1, \dfrac{1}{2}\right)\), \(\overrightarrow{E F}=\left(-1,1, \dfrac{1}{2}\right)\),
    平面\(ABB_1 A_1\)的法向量\(\vec{n}=(0,1,0)\),
    \(∴\)当三棱锥\(F-AED\)的体积最大时,\(EF\)与平面\(ABB_1 A_1\)所成角的正弦值为:
    \(|\cos <\overrightarrow{E F}, \vec{n}>|=\dfrac{|\overrightarrow{E F} \cdot \vec{n}|}{|\overrightarrow{E F}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{1+1+\dfrac{1}{4}}}=\dfrac{2}{3}\).

  6. 答案 (1)略 (2) \(\dfrac{\sqrt{10}}{5}\)
    解析 (1)取\(AB\)的中点\(O\),连接\(OC\),\(OA_1\),\(A_1 B\),
    因为\(CA=CB\),所以\(OC⊥AB\),由于\(AB=AA_1\),\(∠BAA_1=60^∘\),
    所以\(△AA_1 B\)为等边三角形,
    所以\(OA_1⊥AB\),
    又因为\(OC∩OA_1=O\),
    所以\(AB⊥\)平面\(OA_1 C\),
    又\(A_1 C⊂\)平面\(OA_1 C\),故\(AB⊥A_1 C\);
    (2)由(1)知\(OC⊥AB\),\(OA_1⊥AB\),又平面\(ABC⊥\)平面\(AA_1 B_1 B\),交线为\(AB\),
    所以\(OC⊥\)平面\(AA_1 B_1 B\),故\(OA,OA_1,OC\)两两垂直.
    以\(O\)为坐标原点, \(\overrightarrow{O A}\)的方向为\(x\)轴的正向,\(|\overrightarrow{O A}|\)为单位长,建立如图所示的坐标系,
    可得\(A(1,0,0)\), \(A_{1}(0, \sqrt{3}, 0)\), \(C(0,0, \sqrt{3})\),\(B(-1,0,0)\),
    则 \(\overrightarrow{B C}=(1,0, \sqrt{3})\), \(\overrightarrow{B B}_{1}=\overrightarrow{A A_{1}}=(-1, \sqrt{3}, 0)\), \(\overrightarrow{A_{1}} C=(0,-\sqrt{3}, \sqrt{3})\),
    设\(\vec{n}=(x,y,z)\)为平面\(BB_1 C_1 C\)的法向量,
    则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{B C}=0 \\ \vec{n} \cdot \vec{n} B_{1}=0 \end{array}\right.\),即 \(\left\{\begin{array}{l} x+\sqrt{3} z=0 \\ -x+\sqrt{3} y=0 \end{array}\right.\),
    可取\(y=1\),可得\(\vec{n}=(\sqrt{3},1,-1)\),
    故 \(\cos \left\langle\vec{n}, \overrightarrow{A_{1} C}\right\rangle=\dfrac{\vec{n} \cdot \overrightarrow{A_{1} C}}{|\vec{n}|\left|\overrightarrow{A_{1} C}\right|}=-\dfrac{\sqrt{10}}{5}\),
    又因为直线与法向量的余弦值的绝对值等于直线与平面的正弦值,
    故直线\(A_1 C\)与平面\(BB_1 C_1 C\)所成角的正弦值为: \(\dfrac{\sqrt{10}}{5}\).
    image.png

  7. 答案 (1)略 (2) \(\dfrac{5 \sqrt{7}}{14}\)
    解析 证明: 由于\(CD⊥CB\),\(CD⊥CF\),
    平面\(ABCD∩\)平面\(CDEF=CD\),\(CF⊂\)平面\(CDEF\),\(CB⊂\)平面\(ABCD\),
    所以\(∠FCB\)为二面角\(F-DC-B\)的平面角,
    则\(∠FCB=60^∘\),\(CD⊥\)平面\(CBF\),则\(CD⊥FN\).
    又 \(C F=\sqrt{3}(C D-E F)=2 \sqrt{3}\), \(C B=\sqrt{3}(A B-C D)=2 \sqrt{3}\),
    则\(△BCF\)是等边三角形,则\(CB⊥FN\),
    因为\(DC⊥FC\),\(DC⊥BC\),\(F C \cap B C=C\),\(FC⊂\)平面\(FCB\),\(BC⊂\)平面\(FCB\),
    所以\(DC⊥\)平面\(FCB\),
    因为\(FN⊂\)平面\(FCB\),所以\(DC⊥FN\),
    又因为\(DC\cap CB=C\),\(DC⊂\)平面\(ABCD\),\(CB⊂\)平面\(ABCD\),
    所以\(FN⊥\)平面\(ABCD\),因为\(AD⊂\)平面\(ABCD\),故\(FN⊥AD\);
    解:(Ⅱ)由于\(FN⊥\)平面\(ABCD\),如图建系:
    image.png
    于是 \(B(0, \sqrt{3}, 0)\), \(A(5, \sqrt{3}, 0)\),\(F(0,0,3)\),\(E(1,0,3)\), \(D(3,-\sqrt{3}, 0)\),
    则 \(M\left(3, \dfrac{\sqrt{3}}{2}, \dfrac{3}{2}\right)\), \(\overrightarrow{B M}=\left(3,-\dfrac{\sqrt{3}}{2}, \dfrac{3}{2}\right)\), \(\overrightarrow{D A}=(2,2 \sqrt{3}, 0)\), \(\overrightarrow{D E}=(-2, \sqrt{3}, 3)\),
    设平面\(ADE\)的法向量\(\vec{n}=(x,y,z)\),
    则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{D A}=0 \\ \vec{n} \cdot \overrightarrow{D E}=0 \end{array}\right.\), \(\therefore\left\{\begin{array}{l} 2 x+2 \sqrt{3} y=0 \\ -2 x+\sqrt{3} y+3 z=0 \end{array}\right.\),
    令 \(x=\sqrt{3}\),则 \(y=-1, z=\sqrt{3}\),
    \(∴\)平面\(ADE\)的法向量 \(\vec{n}=(\sqrt{3},-1, \sqrt{3})\),
    设\(BM\)与平面\(ADE\)所成角为\(θ\),则 \(\sin \theta=\dfrac{|\overrightarrow{B M} \cdot \vec{n}|}{|\overrightarrow{B M}| \vec{n} \mid}=\dfrac{5 \sqrt{7}}{14}\).
     

【B组---提高题】

1如图示,三棱锥\(P-ABC\)的底面\(ABC\)是等腰直角三角形,\(∠ACB=90°\),且 \(P A=P B=A B=\sqrt{2}\), \(P C=\sqrt{3}\),则\(PC\)与面\(PAB\)所成角的正弦值等于(  )
image.png
 A. \(\dfrac{1}{3}\) \(\qquad \qquad\) B. \(\dfrac{\sqrt{6}}{3}\) \(\qquad \qquad\) C. \(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{2}}{3}\)
 

2 如图,在正四棱柱\(ABCD-A_1 B_1 C_1 D_1\)中,\(AB=AD=3\),\(AA_1=4\),\(P\)是侧面\(BCC_1 B_1\)内的动点,且\(AP⊥BD_1\),记\(AP\)与平面\(BCC_1 B_1\)所成的角为\(θ\),则\(\tanθ\)的最大值为(  )
image.png
 A. \(\dfrac{4}{3}\) \(\qquad \qquad\) B. \(\dfrac{5}{3}\) \(\qquad \qquad\) C.\(2\)\(\qquad \qquad\) D. \(\dfrac{25}{9}\)
 

3如图①,在\(Rt△ABC\)中,\(B\)为直角,\(AB=BC=6\),\(EF∥BC\),\(AE=2\),沿\(EF\)将\(△AEF\)折起,使 \(\angle A E B=\dfrac{\pi}{3}\),得到如图②的几何体,点\(D\)在线段\(AC\)上.
(1)求证:平面\(AEF⊥\)平面\(ABC\);
(2)若\(AE∥\)平面\(BDF\),求直线\(AF\)与平面\(BDF\)所成角的正弦值.
image.png
 

参考答案

  1. 答案 \(A\)
    解析 \(∵\)三棱锥\(P-ABC\)的底面\(ABC\)是等腰直角三角形,\(∠ACB=90°\),且 \(P A=P B=A B=\sqrt{2}\), \(P C=\sqrt{3}\),
    \(∴\)可以把三棱椎\(P-ABC\)补成棱长为\(1\)的正方体,如图所示.
    如图,以\(A\)为原点建立空间直角坐标系,
    image.png
    则\(A(0,0,0)\),\(B(1,1,0)\),\(C(0,1,0)\),\(P(1,0,1)\).
    \(\overrightarrow{A P}=(1,0,1)\), \(\overrightarrow{A B}=(1,1,0)\), \(\overrightarrow{P C}=(-1,1,-1)\),
    设面\(ABP\)的法向量为\(\vec{m}=(x,y,z)\),
    \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{A P}=x+z=0 \\ \vec{m} \cdot \overrightarrow{A B}=x+y=0 \end{array} \Rightarrow \vec{m}=(1,-1,-1)\right.\).
    \(\cos <\vec{m}, \overrightarrow{P C}>=\dfrac{\overrightarrow{P C} \cdot \vec{m}}{|\vec{m}| \cdot|\overrightarrow{P C}|}=\dfrac{-1}{\sqrt{3} \times \sqrt{3}}=-\dfrac{1}{3}\).
    则\(PC\)与面\(PAB\)所成角的正弦值等于 \(\dfrac{1}{3}\).
    故选:\(A\).

  2. 答案 \(B\)
    解析 以\(D\)为原点,\(DA\)为\(x\)轴,\(DC\)为\(y\)轴,\(DD_1\)为\(z\)轴,建立空间直角坐标系,
    image.png
    设\(P(a,3,c)\),\((0≤a≤3,0≤c≤4)\),
    则\(A(3,0,0)\),\(B(3,3,0)\),\(D_1 (0,0,4)\),
    \(\overrightarrow{A P}=(a-3,3, c)\), \(\overrightarrow{B D}_{1}=(-3,-3,4)\),平面\(BCC_1 B_1\)的法向量\(\vec{n}=(0,1,0)\),
    \(∵AP⊥BD_1\),
    \(\therefore \overrightarrow{A P} \cdot \overrightarrow{B D}=-3(a-3)-9+4 c=0\),解得 \(c=\dfrac{3}{4} a\),
    \(\therefore \overrightarrow{A P}=\left(a-3,3, \dfrac{3}{4} a\right)\),
    \(∵AP\)与平面\(BCC_1 B_1\)所成的角为\(θ\),
    \(\therefore \sin \theta=\dfrac{|\overrightarrow{A P} \cdot \vec{n}|}{|\overrightarrow{A P}| \cdot|\vec{n}|}=\dfrac{3}{\sqrt{(a-3)^{2}+9+\dfrac{9}{16} a^{2}}}=\dfrac{12}{\sqrt[5]{\left(a-\dfrac{48}{25}\right)^{2}+\dfrac{4896}{625}}}\),
    \(∴\)当 \(a=\dfrac{48}{25}\)时,\(\sinθ\)取最大值为 \(\dfrac{5}{\sqrt{34}}\).此时 \(\cos \theta=\sqrt{1-\left(\dfrac{5}{\sqrt{34}}\right)^{2}}=\dfrac{3}{\sqrt{34}}\),
    \(∴\tanθ\)的最大值为: \(\dfrac{\dfrac{5}{\sqrt{34}}}{\dfrac{3}{\sqrt{34}}}=\dfrac{5}{3}\).
    故选:\(B\).
    image.png

  3. 答案 (1)略 (2) \(\frac{\sqrt{6}}{4}\)
    解析 (1)证明:\(AB⊥BC\),\(EF∥BC\),\(∴EF⊥AE\),\(EF⊥BE\),
    又\(EA∩EB=E\),\(∴EF⊥\)平面\(AEB\),则\(EF⊥AB\),
    在\(△AEB\)中,由\(AE=2\),\(EB=4\), \(\angle A E B=\dfrac{\pi}{3}\),
    得 \(A B^{2}=4+16-2 \times 2 \times 4 \times \dfrac{1}{2}=12\).
    \(∴AE^2+AB^2=BE^2\),即\(AE⊥AB\),
    又\(AE∩EF=E\),\(∴AB⊥\)平面\(AEF\),
    而\(AB⊂\)平面\(ABC\),则平面\(AEF⊥\)平面\(ABC\);
    (2)解:以\(A\)为坐标原点,分别以\(AB\),\(AE\)所在直线为\(x,y\)轴,
    以过\(A\)垂直于平面\(AEB\)的直线为\(z\)轴建立如图所示空间直角坐标系.
    则\(A(0,0,0)\),\(E(0,2,0)\), \(B(2 \sqrt{3}, 0,0)\),\(F(0,2,2)\),\(C(2\sqrt{3},0,6)\).
    \(\overrightarrow{A E}=(0,2,0)\), \(\overrightarrow{A F}=(0,2,2)\), \(\overrightarrow{A C}=(2 \sqrt{3}, 0,6)\), \(\overrightarrow{B F}=(-2 \sqrt{3}, 2,2)\).
    设 \(\overrightarrow{A D}=\lambda \overrightarrow{A C}=(2 \sqrt{3} \lambda, 0,6 \lambda)(0<\lambda \leq 1)\).
    则 \(\overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=(2 \sqrt{3} \lambda-2 \sqrt{3}, 0,6 \lambda)\),
    设平面\(BDF\)的一个法向量为\(\vec{n}=(x,y,z)\).
    由 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{B F}=-2 \sqrt{3} x+2 y+2 z=0 \\ \vec{n} \cdot \overrightarrow{B D}=(2 \sqrt{3} \lambda-2 \sqrt{3}) x+6 \lambda z=0 \end{array}\right.\),
    取\(x=1\),得 \(\vec{n}=\left(1, \dfrac{4 \sqrt{3} \lambda-\sqrt{3}}{3 \lambda}, \dfrac{\sqrt{3}-\sqrt{3} \lambda}{3 \lambda}\right)\).
    由\(AE∥\)平面\(BDF\),得 \(\overrightarrow{A E} \cdot \vec{n}=\dfrac{4 \sqrt{3} \lambda-\sqrt{3}}{3 \lambda}=0\),即 \(\lambda=\dfrac{1}{4}\).
    \(\therefore \vec{n}=(1,0, \sqrt{3})\).
    设直线\(AF\)与平面\(BDF\)所成角为\(θ\),
    则 \(\sin \theta=|\cos <\vec{n}, \overrightarrow{A F}>|=\dfrac{|\vec{n} \cdot \overrightarrow{A F}|}{|\vec{n}| \cdot|\overrightarrow{A F}|}=\dfrac{2 \sqrt{3}}{2 \times 2 \sqrt{2}}=\dfrac{\sqrt{6}}{4}\).
    \(∴\)直线\(AF\)与平面\(BDF\)所成角的正弦值为 \(\dfrac{\sqrt{6}}{4}\).
    image.png

【C组---拓展题】

1如图正四面体\(P-ABC\)中,\(E\)在边\(PA\)上,且\(AE=2EP\),\(F\)是\(BC\)边中点,记\(EF\)与平面\(ABC\)、平面\(PAB\)、平面\(PBC\)所成的角分别为\(α,β,γ\),则以下选项正确的是(  )
 A.\(α>β>γ\) \(\qquad \qquad\) B . \(γ>β>α\) \(\qquad \qquad\) C. \(β>α>γ\) \(\qquad \qquad\) D. \(β>γ>α\)
 

2在梯形\(ABCD\)中,\(AB∥CD\), \(\angle B A D=\dfrac{\pi}{3}\),\(AB=2AD=2CD=4\),\(P\)为\(AB\)的中点,线段\(AC\)与\(DP\)交于\(O\)点(如图\(1\)).将\(△ACD\)沿\(AC\)折起到\(△ACD'\)的位置,使得二面角\(AB-AC-D'\)为直二面角(如图\(2\)).
(1)求证:\(BC∥\)平面\(POD'\);
(2)线段\(PD'\)上是否存在点\(Q\),使得\(CQ\)与平面\(BCD'\)所成角的正弦值为 \(\dfrac{\sqrt{6}}{8}\)?若存在,求出 \(\dfrac{P Q}{P D^{\prime}}\)的值;若不存在,请说明理由.
image.png
 
 

参考答案

  1. 答案 \(A\)
    解析 将正四面体\(P-ABC\)补成正方体\(PMAN-DBGC\),设该正方体的棱长为\(6\),
    以点\(P\)为坐标原点,\(PM、PN、PD\)所在直线分别为\(x,y,z\)轴建立如下图所示的空间直角坐标系,
    image.png
    则\(P(0,0,0)\),\(A(6,6,0)\),\(B(6,0,6)\),\(C(0,6,6)\),\(F(3,3,6)\),\(E(2,2,0)\),
    设平面\(ABC\)的法向量为 \(\vec{a}=\left(x_{1}, y_{1}, z_{1}\right)\) , \(\overrightarrow{A B}=(0,-6,6)\), \(\overrightarrow{A C}=(-6,0,6)\),
    由 \(\left\{\begin{array}{l} \vec{a} \cdot \overrightarrow{A B}=-6 y_{1}+6 z_{1}=0 \\ \vec{a} \cdot \overrightarrow{A C}=-6 x_{1}+6 z_{1}=0 \end{array}\right.\),取\(z_1=1\),可得\(\vec{a}=(1,1,1)\),
    设平面\(PAB\)的法向量为\(\vec{b}=(x_2,y_2,z_2 )\), \(\overrightarrow{P A}=(6,6,0)\), \(\overrightarrow{P B}=(6,0,6)\),
    由 \(\left\{\begin{array}{l} \vec{b} \cdot \overrightarrow{P A}=6 x_{2}+6 y_{2}=0 \\ \vec{b} \cdot \overrightarrow{P B}-6 x_{2}+6 z_{2}=0 \end{array}\right.\),取\(x_2=1\),可得\(\vec{b}=(1,-1,-1)\),
    设平面\(PBC\)的法向量为\(\vec{c}=(x_3,y_3,z_3)\) , \(\overrightarrow{P B}=(6,0,6)\), \(\overrightarrow{P C}=(0,6,6)\),
    由 \(\left\{\begin{array}{l} \vec{c} \cdot \overrightarrow{P B}=6 x_{3}+6 z_{3}=0 \\ \vec{c} \cdot \overrightarrow{P C}=6 y_{3}+6 z_{3}=0 \end{array}\right.\),取\(z_3=-1\),可得\(\vec{c}=(1,1,-1)\),
    所以 \(\overrightarrow{E F}=(1,1,6)\), \(\sin \alpha=\dfrac{|\overrightarrow{E F} \cdot \vec{a}|}{|\overrightarrow{E F}| \cdot|\vec{a}|}=\dfrac{8}{\sqrt{38} \times \sqrt{3}}=\dfrac{8}{\sqrt{114}}\),
    所以 \(\sin \beta=\dfrac{|\overrightarrow{E F} \cdot \vec{b}|}{|\overrightarrow{E F}| \cdot|\vec{b}|}=\dfrac{6}{\sqrt{38} \times \sqrt{3}}=\dfrac{6}{\sqrt{114}}\), \(\sin \gamma=\dfrac{|\overrightarrow{E F} \cdot \vec{c}|}{|\overrightarrow{E F}| \cdot|\vec{c}|}=\dfrac{4}{\sqrt{38} \times \sqrt{3}}=\dfrac{4}{\sqrt{114}}\),
    故选:\(A\).

  2. 答案 (1)略 (2) 线段\(PD'\)上存在点\(Q\),且 \(\dfrac{P Q}{P D^{\prime}}=\dfrac{1}{3}\).
    解析 (1)证明:因为在梯形\(ABCD\)中,\(AB∥CD\),\(AB=2CD=4\),\(P\)为\(AB\)的中点,
    所以\(CD∥AP\),\(CD=AP\),所以四边形\(APCD\)为平行四边形,
    因为线段\(AC\)与\(DP\)交于\(O\)点,
    所以\(O\)为线段\(AC\)的中点,
    所以\(△ABC\)中\(OP∥BC\),
    因为\(OP⊂\)平面\(POD'\),\(BC⊄\)平面\(POD'\),所以\(BC∥\)平面\(POD'\).
    (2)解:平行四边形\(APCD\)中,\(AP=AD=2\),
    所以四边形\(APCD\)是菱形,\(AC⊥DP\),垂足为\(O\),
    所以\(AC⊥OD'\),\(AC⊥OP\),
    因为\(OD'⊂\)平面\(ACD'\),\(OP⊂\)平面\(ACB\),
    所以\(∠D'OP\)是二面角\(B-AC-D'\)的平面角,
    因为二面角\(B-AC-D'\)为直二面角,
    所以 \(\angle D^{\prime} O P=\dfrac{\pi}{2}\),即\(OP⊥OD'\).
    可以如图建立空间直角坐标系\(O-xyz\),其中\(O(0,0,0)\),
    因为在菱形\(APCD\)中, \(\angle B A D=\dfrac{\pi}{3}\),
    所以\(OD-OP-1\), \(O A=O C=\sqrt{3}\).
    所以 \(B(-\sqrt{3}, 2,0)\), \(P(0,1,0)\), \(C(-\sqrt{3}, 0,0)\), \(D^{\prime}(0,0,1)\)
    所以 \(\overrightarrow{B D^{\prime}}=(\sqrt{3},-2,1)\), \(\overrightarrow{C B}=(0,2,0)\).
    设\(\vec{n} =(x,y,z)\)为平面\(BCD'\)的法向量,
    因为 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{C B}=0 \\ \vec{n} \cdot \overrightarrow{B D^{\prime}}=0 \end{array}\right.\),
    所以 \(\left\{\begin{array}{l} 2 y=0 \\ \sqrt{3} x-2 y+z=0 \end{array}\right.\).取\(x=1\),得 \(\vec{n}=(1,0,-\sqrt{3})\),
    线段\(PD'\)上存在点\(Q\)使得\(CQ\)与平面\(BCD'\)所成角的正弦值为√6/8,
    设 \(\overrightarrow{P Q}=\lambda \overrightarrow{P D^{\prime}}(0 \leq \lambda \leq 1)\),
    因为 \(\overrightarrow{C P}=(\sqrt{3}, 1,0)\), \(\overrightarrow{P D^{\prime}}=(0,-1,1)\),
    所以 \(\overrightarrow{C Q}=\overrightarrow{C P}+\overrightarrow{P Q}=\overrightarrow{C P}+\lambda \overrightarrow{P D^{\prime}}=(\sqrt{3}, 1-\lambda, \lambda)\).
    因为 \(\cos <\overrightarrow{C Q}, \vec{n}\rangle=\dfrac{\overrightarrow{C Q} \cdot \vec{n}}{|\overrightarrow{C Q}| \cdot|\vec{n}|}=\dfrac{\sqrt{3}(1-\lambda)}{2 \sqrt{2 \lambda^{2}-2 \lambda+4}}=\dfrac{\sqrt{6}}{8}\),
    所以\(3λ^2-7λ+2=0\),因为\(0≤λ≤1\),所以 \(\lambda=\dfrac{1}{3}\).
    所以线段\(PD'\)上存在点\(Q\),且 \(\dfrac{P Q}{P D r}=\dfrac{1}{3}\),使得\(CQ\)与平面\(BCD'\)所成角的正弦值为 \(\dfrac{\sqrt{6}}{8}\).

标签:1.4,overrightarrow,异面,cdot,dfrac,sqrt,角和线,vec,平面
来源: https://www.cnblogs.com/zhgmaths/p/16651106.html

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