1.4.2 空间向量的应用---所成角
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模块导图
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1642650116223-23060532-57d6-444d-a8b6-8f4f3f257b15.png)
知识剖析
1 求空间角
\((1)\)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
\({\color{Red}{解释}}\)
① 向量\(\overrightarrow{A C}\),\(\overrightarrow{B D}\)所成角\(\langle\overrightarrow{A C}, \overrightarrow{B D}\rangle\)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\(\langle\overrightarrow{A C}, \overrightarrow{B D}\rangle\)与\(θ\)的关系相等或互补;
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230519131759692-85933006.png)
(2)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\(α\)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\(\sin \theta=|\cos \alpha|=\dfrac{|\vec{a} \cdot \vec{n}|}{|\vec{a}||\vec{n}|}\).
\({\color{Red}{解释}}\)
当\(\theta=\dfrac{\pi}{2}-\alpha\)时,\(\sin \theta=\cos \alpha\);当\(\theta=\alpha-\dfrac{\pi}{2}\)时,\(\sin \theta=-\cos \alpha\);
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230519131856339-1496909058.png )
(3)求二面角\(α-l-β\)
二面角的平面角是指在二面角\(α-l-β\)的棱上任取一点\(O\),分别在两个半平面内作射线
\(AO⊥ l\),\(BO⊥ l\),则\(∠AOB\)为二面角\(α-l-β\)的平面角,二面角的取值范围是\(\left[0, \dfrac{\pi}{2}\right]\).
如图:
求法:设二面角\(α-l-β\)的两个半平面的法向量分别为\(\vec{m}\),\(\vec{n}\),
再设\(\vec{m}\),\(\vec{n}\)的夹角为\(φ\),二面角\(α-l-β\)的平面角为\(θ\),则二面角\(θ\)为\(φ\)或\(π-φ\),
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230519132000205-1071886239.png )
经典例题
【题型一】求异面直线所成的角
【典题1】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\(\underline{\quad \quad}\).
【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\(SA=SB=SC=2\),(题中没有确定线段,可设任意长度)
则\(S(0 ,0 ,0)\),\(B(0 ,2 ,0)\),\(M(0 ,1 ,1)\),\(N(1 ,0 ,0)\),
则\(\overrightarrow{S M}=(0,1,1)\),\(\overrightarrow{B N}=(1,-2,0)\),
\(\therefore \cos <\overrightarrow{S M}, \overrightarrow{B N}>=\dfrac{|\overrightarrow{S M} \cdot \overrightarrow{B N}|}{|\overrightarrow{S M}| \cdot|\overrightarrow{B N}|}=\dfrac{2}{\sqrt{2} \cdot \sqrt{5}}=\dfrac{\sqrt{10}}{5}\).
\(∴\)异面直线\(SM\)与\(BN\)所成角的余弦值为\(\dfrac{\sqrt{10}}{5}\).
【点拨】
向量法求异面直线\(SM\)与\(BN\)所成角的步骤
① 建系求出涉及的\(S\)、\(M\)、\(B\)、\(N\)四点坐标;
② 求\(\overrightarrow{S M}\),\(\overrightarrow{B N}\)得到\(\mid \cos <\overrightarrow{S M}, \overrightarrow{B N}>=\dfrac{|\overrightarrow{S M} \cdot \overrightarrow{B N}|}{|\overrightarrow{S M}| \cdot|\overrightarrow{B N}|}\);
③ 由公式\(\cos \theta=|\cos <\overrightarrow{S M}, \overrightarrow{B N}>|\)得到异面直线\(SM\)与\(BN\)所成角.
【典题2】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.
【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,
则\(A(\sqrt{2}, 0,0)\),\(B(0, \sqrt{2}, 0)\),\(C(-\sqrt{2}, 0,0)\),\(D(0,-\sqrt{2}, 0)\),
\(V(0,0,4)\),\(E\left(\dfrac{\sqrt{2}}{2}, 0,2\right)\),\(F\left(-\dfrac{\sqrt{2}}{2}, 0,2\right)\),
\(\therefore \overrightarrow{B E}=\left(\dfrac{\sqrt{2}}{2},-\sqrt{2}, 2\right)\),\(\overrightarrow{D F}=\left(-\dfrac{\sqrt{2}}{2}, \sqrt{2}, 2\right)\)
\(\cos <\overrightarrow{B E}, \overrightarrow{D F}>=\dfrac{\overrightarrow{B E} \cdot \overrightarrow{D F}}{|\overrightarrow{B E}| \cdot|\overrightarrow{D F}|}\)\(=\dfrac{-\dfrac{1}{2}-2+4}{\sqrt{\dfrac{1}{2}+2+4} \cdot \sqrt{\dfrac{1}{2}+2+4}}=-\dfrac{3}{13}\),
设向量\(BE\)和\(DF\)成角为\(θ\),\(\therefore \cos \theta=|\cos <\overrightarrow{B E}, \overrightarrow{D F}>|=\dfrac{3}{13}\),
\(\therefore \sin \theta=\sqrt{1-\left(\dfrac{3}{13}\right)^{2}}=\dfrac{4 \sqrt{10}}{13}\),
\(\therefore \tan \theta=\dfrac{\sin \theta}{\cos \theta}=\dfrac{4 \sqrt{10}}{3}\).
\(∴\)异面直线\(BE\)与\(DF\)所成角的正切值为\(\dfrac{4 \sqrt{10}}{3}\).
【点拨】向量\(\overrightarrow{B E}\),\(\overrightarrow{D F}\)所成角\(<\overrightarrow{B E}, \overrightarrow{D F}>\)是个钝角,而异面直线BE与DF所成角是锐角,它们之间是互补,所以\(\cos \theta=|\cos <\overrightarrow{B E}, \overrightarrow{D F}>|=\dfrac{4 \sqrt{10}}{13}\).
巩固练习
1 (★) 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\(\underline{\quad \quad}\).
答案
1.\(\dfrac{\sqrt{30}}{10}\)
【题型二】求线面角
【典题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\(\underline{\quad \quad}\) .
【解析】\(∵\)三棱椎\(P-ABC\)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,以\(A\)为原点建立空间直角坐标系,
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\(\vec{m}=(x, y, z)\),
\(\left\{\begin{array}{l}
\vec{m} \cdot \overrightarrow{A P}=x+z=0 \\
\vec{m} \cdot \overrightarrow{A B}=x+y=0
\end{array} \Rightarrow \vec{m}=(1,-1,-1)\right.\).
\(\therefore \cos <\vec{m}, \overrightarrow{P C}>=\dfrac{\overrightarrow{P C} \cdot \vec{m}}{|\vec{m}| \cdot|\overrightarrow{P C}|}=\dfrac{-1}{\sqrt{3} \times \sqrt{3}}=-\dfrac{1}{3}\).
则\(PC\)与面\(PAB\)所成角的正弦值等于\(\dfrac{1}{3}\).
【点拨】
① 本题根据“墙角模型”巧妙的构造一个长方体进而建系;
② 向量法求直线\(PC\)与面\(PAB\)所成角的步骤
(1) 求直线\(PC\)的方向向量\(\overrightarrow{P C}\)和平面PAB的法向量\(\vec{m}\);
(2) 求\(\cos <\vec{m}, \overrightarrow{P C}>\);
(3) 求\(PC\)与面\(PAB\)所成角的正弦值\(\sin \theta=|\cos <\vec{m}, \overrightarrow{P C}>|\).
【典题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.
【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\(\dfrac{\sqrt{6}}{8}\),
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\(\overrightarrow{C Q}=\overrightarrow{C P}+\overrightarrow{P Q}=\overrightarrow{C P}+\lambda \overrightarrow{P D^{\prime}}=(\sqrt{3}, 1-\lambda, \lambda)\).
因为\(\cos \langle\overrightarrow{C Q}, \vec{n}\rangle=\dfrac{\overrightarrow{C Q} \cdot \vec{n}}{|\overrightarrow{C Q}| \cdot|\vec{n}|}=\dfrac{\sqrt{3}(1-\lambda)}{2 \sqrt{2 \lambda^{2}-2 \lambda+4}}=\dfrac{\sqrt{6}}{8}\),
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\(\dfrac{P Q}{P D^\prime}=\dfrac{1}{3}\),使得\(CQ\)与平面\(BCD'\)所成角的正弦值为\(\dfrac{\sqrt{6}}{8}\).
【点拨】
① 本题属于折叠问题,需要通过平几知识点确定各个量之间的关系,明确哪些量在折叠前后是否发生变化;
② 本题第二问已知直线\(CQ\)与平面\(BCD'\)所成角正弦值为\(\dfrac{\sqrt{6}}{8}\),由线面角公式\(\sin \theta=|\cos \alpha|=\dfrac{|\vec{a} \cdot \vec{n}|}{|\vec{a}||\vec{n}|}\),可知\(|\cos <\overrightarrow{C Q}, \vec{n}>|=\dfrac{|\overrightarrow{C Q} \cdot \vec{n}|}{|\overrightarrow{C Q}| \cdot|\vec{n}|}=\dfrac{\sqrt{6}}{8}\),先求出平面\(BCD'\)的法向量\(\vec{n}=(1,0,-\sqrt{3})\),再求\(\overrightarrow{C Q}\),那关键点在于\(Q\)的位置;
③ 求向量\(\overrightarrow{C Q}\)的坐标,最直接的想法是设\(Q(0 ,y ,z)\),这里有两种方法提供
(1) 几何法
由图可知\(\dfrac{1-y}{z}=\dfrac{O P}{O D^{\prime}}=1 \Rightarrow z=1-y\),\(∴Q(0 ,y ,1-y)\),
即\(\overrightarrow{C Q}=(\sqrt{3}, y, 1-y)\).
(2) 代数法
设\(\overrightarrow{P Q}=\lambda \overrightarrow{P D^{\prime}}\),则\((0, y-1, z)=\lambda(0,-1,1) \Rightarrow z=1-y\),
\(∴Q(0 ,y ,1-y)\),即\(\overrightarrow{C Q}=(\sqrt{3}, y, 1-y)\).
但本题给到的解法,并没求点\(Q\)的坐标,而是先设\(\overrightarrow{P Q}=\lambda \overrightarrow{P D^{\prime}}\)再由“首尾相接法”\(\overrightarrow{C Q}=\overrightarrow{C P}+\overrightarrow{P Q}\)得到\(\overrightarrow{C Q}=(\sqrt{3}, 1-\lambda, \lambda)\),这样来得也很简单.
巩固练习
1(★★)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\(\underline{\quad \quad}\) .
2(★★★)四棱柱\(ABCD-A_1 B_1 C_1 D_1\)中,底面\(ABCD\)是正方形,\(∠A_1 AB=∠A_1 AD=60°\),\(AA_1=AB\).
(1)求证:平面\(A_1 BD⊥\)平面\(ABCD\);
(2)求\(BD_1\)与平面\(ABB_1 A_1\)所成角的正弦值.
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.
答案
1.\(\dfrac{5}{3}\)
2.\((1)\)略\((2)\dfrac{\sqrt{2}}{3}\)
3. 不存在\(M\)
【题型三】求二面角
【典题1】在底面为锐角三角形的直三棱柱\(ABC-A_1 B_1 C_1\)中,\(D\)是棱\(BC\)的中点,记直线\(B_1 D\)与直线\(AC\)所成角为\(θ_1\),直线\(B_1 D\)与平面\(A_1 B_1 C_1\)所成角为\(θ_2\),二面角\(C_1-A_1 B_1-D\)的平面角为\(θ_3\),则( )
A.\(θ_2<θ_1\),\(θ_2<θ_3\) \(\qquad \qquad \qquad \qquad\) B.\(θ_2>θ_1\),\(θ_2<θ_3\)
C.\(θ_2<θ_1\),\(θ_2>θ_3\) \(\qquad \qquad \qquad \qquad\) D.\(θ_2>θ_1\),\(θ_2>θ_3\)
【解析】由选项可知,角\(θ_1\)与\(θ_2\),\(θ_2\)与\(θ_3\)的大小确定,且三棱柱的底面为锐角三角形.
\(∴\)设三棱柱\(ABC-A_1 B_1 C_1\)是棱长均为\(2\)的正三棱柱,
\({\color{Red}{(加强条件处理,选择题的作法) }}\)
如图,以\(A\)为原点,在平面\(ABC\)中,过\(A\)作\(AC\)的垂线为\(x\)轴,\(AC\)为\(y\)轴,\(AA_1\)为\(z\)轴,建立空间直角坐标系,
则\(A(0 ,0 ,0)\),\(A_1 (0 ,0 ,2)\),\(B_{1}(\sqrt{3}, 1,2)\),\(C(0,2,0)\),\(D\left(\dfrac{\sqrt{3}}{2}, \dfrac{3}{2}, 0\right)\),
\(\therefore \overrightarrow{A C}=(0,2,0)\),\(\overrightarrow{B_{1} D}=\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2},-2\right)\),\(\overrightarrow{A_{1} B_{1}}=(\sqrt{3}, 1,0)\),
\(∵\)直线\(B_1 D\)与直线\(AC\)所成的角为\(θ_1\),
\(\therefore \cos \theta_{1}=\left|\cos <\overrightarrow{B_{1} D}, \overrightarrow{A C}>\right|=\dfrac{\left|\overrightarrow{B_{1} D} \cdot \overrightarrow{A C}\right|}{\left|\overrightarrow{B_{1} D}\right| \cdot|\overrightarrow{A C}|}=\dfrac{1}{2 \cdot \sqrt{5}}=\dfrac{\sqrt{5}}{10}\),
\(∵\)直线\(B_1 D\)与平面\(A_1 B_1 C_1\)所成的角为\(θ_2\),平面\(A_1 B_1 C_1\)的法向量\(\vec{n}=(0,0,1)\),
\(\therefore \sin \theta_{2}=\dfrac{\left|\overrightarrow{B_{1} D} \cdot \vec{n}\right|}{\left|\overrightarrow{B_{1} D}\right| \cdot|\vec{n}|}=\dfrac{|-2|}{\sqrt{5} \cdot 1}=\dfrac{2}{\sqrt{5}}\),\(\therefore \cos \theta_{2}=\sqrt{1-\left(\dfrac{2}{\sqrt{5}}\right)^{2}}=\dfrac{\sqrt{5}}{5}\)
设平面\(A_1 B_1 D\)的法向量\(\vec{m}=(x, y, z)\),
由\(\left\{\begin{array}{l}
\vec{m} \cdot \overrightarrow{A_{1} B_{1}}=\sqrt{3} x+y=0 \\
\vec{m} \cdot \overrightarrow{B_{1} D}=-\dfrac{\sqrt{3}}{2} x+\dfrac{1}{2} y-2 z=0
\end{array}\right.\),
取\(x=\sqrt{3}\),得\(\vec{m}=\left(\sqrt{3},-3,-\dfrac{3}{2}\right)\),
\(\therefore \cos <\vec{m}, \vec{n}>=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=-\dfrac{\sqrt{57}}{19}\),
\(∵\)二面角\(C_1-A_1 B_1-D\)的平面角为\(θ_3\),由于\(θ_3\)是锐角,
\(\therefore \cos \theta_{3}=\dfrac{\sqrt{57}}{19}\).
\(∵θ_1\),\(θ_2\)与\(θ_3\)均为锐角,
结合余弦函数\(y=\cosθ\)在\(\left[0, \dfrac{\pi}{2}\right]\)上为减函数,则\(θ_2<θ_1\),\(θ_2<θ_3\),
故选:\(A\).
【点拨】
① 本题采取了“小题小作”,假设图形是正三棱柱,得到\(θ_2<θ_1\),\(θ_2<θ_3\),那可排除\(B\)、\(C\)、\(D\),故答案就是\(A\);
② 求二面角\(C_1-A_1 B_1-D\)的步骤
(1) 求出两个平面的法向量:平面\(A_1 B_1 D\)的法向量\(\vec{m}\)和平面\(A_1 B_1 C_1\)的法向量\(\vec{n}\);
(2) 求出\(\cos <\vec{m}, \vec{n}>\);
(3) 由图确定\(θ_3\)是锐角还是钝角求出\(\cosθ_3\).
【典题2】如图,四棱锥\(P-ABCD\)中,侧面\(PAB⊥\)底面\(ABCD\),\(CD∥AB\),\(AD⊥AB\),\(AD=AB=2\),\(C F=\dfrac{1}{3} C D=\dfrac{1}{2}\),\(P A=P B=\sqrt{5}\),\(E\),\(N\)分别为\(AB\),\(PB\)的中点.
(1)求证:\(CN∥\)平面\(PEF\);
(2)求二面角\(N-CD-A\)的余弦值;
(3)在线段\(BC\)上是否存在一点\(Q\),使\(NQ\)与平面\(PEF\)所成角的正弦值为\(\dfrac{\sqrt{14}}{14}\),若存在求出\(BQ\)的长,若不存在说明理由.
【解析】(Ⅰ)证明:取\(PE\)中点\(G\),连接\(GN\),\(FN\),\(GN∥BE\),\(G N=\dfrac{1}{2}\),
即\(GN∥CF\),\(GN=CF\),所以\(GNCF\)为平行四边形,所以\(CN∥FG\),
因为\(CN⊄\)平面\(PEF\),\(FG⊂\)平面\(PEF\),所以\(CN∥\)平面\(PEF\).
(Ⅱ)解:因为\(PA=PB\),\(E\)为\(AB\)的中点,所以\(PE⊥AB\),
因为\(AD=AB=2\),\(C F=\dfrac{1}{3} C D=\dfrac{1}{2}\),
所以\(C D=\dfrac{3}{2}\),\(DF=AE=1\),所以\(EF⊥AB\),
又因为侧面\(PAB⊥\)底面\(ABCD\),且它们的交线为\(AB\),
所以\(PE⊥\)平面\(ABCD\),
又\(CD∥AB\),\(AD⊥AB\),
分别以\(EB\),\(EF\),\(EP\)为\(x\),\(y\),\(z\)轴建立空间直角坐标系.
\(P(0 ,0 ,2)\),\(C\left(\dfrac{1}{2}, 2,0\right)\),\(D(-1 ,2 ,0)\),\(A(-1 ,0 ,0)\),\(B(1 ,0 ,0)\),\(N\left(\dfrac{1}{2}, 0,1\right)\),
平面\(CDA\)的法向量\(\vec{m}=(0,0,1)\),
\(\because \overrightarrow{C D}=\left(-\dfrac{3}{2}, 0,0\right)\),\(\overrightarrow{C N}=(0,-2,1)\),
设平面\(CDN\)的法向量\(\vec{n}=(x, y, z)\),
则\(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{C D}=0 \\
\vec{n} \cdot \overrightarrow{C N}=0
\end{array}\right.\),即\(\left\{\begin{array}{l}
\dfrac{3}{2} x=0 \\
-2 y+z=0
\end{array}\right.\)
令\(y=1\),得\(\vec{n}=(0,1,2)\).
所以\(\cos <\vec{m}, \vec{n}>=\dfrac{2}{\sqrt{5}}=\dfrac{2 \sqrt{5}}{5}\),
由图可知二面角\(N-CD-A\)为锐角,
所以二面角\(N-CD-A\)的余弦值为\(\dfrac{2 \sqrt{5}}{5}\).
(Ⅲ)解:设\(\overrightarrow{B Q}=\lambda \overrightarrow{B C}=\left(-\dfrac{1}{2} \lambda, 2 \lambda, 0\right)\),\(\lambda \in[0,1]\),
\(\therefore Q\left(-\dfrac{1}{2} \lambda+1,2 \lambda, 0\right)\),\(\overrightarrow{N Q}=\left(-\dfrac{1}{2} \lambda+\dfrac{1}{2}, 2 \lambda,-1\right)\),
平面\(PEF\)的法向量\(\vec{p}=(1,0,0)\),
设\(NQ\)与平面\(PEF\)所成角为\(θ\),则\(\sin \theta=\dfrac{\sqrt{14}}{14}\),
\(\therefore|\cos <\overrightarrow{N Q}, \vec{p}>|=\dfrac{\sqrt{14}}{14}\)\(\Rightarrow \dfrac{\left|-\dfrac{1}{2} \lambda+\dfrac{1}{2}\right|}{\sqrt{\left(-\dfrac{1}{2} \lambda+\dfrac{1}{2}\right)^{2}+(2 \lambda)^{2}+1}}=\dfrac{\sqrt{14}}{14}\),
解得\(\lambda=\dfrac{1}{3}\)或\(λ=-9\)(舍),
\(\therefore Q\left(\dfrac{5}{6}, \dfrac{2}{3}, 0\right)\), 又由\(B(1 ,0 ,0)\),
\(\therefore \overrightarrow{B Q}=\left(-\dfrac{1}{6}, \dfrac{2}{3}, 0\right)\)
\(\therefore B Q=|\overrightarrow{B Q}|=\dfrac{\sqrt{17}}{6}\).
【点拨】本题的难点在于由平几知识点确定垂直关系建立直角坐标系,二面角的求法和线面角的运用按照一般的解题套路来就可以,第三问关于\(Q\)设元的常见方法多消化下!
【典题3】如图所示的几何体中,四边形\(ABCD\)是菱形,\(ADNM\)是矩形,\(ND⊥\)平面\(ABCD\),\(∠DAB=60°\),\(AD=2\),\(AM=1\),\(E\)为\(AB\)的中点.
(1)求证:\(NA∥\)平面\(MEC\);
(2)求直线\(MB\)与平面\(MEC\)所成角的正弦值;
(3)设\(P\)为线段\(AM\)上的动点,二面角\(P-EC-D\)的平面角的大小为\(30°\),求线段\(AP\)的长.
【解析】由题意可知四边形\(ABCD\)为菱形,\(E\)为\(AB\)的中点,\(∠DAB=60°\),\(∴DE⊥AB\).
又\(ND⊥\)平面\(ABCD\),以\(D\)为原点建立空间直角坐标系,
则\(A(\sqrt{3},-1,0)\),\(B(\sqrt{3}, 1,0)\),\(C(0 ,2 ,0)\),\(D(0 ,0 ,0)\)
\(E(\sqrt{3}, 0,0)\),\(M(\sqrt{3},-1,1)\),\(N(0 ,0 ,1)\).
(Ⅰ)证明:由题意\(\overrightarrow{M E}=(0,1,-1)\),\(\overrightarrow{M C}=(-\sqrt{3}, 3,-1)\),
设\(\vec{n}=(x, y, z)\)为平面\(MEC\)的法向量.
所以\(\left\{\begin{array} { l }
{ \vec { n } \cdot \vec { M E } = 0 } \\
{ \vec { n } \cdot \vec { M C } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y-z=0 \\
-\sqrt{3} x+3 y-z=0
\end{array}\right.\right.\),
令\(y=\sqrt{3}\), 可得\(\vec{n}=(2, \sqrt{3}, \sqrt{3})\).
又\(\overrightarrow{N A}=(\sqrt{3},-1,-1)\).
\(\therefore \overrightarrow{N A} \cdot \vec{n}=2 \sqrt{3}-\sqrt{3}-\sqrt{3}=0\).
\(∵NA⊄\)平面\(MEC\),\(∴NA∥\)平面\(MEC\).
(Ⅱ)可得\(\overrightarrow{M B}=(0,2,-1)\),平面\(MEC\)的法向量为\(\vec{n}=(2, \sqrt{3}, \sqrt{3})\),
\(\therefore \cos <\vec{n}, \overrightarrow{M B}>=\dfrac{\overrightarrow{M B} \cdot \vec{n}}{|\vec{n}| \cdot|\overrightarrow{M B}|}=\dfrac{\sqrt{6}}{10}\).
\(∴\)直线\(MB\)与平面\(MEC\)所成角的正弦值为\(\dfrac{\sqrt{6}}{10}\);
(Ⅲ)设\(P(\sqrt{3},-1, h)\),\(h∈[0 ,1]\),
\(\overrightarrow{E C}=(-\sqrt{3}, 2,0)\),\(\overrightarrow{E P}=(0,-1, h)\),
设\(\vec{m}=(x, y, z)\)为平面\(PEC\)的法向量,
则\(\left\{\begin{array}{l}
\vec{m} \cdot \overrightarrow{E P}=-y+h z=0 \\
\vec{m} \cdot \overrightarrow{E C}=-\sqrt{3} x+2 y=0
\end{array}\right.\),
令\(z=\sqrt{3}\),可得\(\vec{m}=(2 h, \sqrt{3} h, \sqrt{3})\),
又\(\overrightarrow{D N}=(0,0,1)\)是平面\(DEC\)的法向量.
\(\therefore \cos <\vec{m}, \overrightarrow{D N}>=\dfrac{\overrightarrow{D N} \cdot \vec{m}}{|\vec{m}||\overrightarrow{D N}|}=\dfrac{\sqrt{3}}{\sqrt{4 h^{2}+3 h^{2}+3}}\),
又二面角\(P-EC-D\)的平面角的大小为\(30°\),
\(\therefore \dfrac{\sqrt{3}}{\sqrt{4 h^{2}+3 h^{2}+3}}=\dfrac{\sqrt{3}}{2},\),解得\(h=\dfrac{\sqrt{7}}{7}\),
\(∴\)线段\(AP\)的长为\(\dfrac{\sqrt{7}}{7}\).
【点拨】
① 第一问也可用非向量法求解:连接\(NB\)交\(MC\)于\(O\),易得\(MNCB\)是平行四边形,则点\(O\)是\(NB\)的中点,所以\(OE//AN\),则\(NA∥\)平面\(MEC\);
② 作立体几何题是否都用“空间向量法”去思考呢?类似本题有\(3\)问,拿到题目时要把全部内容审完,把\(3\)个问题作个整体的思考,较容易发现\(2\)、\(3\)问用向量法较为容易,而发现第\(1\)、\(2\)问均与平面\(MEC\)的法向量有关,则第1问就开始利用向量法求解;若你觉得第\(2\)、\(3\)问你没有思路,选择放弃它们,那第\(1\)问“非向量法”在思考量和时间上都来得更“实惠”些;
③ 第\(2\)问中的线面角和第3问的二面角均很难确定具体位置,故用空间向量法更容易.
巩固练习
1(★★)在正方体\(ABCD-A_1 B_1 C_1 D_1\)中,平面\(A_1 BD\)与平面\(ABCD\)所成二面角的正弦值为( )
A.\(\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\)B.\(\dfrac{\sqrt{2}}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{\sqrt{6}}{3}\)\(\qquad \qquad \qquad \qquad\) D.\(\dfrac{1}{3}\)
2(★★★)如图.正四面体\(ABCD\)的顶点\(A\),\(B\),\(C\)分别在两两垂直的三条射线\(OX\),\(OY\),\(OZ\)上,则在下列命题中,错误的为( )
A.\(O-ABC\)是正三棱锥
B.二面角\(D-OB-A\)的平面角为\(\dfrac{\pi}{3}\)
C.直线\(AD\)与直线\(OB\)所成角为\(\dfrac{\pi}{4}\)
D.直线\(OD⊥\)平面\(ABC\)
3(★★★)如图,四棱锥\(P-ABCD\)中,底面\(ABCD\)为矩形,\(PA⊥\)平面\(ABCD\),\(E\)为\(PD\)的中点.
(1)证明:\(PB∥\)平面\(AEC\);
(2)若\(AB=1\),\(AD=2\),\(AP=2\),求二面角\(D-AE-C\)的平面角的余弦值.
4(★★★★)在如图所示的几何体中,四边形\(ABCD\)是正方形,四边形\(ADPQ\)是梯形,\(PD∥QA\),\(\angle P D A=\dfrac{\pi}{2}\),平面\(ADPQ⊥\)平面\(ABCD\),且\(AD=PD=2QA=2\).
(1)求证:\(QB∥\)平面\(PDC\);
(2)求二面角\(C-PB-Q\)的大小;
(3)已知点\(H\)在棱\(PD\)上,且异面直线\(AH\)与\(PB\)所成角的余弦值为\(\dfrac{7 \sqrt{3}}{15}\),求线段\(DH\)的长.
5(★★★★)四棱锥\(P-ABCD\)中,\(PA⊥\)平面\(ABCD\),四边形\(ABCD\)是矩形,且\(PA=AB=2\),\(AD=3\),\(E\)是线段\(BC\)上的动点,\(F\)是线段\(PE\)的中点.
(1)求证:\(PB⊥\)平面\(ADF\);
(2)若直线\(DE\)与平面\(ADF\)所成角为\(30°\),
①求线段\(CE\)的长;\(\qquad \qquad\)②求二面角\(P-ED-A\)的余弦值.
答案
1.\(C\)
2.\(B\)
3.\((1)\)略\(\text { (2) } \dfrac{\sqrt{6}}{3}\)
4.\((1)\)略\(\text { (2) } \dfrac{5 \pi}{6}\)\(\text { (3) } \dfrac{3}{2}\)
5.\((1)\)略 \((2)\) ①\(2\) ②\(\dfrac{3 \sqrt{17}}{17}\)