1.4.2(2) 用空间向量研究平面间的夹角
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基础知识
二面角
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\([0 ,π]\).
如图:
平面α与平面β的夹角
平面\(α\)与平面\(β\)相交,形成四个二面角,我们把这四个二面角不大于\(90^o\)的二面角称为平面\(α\)与平面\(β\)的夹角.
空间向量求平面α与平面β的夹角
求法:设平面\(α\)与平面\(β\)的法向量分别为\(\vec{m}\) ,\(\vec{n}\),
再设\(\vec{m}\),\(\vec{n}\)的夹角为\(φ\),平面\(α\)与平面\(β\)的平面角为\(θ\),则\(θ\)为\(φ\)或\(π-φ\) ,
则 \(\cos \theta=|\cos \varphi|=\left|\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}||\vec{n}|}\right|=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}||\vec{n}|}\) .
(与线面所成角的情况一样,均由法向量的方向导致两种情况的出现)
【例1】在正方体\(ABCD-A'B'C'D'\)中,易得平面\(ACD'\)与平面\(ABC\)的法向量分别是 \(\overrightarrow{D B^{\prime}}\)、 \(\overrightarrow{B B^{\prime}}\),其夹角是\(φ\),二面角\(D'-AC-B\)为θ,平面\(ACD'\)与平面\(ABC\)的夹角\(α\),判断三个角之间的关系!
解析 两个法向量的 \(\overrightarrow{D B^{\prime}}\)、 \(\overrightarrow{B B^{\prime}}\)的夹角\(φ=∠DB'B\),二面角\(D'-AC-B\)为\(θ=∠D'OB\)是个钝角,则平面\(ACD'\)与平面\(ABC\)的夹角\(α=π-θ\),而\(φ=π-θ\).所以三角关系是\(α=φ=π-θ\).
【例2】两平面的法向量分别为 \(\vec{m}=(0,1,0)\), \(\vec{n}=(0,1,1)\),则两平面的夹角为\(\underline{\quad \quad}\) .
解析 两平面的法向量分别为 \(\vec{m}=(0,1,0)\), \(\vec{n}=(0,1,1)\),
所以 \(\cos \langle\vec{m}, \vec{n}\rangle=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{n}||\vec{m}|}=\dfrac{1}{1 \times \sqrt{2}}=\dfrac{\sqrt{2}}{2}\),则两平面的夹角为\(45^∘\).
基本方法
【题型1】求平面间的夹角
【典题1】 正方体\(ABCD-A_1 B_1 C_1 D_1\)的棱长为\(1\),点\(E\),\(F\)分别为\(CD\),\(DD_1\)的中点,求平面\(AED\)与平面\(AEF\)夹角的余弦值.
解析 因为\(ABCD-A_1 B_1 C_1 D_1\)是正方体,
以\(A\)为原点,\(AB\)为\(x\)轴,\(AD\)为\(y\)轴,\(AA_1\)为\(z\)轴,建立空间直角坐标系,
依题意得\(A(0,0,0)\), \(F\left(0,1, \dfrac{1}{2}\right)\), \(E\left(\dfrac{1}{2}, 1,0\right)\), \(\overrightarrow{A E}=\left(\dfrac{1}{2}, 1,0\right)\), \(\overrightarrow{A F}=\left(0,1, \dfrac{1}{2}\right)\),
设平面\(AEF\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{A E}=\dfrac{1}{2} x+y=0 \\
\vec{n} \cdot \overrightarrow{A F}=y+\dfrac{1}{2} z=0
\end{array}\right.\),取\(y=-1\),得 \(\vec{n}=(2,-1,2)\),
易知平面\(ADE\)的法向量\(\vec{m}=(0,0,1)\),
设平面\(AED\)与平面\(AEF\)的夹角为\(θ\),则 \(\cos \theta=\dfrac{|\vec{n} \cdot \vec{m}|}{|\vec{n}| \cdot|\vec{m}|}=\dfrac{2}{3}\),
所以平面\(AED\)与平面\(AEF\)夹角的余弦值为 \(\dfrac{2}{3}\).
【典题2】如图,四边形\(ABCD\)是边长为\(3\)的菱形,\(DE⊥\)平面\(ABCD\),\(AB⊥AD\),\(AF∥DE\),\(DE=3AF\).
(1)求证:\(AC⊥\)平面\(BDE\);
(2)若\(BE\)与平面\(ABCD\)所成角为\(60°\),求平面\(FBE\)与平面\(DEB\)夹角的正弦值.
解析 (1)证明:因为\(DE⊥\)平面\(ABCD\),\(AC⊂\)平面\(ABCD\),
所以\(DE⊥AC\).
因为四边形\(ABCD\)是菱形,所以\(AC⊥BD\).
又因为\(BD∩DE=D\),\(BD⊂\)平面\(BDE\),\(DE⊂\)平面\(BDE\),
所以\(AC⊥\)平面\(BDE\).
(2)据题设知\(DA\)、\(DC\)、\(DE\)两两互相垂直.
以\(DA\)、\(DC\)、\(DE\)分别为\(x\)轴,\(y\)轴,\(z\)轴建立空间直角坐标系\(D-xyz\),如图所示,
因为\(BE\)与平面\(ABCD\)所成角为\(60°\),即\(∠DBE\),所以 \(\dfrac{D E}{B E}=\sqrt{3}\),
又\(AD=3\),\(DE=3AF\),
所以 \(D E=3 \sqrt{6}\), \(A F=\sqrt{6}\),
所以\(A(3,0,0)\),\(B(3,3,0)\), \(F(3,0, \sqrt{6})\), \(E(0,0,3 \sqrt{6})\),\(C(0,3,0)\),
所以 \(\overrightarrow{B F}=(0,-3, \sqrt{6})\), \(\overrightarrow{E F}=(3,0,-2 \sqrt{6})\),
设平面\(FBE\)的一个法向量\(\vec{m}=(x,y,z)\),
则 \(\left\{\begin{array}{l}
\vec{m} \cdot \overrightarrow{B F}=-3 y+\sqrt{6} z=0 \\
\vec{m} \cdot \overrightarrow{E F}=3 x-2 \sqrt{6} z=0
\end{array}\right.\),令 \(z=\sqrt{6}\),得 \(\vec{m}=(4,2, \sqrt{6})\).
因为\(AC⊥\)平面\(DEB\),所以 \(\overrightarrow{C A}\)为平面DEB的一个法向量,且 \(\overrightarrow{C A}=(3,-3,0)\),
所以 \(\cos \langle\vec{m}, \overrightarrow{C A}\rangle=\dfrac{\vec{m} \cdot \overrightarrow{C A}}{|\vec{m}| \cdot|\overrightarrow{C A}|}=\dfrac{6}{\sqrt{26} \cdot \sqrt{18}}=\dfrac{\sqrt{13}}{13}\),
所以 \(\sin \langle\vec{m}, \overrightarrow{C A}\rangle=\sqrt{1-\left(\dfrac{\sqrt{13}}{13}\right)^{2}}=\dfrac{2 \sqrt{39}}{13}\).
所以平面\(FBE\)与平面\(DEB\)夹角的正弦值为 \(\dfrac{2 \sqrt{39}}{13}\).
巩固练习
1如图,在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,已知 \(A D=\sqrt{6}\), \(A B=\sqrt{3}\), \(A A_{1}=\sqrt{2}\),求平面\(A_1 BD\)与平面\(ABCD\)夹角的大小和二面角\(A_1-BD-C\)的大小.
2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\(AB=1\),\(AD=2\),\(AP=2\),求平面\(AEC\)与平面\(DAE\)夹角的余弦值.
3 如图,四边形\(ABCD\)为菱形,\(DE⊥\)平面\(ABCD\),\(FC⊥\)平面\(ABCD\),\(DE=2FC\),\(∠DAB=60^∘\),\(DE=DC=2\).
(1)设\(BE\)的中点为\(H\),证明:\(FH⊥\)平面\(EDB\);
(2)求二面角\(A-EB-D\)的平面角的正弦值.
参考答案
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答案 平面\(A_1 BD\)与平面\(ABCD\)夹角为\(45^∘\),二面角\(A_1-BD-C\)为\(135^∘\).
解析 以\(A\)为原点,\(AB\)为\(x\)轴,\(AD\)为\(y\)轴,\(AA_1\)为\(z\)轴,建立空间直角坐标系,
\(A_{1}(0,0, \sqrt{2})\), \(B(\sqrt{3}, 0,0)\), \(D(0, \sqrt{6}, 0)\),
\(\overrightarrow{A_{1} B}=(\sqrt{3}, 0,-\sqrt{2})\), \(\overrightarrow{A_{1} D}=(0, \sqrt{6},-\sqrt{2})\),
设平面\(A_1 BD\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A_{1} B}=\sqrt{3} x-\sqrt{2} z=0 \\ \vec{n} \cdot \overrightarrow{A_{1} D}=\sqrt{6} y-\sqrt{2} z=0 \end{array}\right.\),取 \(x=\sqrt{2}\),得 \(\vec{n}=(\sqrt{2}, 1, \sqrt{3})\),
易知平面\(ABCD\)的法向量\(\vec{m}=(0,0,1)\),
设平面\(A_1 BD\)与平面\(ABCD\)夹角为\(θ\),
则 \(\cos \theta=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{\sqrt{3}}{\sqrt{2+1+3}}=\dfrac{\sqrt{2}}{2}\),\(∴θ=45^∘\).
\(∴\)平面\(A_1 BD\)与平面\(ABCD\)夹角为\(45^∘\),二面角\(A_1-BD-C\)为\(135^∘\). -
答案 (1) 略 (2) \(\dfrac{\sqrt{6}}{3}\)
解析 (1)证明:连接\(BD\),设\(BD∩AC=O\),连接\(EO\),
\(∵E\)是\(PD\)的中点,\(O\)为\(BD\)的中点,\(∴EO∥PB\),
又\(∵EO⊂\)平面\(AEC\),\(PB⊄\)平面\(AEC\),
\(∴PB∥\)平面\(AEC\);
(2)以\(A\)为坐标原点,分别以\(AB\),\(AD\),\(AP\)所在直线为\(x,y,z\)轴建立空间直角坐标系.
则\(D(0,2,0)\),\(E(0,1,1)\),\(C(1,2,0)\), \(\overrightarrow{A E}=(0,1,1)\), \(\overrightarrow{A C}=(1,2,0)\),
设\(\vec{n}=(x,y,z)\)为平面\(AEC\)的一个法向量,
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A C}=x+2 y=0 \\ \vec{n} \cdot \overrightarrow{A E}=y+z=0 \end{array}\right.\),取\(z=1\),则\(\vec{n}=(2,-1,1)\),
又平面\(DAE\)的一个法向量为\(\vec{m}=(1,0,0)\).
\(\cos <\vec{m}, \vec{n}>=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}\).
则平面\(AEC\)与平面\(DAE\)夹角的余弦值为 \(\dfrac{\sqrt{6}}{3}\). -
答案 (1) 略 (2) \(\dfrac{\sqrt{42}}{7}\)
解析 (1)证明:连接\(AC\)交\(BD\)于点\(O\),连接\(HO\).
\(∵\)四边形\(ABCD\)为菱形,点\(H\)是\(EB\)的中点,
所以\(OH\)为\(△BDE\)的中位线.
所以\(OH//DE\), \(O H=\dfrac{1}{2} D E\),而\(FC//DE\),\(FC=\dfrac{1}{2} DE\),
所以\(OH//FC\),\(OH=FC\),
\(∴\)四边形\(CFHO\)为平行四边形,\(∴FH//CO\),
\(∵DE⊥\)平面\(ABCD\),\(CO⊂\)平面\(ABCD\),
\(∴DE⊥CO\).
又\(∵CO⊥BD\),\(ED\cap BD=D\),
\(∴CO⊥\)平面\(EDB\),
\(∴FH⊥\)平面\(EDB\)
(2)解:因为\(OB,OC,OH\)两两垂直,
故以点\(O\)为坐标原点,分别 \(\overrightarrow{O B}, \overrightarrow{O C}, \overrightarrow{O H}\)的方向为\(x\)轴,\(y\)轴,\(z\)轴的正方向,
建立如图所示的空间直角坐标系\(O-xyz\).
由题意得 \(A(0,-\sqrt{3}, 0)\),\(C(0,\sqrt{3},0)\),\(B(1,0,0)\),\(E(-1,0,2)\),
则 \(\overrightarrow{E B}=(2,0,-2)\), \(\overrightarrow{A B}=(1, \sqrt{3}, 0)\), \(\overrightarrow{B C}=(-1, \sqrt{3}, 0)\).
设平面\(AEB\)的一个法向量为\(\vec{m}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{E B}=2 x-2 z=0 \\ \vec{m} \cdot \overrightarrow{A B}=x+\sqrt{3} y=0 \end{array}\right.\),取 \(\vec{m}=(3,-\sqrt{3}, 3)\),
由第(1)问可知\(OC⊥\)平面\(DEB\),
故平面\(DEB\)的一个法向量为\(\vec{n}=(0,1,0)\),
所以 \(\cos \langle\vec{m}, \vec{n}\rangle=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{3 \times 0+(-\sqrt{3}) \times 1+3 \times 0}{\sqrt{9+3+9} \times \sqrt{0+1+0}}=-\dfrac{\sqrt{7}}{7}\),
设二面角\(A-EB-D\)的平面角为\(θ\),
则 \(\sin \theta=\sqrt{1-\cos ^{2}<\vec{m}, \vec{n}>}=\dfrac{\sqrt{42}}{7}\).
【题型2】空间角问题的综合题
【典题1】 如图,在四棱锥\(P-ABCD\)中,平面\(PAB⊥\)平面\(ABCD\),\(AD∥BC\),\(AB⊥AD\),\(AB⊥PA\),点\(E\)为\(BC\)上一点且\(BC=2AB=2AD=4BE\).
(1)求证:平面\(PED⊥\)平面\(PAC\);
(2)若直线\(PE\)与平面\(PAC\)所成的角的正弦值为 \(\dfrac{\sqrt{5}}{5}\),求二面角\(A-PC-D\)的余弦值.
解析 (1)证明:\(∵\)平面\(PAB⊥\)平面\(ABCD\),平面\(PAB∩\)平面\(ABCD=AB\),\(AB⊥PA\),
\(∴PA⊥\)平面\(ABCD\),又\(AB⊥AD\).
分别以\(AB\)、\(AD\)、\(AP\)为\(x\)轴、\(y\)轴、\(z\)轴,建立空间直角坐标系\(o-xyz\),
可得\(A(0,0,0)\),\(D(0,2,0)\),\(E(2,1,0)\),\(C(2,4,0)\),\(P(0,0,λ)(λ>0)\).
\(\therefore \overrightarrow{A C}=(2,4,0)\), \(\overrightarrow{A P}=(0,0, \lambda)\), \(\overrightarrow{D E}=(2,-1,0)\).
由 \(\overrightarrow{D E} \cdot \overrightarrow{A C}=4-4+0=0\), \(\overrightarrow{D E} \cdot \overrightarrow{A P}=0\),
\(∴DE⊥AC\)且\(DE⊥AP\),
\(∵AC\)、\(AP\)是平面\(PAC\)内的相交直线,\(∴ED⊥\)平面\(PAC\).
\(∵ED⊂\)平面\(PED\),
\(∴\)平面\(PED⊥\)平面\(PAC\);
(2)解:由(1)得平面\(PAC\)的一个法向量是 \(\overrightarrow{D E}=(2,-1,0)\),而 \(\overrightarrow{P E}=(2,1,-\lambda)\).
设直线\(PE\)与平面\(PAC\)所成的角为\(θ\),
则 \(|\sin \theta=| \cos <\overrightarrow{P E}, \overrightarrow{D E}>\mid=\dfrac{|\overrightarrow{P E} \cdot \overrightarrow{D E}|}{|\overrightarrow{P E}| \cdot|\overrightarrow{D E}|}=\dfrac{|4-1|}{\sqrt{5} \cdot \sqrt{5+\lambda^{2}}}=\dfrac{\sqrt{5}}{5}\),解得\(λ=±2\).
\(∵λ>0\),
\(∴λ=2\),可得\(P\)的坐标为\((0,0,2)\).
设平面\(PCD\)的一个法向量为\(\vec{n}=(x,y,z)\),
\(\overrightarrow{D C}==(2,2,0)\), \(\overrightarrow{D P}=(0,-2,2)\),
由 \(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{D C}=2 x+2 y=0 \\
\vec{n} \cdot \overrightarrow{D P}=-2 y+2 z=0
\end{array}\right.\),令\(x=1\),得\(\vec{n}=(1,-1,-1)\).
\(\therefore \cos <\vec{n}, \overrightarrow{D E}>=\dfrac{\vec{n} \cdot \overrightarrow{D E}}{|\vec{n}| \cdot|\overrightarrow{D E}|}=\dfrac{\sqrt{15}}{5}\).
由图形可得二面角\(A-PC-D\)的平面角是锐角,
\(∴\)二面角\(A-PC-D\)的平面角的余弦值为 \(\dfrac{\sqrt{15}}{5}\).
巩固练习
1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 \(\qquad \qquad \qquad \qquad\) 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 \(\dfrac{\pi}{4}\) \(\qquad \qquad\) D.直线\(OD⊥\)平面\(ABC\)
2 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.
(Ⅰ)求证:\(PB⊥\)平面\(ADF\);
(Ⅱ)若直线\(DE\)与平面\(ADF\)所成角为\(30°\),
(1)求线段\(CE\)的长;(2)求二面角\(P-ED-A\)的余弦值.
参考答案
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答案 \(B\)
解析 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,
在\(A\)中,\(∵AC=AB=BC\),\(OA=OB=OC\),
\(∴O-ABC\)是正三棱锥,故\(A\)正确;
在\(B\)中,设\(OB=1\),则\(A(1,0,0)\),\(B(0,1,0)\),\(D(1,1,1)\),\(O(0,0,0)\),
\(\overrightarrow{O D}=(1,1,1)\), \(\overrightarrow{O B}=(0,1,0)\),
设平面\(OBD\)的法向量\(\vec{m}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{O B}=y=0 \\ \vec{m} \cdot \overrightarrow{O D}=x+y+z=0 \end{array}\right.\),取\(x=1\),得\(\vec{m}=(1,0,-1)\),
而平面\(OAB\)的法向量\(\vec{n}=(0,0,1)\),
\(\cos \langle\vec{m}, \vec{n}\rangle=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\),
二面角\(D-OB-A\)的平面角为 \(\dfrac{\pi}{4}\),故\(B\)错误;
在\(C\)中,设\(OB=1\),则\(A(1,0,0)\),\(B(0,1,0)\),\(D(1,1,1)\),\(O(0,0,0)\),
\(\overrightarrow{A D}=(0,1,1)\), \(\overrightarrow{A D}=(0,1,1), \quad \overrightarrow{O B}=(0,1,0)\),
\(\cos <\overrightarrow{A D}, \overrightarrow{O B}>=\dfrac{|\overrightarrow{A D} \cdot \overrightarrow{O B}|}{|\overrightarrow{A D}| \cdot|\overrightarrow{O B}|}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\),
\(∴\)直线\(AD\)与直线\(OB\)所成角为 \(\dfrac{\pi}{4}\),故\(C\)正确;
在\(D\)中,设\(OB=1\),则\(A(1,0,0)\),\(B(0,1,0)\),\(D(1,1,1)\),\(O(0,0,0)\),\(C(0,0,1)\),
\(\overrightarrow{O D}=(1,1,1)\), \(\overrightarrow{A B}=(-1,1,0)\), \(\overrightarrow{A C}=(-1,0,1)\),
\(\overrightarrow{O D} \cdot \overrightarrow{A B}=0\), \(\overrightarrow{O D} \cdot \overrightarrow{A C}=0\),\(∴OD⊥AB\),\(OD⊥AC\),
\(∵AB∩AC=A\),\(∴\)直线\(OD⊥\)平面\(ABC\),故\(D\)正确.
故选:\(B\).
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答案 (Ⅰ) 略 (Ⅱ)(1)略 (2) \(\dfrac{3 \sqrt{17}}{17}\)
解析 (Ⅰ)证明:依题意,以点\(A\)为原点建立空间直角坐标系(如图),
可得\(A(0,0,0)\),\(B(2,0,0)\),\(C(2,3,0)\),\(D(0,3,0)\),\(E(2,y,0)\),\(F(1,\dfrac{m}{2},1)\),\(P(0,0,2)\).
\(\overrightarrow{P B}=(2,0,-2)\), \(\overrightarrow{A D}=(0,3,0)\), \(\overrightarrow{A F}=\left(1, \dfrac{m}{2}, 1\right)\),
\(\overrightarrow{P B} \cdot \overrightarrow{A D}=0\), \(\overrightarrow{P B} \cdot \overrightarrow{A F}=0\),
即\(PB⊥AD\),\(PB⊥AF\),\(AF∩AD=A\),所以\(PB⊥\)平面\(ADF\).
(Ⅱ)解:(1)设\(\vec{n}=(x,y,z)\)为平面\(ADF\)的法向量,
则 \(\left\{\begin{array}{l} \overrightarrow{A D} \cdot \vec{n}=3 y=0 \\ \overrightarrow{A F} \cdot \vec{n}=x+\dfrac{m}{2} y+z=0 \end{array}\right.\),
不妨令\(v\),可得\(\vec{n}=(1,0,-1)\)为平面\(ADF\)的一个法向量,
向量 \(\overrightarrow{D E}=(2, y-3,0)\)
\(∵\)直线\(DE\)与平面\(ADF\)所成角为\(30°\),于是有 \(\cos \langle\vec{n} \cdot \overrightarrow{D E}\rangle=\dfrac{\vec{n} \cdot \overrightarrow{D E}}{|\vec{n}| \cdot|\overrightarrow{D E}|}=\dfrac{1}{2}\),
所以 \(\dfrac{1 \times 2+0 \times(y-3)+0 \times(-1)}{\sqrt{1+0+1} \cdot \sqrt{2^{2}+(y-3)^{2}+0}}=\dfrac{1}{2}\),得\(y=1\),\(y=5\)(舍)
\(E(2,1,0)\),\(C(2,3,0)\),线段\(CE\)的长为\(2\).
(2)设\(\vec{n}=(a,b,c)\)为平面\(PED\)的法向量,
\(\overrightarrow{P E}==(2,1,-2)\), \(\overrightarrow{P D}=(0,3,-2)\)
则 \(\left\{\begin{array}{l} \overrightarrow{P E} \cdot \vec{m}=2 a+y-2 c=0 \\ \overrightarrow{P D} \cdot \vec{m}=3 b-2 c=0 \end{array}\right.\),
不妨令\(a=2\),可得\(\vec{n}=(2,2,3)\)为平面\(ADF\)的一个法向量,
又 \(\overrightarrow{A P}=(0,0,2)\)为平面\(ADE\)的一个法向量,
\(∴\)二面角\(P-ED-A\)的余弦值为 \(\cos <\vec{n}, \overrightarrow{A P}>=\dfrac{\vec{n} \cdot \overrightarrow{A P}}{\vec{n} \cdot \overrightarrow{A P}}=\dfrac{3 \sqrt{17}}{17}\).
分层练习
【A组---基础题】
1 平面\(α\),\(β\)的法向量分别是 \(\overrightarrow{n_{1}}=(1,1,1)\), \(\overrightarrow{n_{2}}=(-1,0,-1)\),则平面\(α\),\(β\)所成角的正弦值是( )
A. \(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\) B. \(\dfrac{1}{2}\) \(\qquad \qquad\) C. \(\dfrac{\sqrt{6}}{3}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{2}}{2}\)
2 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. \(\dfrac{\sqrt{3}}{2}\)\(\qquad \qquad\) B. \(\dfrac{\sqrt{2}}{2}\) \(\qquad \qquad\) C. \(\dfrac{\sqrt{6}}{3}\) \(\qquad \qquad\) D. \(\dfrac{1}{3}\)
3 若面\(α\)的法向量\(\vec{n}=(1,λ,1)\),面\(β\)的法向量\(\vec{m}=(2,-1,-2)\),两面夹角的正弦值为 \(\dfrac{\sqrt{34}}{6}\),则\(λ=\underline{\quad \quad}\).
4 如图,在三棱柱\(ABC-A_1 B_1 C_1\)中,\(AB\),\(AC\),\(AA_1\)两两互相垂直,\(AB=AC=AA_1\),\(M\),\(N\)分别是侧棱\(BB_1\),\(CC_1\)上的点,平面\(AMN\)与平面\(ABC\)夹角为 \(\dfrac{\pi}{6}\),则当\(|B_1 M|\)最小时\(∠AMB=\underline{\quad \quad}\).
5 如图,在棱长为\(2\)的正方体\(ABCD-A_1 B_1 C_1 D_1\)中,点\(E\)是棱\(D_1 D\)的中点,点\(F\)在棱\(B_1 B\)上,且满足\(B_1 F=2FB\).
(1)求证:\(A_1 C_1⊥D_1 D\);
(2)求平面\(AEF\)与平面\(AA_1 D_1 D\)所成锐二面角的余弦值.
6 如图,四棱锥\(P-ABCD\)的底面是正方形,\(PD⊥\)底面\(ABCD\),\(PD=DC\),\(E\)是\(PC\)的中点.
(1)证明:平面\(PAB⊥\)平面\(PAD\);(2)求二面角\(P-AB-D\)的大小.
7 如图,在四棱锥\(P-ABCD\)中,底面\(ABCD\)是边长为\(2\)的菱形,\(∠ABC=60°\),\(△PAB\)为正三角形, \(P C=\sqrt{6}\),\(E\)为线段\(AB\)的中点.
(1)证明:\(PE⊥\)平面\(ABCD\);
(2)若 \(3 \overrightarrow{P M}=\overrightarrow{P D}\),求二面角\(M-EC-D\)的大小.
8 如图,菱形\(ABCD\)中\(∠BAD=60^∘\),把\(△BDC\)沿\(BD\)折起,使得点\(C\)至\(P\)处.
(1)证明:平面\(PAC⊥\)平面\(ABCD\);
(2)若\(PA\)与平面\(ABD\)所成角的余弦值为 \(\dfrac{\sqrt{3}}{3}\),求二面角\(B-PA-D\)的余弦值.
参考答案
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答案 \(A\)
解析 设向量 \(\overrightarrow{n_{1}}=(1,1,1)\), \(\overrightarrow{n_{2}}=(-1,0,-1)\)所成角为\(θ\),
则 \(\cos \theta=\dfrac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}=\dfrac{-2}{\sqrt{3} \sqrt{2}}=-\dfrac{2}{\sqrt{6}}\),
\(\therefore \sin \theta=\sqrt{1-\dfrac{2}{3}}=\dfrac{\sqrt{3}}{3}\),故选:\(A\). -
答案 \(C\)
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\(1\),
则\(A_1 (1,0,1)\),\(B(1,1,0)\),\(D(0,0,0)\), \(\overrightarrow{D A}_{1}=(1,0,1)\), \(\overrightarrow{D B}=(1,1,0)\),
设平面\(A_1 BD\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{D A_{1}}=x+z=0 \\ \vec{n} \cdot \overrightarrow{D B}=x+y=0 \end{array}\right.\),取\(x=1\),得\(\vec{n}=(1,-1,-1)\),
平面\(ABCD\)的法向量\(\vec{m}=(0,0,1)\),
设平面\(A_1 BD\)与平面\(ABCD\)所成二面角的平面角为\(α\),
则 \(\cos \alpha=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\),
\(\therefore \sin \alpha=\sqrt{1-\left(\dfrac{\sqrt{3}}{3}\right)^{2}}=\dfrac{\sqrt{6}}{3}\).
\(∴\)平面\(A_1 BD\)与平面\(ABCD\)所成二面角的正弦值为 \(\dfrac{\sqrt{6}}{3}\).
故选:\(C\).
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答案 \(\pm \sqrt{2}\)
解析 设面\(α\)与面\(β\)的夹角为\(θ\),
则 \(|\cos \theta=| \cos <\vec{n}, \vec{m}>|=| \dfrac{-\lambda}{\sqrt{2+\lambda^{2}} \cdot \sqrt{9}} \mid=\dfrac{|\lambda|}{3 \sqrt{2+\lambda^{2}}}\),
所以 \(\sin 2 \theta=1-\cos 2 \theta=1-\dfrac{\lambda^{2}}{9\left(2+\lambda^{2}\right)}=\dfrac{34}{36}\),解得 \(\lambda=\pm \sqrt{2}\). -
答案 \(\dfrac{\pi}{3}\)
解析 以点\(A\)为坐标原点,建立空间直角坐标系如图所示,
设\(CN=b\),\(BM=a\),
则\(N(0,1,b)\),\(M(1,0,a)\),\(A(0,0,0)\),\(B(1,0,0)\),
所以 \(\overrightarrow{A M}=(1,0, a)\), \(\overrightarrow{A N}=(0,1, b)\),
设平面\(AMN\)的法向量为\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A M}=x+a z=0 \\ \vec{n} \cdot \overrightarrow{A N}=y+b z=0 \end{array}\right.\),令\(z=1\),则\(\vec{n}=(-a,-b,1)\),
平面\(ABC\)的法向量为\(\vec{m}=(0,0,1)\),
因为平面\(AMN\)与平面\(ABC\)夹角为 \(\dfrac{\pi}{6}\),
所以 \(\cos <\vec{m}, \vec{n}>=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}||\vec{n}|}=\dfrac{1}{\sqrt{a^{2}+b^{2}+1}}=\dfrac{\sqrt{3}}{2}\),
化简可得\(3a^2+3b^2=1\),
当\(|B_1 M|\)最小时,则\(b=0\), \(B M=a=\dfrac{\sqrt{3}}{3}\),
所以 \(\tan \angle A M B=\dfrac{A B}{B M}=\dfrac{1}{\dfrac{\sqrt{3}}{3}}=\sqrt{3}\),
则 \(\angle A M B=\dfrac{\pi}{3}\). -
答案 (1)略 (2) \(\dfrac{2}{7}\)
解析 证明:(1)\(∵\)在正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(D_1 D⊥\)平面\(A_1 B_1 C_1 D_1\),\(A_1 C_1⊂\)平面\(A_1 B_1 C_1 D_1\),
\(∴A_1 C_1⊥D_1 D\).
(2)如图,以\(D\)为原点,\(DA\),\(DC\),\(DD_1\)分别为\(x,y,z\)轴,建立空间直角坐标系,
则\(A(2,0,0)\),\(E(0,0,1)\), \(, F\left(2,2, \dfrac{2}{3}\right)\),\(B(2,2,0)\),
\(\therefore \overrightarrow{A E}=(-2,0,1)\), \(\overrightarrow{A F}=\left(0,2, \dfrac{2}{3}\right)\),\(\overrightarrow{A B}=(0,2,0)\)
设\(\vec{n}=(x,y,z)\)是平面\(AEF\)的一个法向量,
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A E}=-2 x+z=0 \\ \vec{n} \cdot \overrightarrow{A F}=2 y+\dfrac{2}{3} z=0 \end{array}\right.\),取\(z=6\),得\(\vec{n}=(3,-2,6)\),
\(∵AB⊥\)平面\(AA_1 D_1 D\),
\(: \overrightarrow{A B}=(0,2,0)\)是平面\(AA_1 D_1 D\)的一个法向量,
\(\therefore \cos <\overrightarrow{A B}, \vec{n}>=\dfrac{\overrightarrow{A B} \cdot \vec{n}}{|\overrightarrow{A B}| \cdot|\vec{n}|}=\dfrac{-4}{2 \sqrt{49}}=-\dfrac{2}{7}\),
\(∴\)平面\(AEF\)与平面\(AA_1 D_1 D\)所成锐二面角的余弦值为 \(\dfrac{2}{7}\). -
答案 (1)略 (2)\(45^∘\)
解析 证明:(1)\(∵\)四棱锥\(P-ABCD\)的底面是正方形,\(PD⊥\)底面\(ABCD\),\(PD=DC\),\(E\)是\(PC\)的中点.
\(∴AB⊥AD\),\(AB⊥PD\),
又\(AD\cap PD=D\),\(∴AB⊥\)平面\(PAD\),
\(∵AB⊂\)平面\(PAB\),\(∴\)平面\(PAB⊥\)平面\(PAD\).
解:(2)以\(D\)为原点,\(DA\)为\(x\)轴,\(DC\)为\(y\)轴,\(DP\)为\(z\)轴,建立空间直角坐标系,
设\(PD=DC=AD=2\),
则\(A(2,0,0)\),\(P(0,0,2)\),\(D(0,0,0)\),\(B(2,2,0)\),
\(\overrightarrow{A P}=(-2,0,2)\), \(\overrightarrow{A B}=(0,2,0)\),
设平面 \(PAB\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A P}=-2 x+2 z=0 \\ \vec{n} \cdot \overrightarrow{A B}=2 y=0 \end{array}\right.\),取\(x=1\),得 \(\vec{n}=(1,0,1)\),
平面\(ABD\)的法向量 \(\vec{m}=\left( 0, 0, 1\right)\),
设二面角\(P-AB-D\)的大小为\(θ\),
则 \(\cos \theta=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\),\(θ=45^∘\),
\(∴\)二面角\(P-AB-D\)的大小为\(45^∘\).
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答案 (1)略 (2)\(60°\)
解析 (1)证明:连接\(CE\),
\(∵△PAB\)是边长为\(2\)的正三角形,且\(E\)是\(AB\)中点, \(\therefore P E=\sqrt{3}\)
又\(∵ABCD\)是边长为\(2\)的菱形,\(∠ABC=60°\),
\(∴△ABC\)是正三角形,\(CE=\sqrt{3}\),
又\(∵PC=\sqrt{6}\),\(∴PC^2=PE^2+CE^2\),即\(PE⊥CE\),
又\(PE⊥AB\),\(CE∩AB=E\),
\(∴PE⊥\)平面\(ABCD\).
(2)解:由(1)可得:以\(E\)为原点,分别以 \(\overrightarrow{E B}, \overrightarrow{E C}, \overrightarrow{E P}\)为\(x,y,z\)轴的正方向,建立空间直角坐标系\(E-xyz\),
则\(E(0,0,0)\),\(B(1,0,0)\),\(P(0,0,\sqrt{3})\),\(C(0,\sqrt{3},0)\),\(D(-2,\sqrt{3},0)\).
设点\(M\)坐标为\((x,y,z)\),由 \(3 \overrightarrow{P M}=\overrightarrow{P D}\),得 \(3(x, y, z-\sqrt{3})=(-2, \sqrt{3},-\sqrt{3})\),
\(\therefore M\left(-\dfrac{2}{3}, \dfrac{\sqrt{3}}{3}, \dfrac{2 \sqrt{3}}{3}\right)\),
\(\therefore \overrightarrow{E M}=\left(-\dfrac{2}{3}, \dfrac{\sqrt{3}}{3}, \dfrac{2 \sqrt{3}}{3}\right)\),\(\overrightarrow{E C}=(0, \sqrt{3}, 0)\),
设平面\(CEM\)的法向量为\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{E M}=-\dfrac{2}{3} x+\dfrac{\sqrt{3}}{3} y+\dfrac{2 \sqrt{3}}{3} z=0 \\ \vec{n} \cdot \overrightarrow{E C}=\sqrt{3} y=0 \end{array}\right.\),取\(z=1\),解得 \(\vec{m}=(\sqrt{3}, 0,1)\).
\(∵PE⊥\)平面\(ABCD\),\(∴\)平面\(ABCD\)的法向量 \(\vec{m}=(0,0,1)\),
\(\therefore|\cos <\vec{n}, \vec{m}>|=\dfrac{|\vec{n} \cdot \vec{m}|}{|\vec{n}| \cdot|\vec{m}|}=\dfrac{1}{2}\),
\(∴\)二面角\(M-EC-D\)的大小为\(60°\). -
答案 (1)略 (2) \(\dfrac{1}{3}\)
解析 (1)证明:设\(AC\cap BD\)于\(O\),连接\(PO\),
在菱形\(ABCD\)中,\(O\)为\(BD\)的中点,且\(BD⊥AC\),
\(∵PB=PD\),\(∴BD⊥PO\),
\(∴PO∩AC=O\),\(∴BD⊥\)平面\(PAC\) ,
\(∵BD⊂\)平面\(ABCD\),
\(∴\)平面\(PAC⊥\)平面\(ABCD\);
(2)在平面\(PAC\)内,作\(OM⊥AC\),以\(O\)为坐标原点,\(OA,OB,OM\)为\(x,y,z\)轴,建立空间直角坐标系,如图,
设\(PB=PD=AB=AD=2\),则\(OA=OP=\sqrt{3}\),\(OR=1\),
由(1)知平面\(PAC⊥\)平面\(ABCD\),
\(∴∠PAO\)是\(PA\)与平面\(ABCD\)所成角,
\(\therefore \cos \angle P A O=\dfrac{\sqrt{3}}{3}\),可得 \(P\left(\dfrac{\sqrt{3}}{3}, 0, \dfrac{2 \sqrt{6}}{3}\right)\),
由\(A(\sqrt{3},0,0)\),\(B(0,1,0)\),\(D(0,-1,0)\),
\(\therefore \overrightarrow{A D}=(-\sqrt{3},-1,0)\), \(\overrightarrow{A B}=(-\sqrt{3}, 1,0)\), \(\overrightarrow{A P}=\left(-\dfrac{2 \sqrt{3}}{3}, 0, \dfrac{2 \sqrt{6}}{3}\right),\),
设平面\(PAB\)的法向量为\(\vec{m}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{A B}=-\sqrt{3} x+y=0 \\ \vec{m} \cdot \overrightarrow{A P}=-\dfrac{2 \sqrt{3}}{3} x+\dfrac{2 \sqrt{6}}{3} z=0 \end{array}\right.\),取\(z=1\),得 \(\vec{m}=(\sqrt{2}, \sqrt{6}, 1)\),
设平面\(PAD\)的法向量为\(\vec{n}=(a,b,c)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A D}=-\sqrt{3} a-b=0 \\ \vec{n} \cdot \overrightarrow{A P}=-\dfrac{2 \sqrt{3}}{3} a+\dfrac{2 \sqrt{6}}{3} c=0 \end{array}\right.\),取\(c=1\),得 \(\vec{n}=(\sqrt{2},-\sqrt{6}, 1)\),
\(\therefore \cos \langle\vec{m}, \vec{n}\rangle=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{2-6+1}{\sqrt{2+6+1} \cdot \sqrt{2+6+1}}=-\dfrac{1}{3}\),
由图知二面角\(B-PA-D\)为锐角,故其余弦值为 \(\dfrac{1}{3}\).
【B组---提高题】
1 如图,三棱锥\(V-ABC\)的侧棱长都相等,底面\(ABC\)与侧面\(VAC\)都是以\(AC\)为斜边的等腰直角三角形,\(E\)为线段\(AC\)的中点,\(F\)为直线\(AB\)上的动点,若平面\(VEF\)与平面\(VBC\)所成锐二面角的平面角为\(θ\),则\(\cosθ\)的最大值是( )
A. \(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\) B. \(\dfrac{2}{3}\) \(\qquad \qquad\) C. \(\dfrac{\sqrt{5}}{3}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{6}}{3}\)
2 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.\(θ_2<θ_1,θ_2<θ_3\) \(\qquad \qquad\) B.\(θ_2>θ_1,θ_2<θ_3\) \(\qquad \qquad\) C.\(θ_2<θ_1,θ_2>θ_3\) \(\qquad \qquad\) D.\(θ_2>θ_1,θ_2>θ_3\)
3如图所示,在等边\(△ABC\)中,\(AB=6\),\(M\),\(N\)分别是\(AB\),\(AC\)上的点,且\(AM=AN=4\),\(E\)是\(BC\)的中点,\(AE\)交\(MN\)于点\(F\).以\(MN\)为折痕把\(△AMN\)折起,使点\(A\)到达点\(P\)的位置\((0<∠PFE<π)\),连接\(PB,PE,PC\).
(1)证明:\(MN⊥PE\);
(2)设点\(P\)在平面\(ABC\)内的射影为点\(Q\),若二面角\(P-MN-B\)的大小为 \(\dfrac{2}{3} \pi\),求直线\(QC\)与平面\(PBC\)所成角的正弦值.
参考答案
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答案 \(D\)
解析 由底面\(ABC\)与侧面\(VAC\)都是以\(AC\)为斜边的等腰直角三角形,
得\(Rt△ABC≌Rt△AVC\),\(∴VA=VC=BA=BC\).
设\(VA=VC=BA=BC=2\),
由\(E\)为线段\(AC\)的中点,可得 \(V E=E B=\sqrt{2}\).
由\(VE^2+BE^2=VB^2\),可得\(VE⊥EB\).
以\(E\)为坐标原点,分别以\(EB\),\(EC\),\(EV\)所在直线为\(x,y,z\)轴建立空间直角坐标系.
则\(C(0,\sqrt{2},0)\),\(B(\sqrt{2},0,0)\),\(V(0,0,\sqrt{2})\),设\(F(x,x-\sqrt{2},0)\),
\(\overrightarrow{V C}=(0, \sqrt{2},-\sqrt{2})\), \(\overrightarrow{V B}=(\sqrt{2}, 0,-\sqrt{2})\),
\(\overrightarrow{E V}=(0,0, \sqrt{2})\), \(\overrightarrow{V F}=(x, x-\sqrt{2},-\sqrt{2})\).
设平面\(VBC\)的一个法向量为\(\vec{m}=(x,y,z)\),
由 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{V C}=\sqrt{2} y-\sqrt{2} z=0 \\ \vec{m} \cdot \overrightarrow{V B}=\sqrt{2} x-\sqrt{2} z=0 \end{array}\right.\),取\(x=1\),得\(\vec{m}=(1,1,1)\);
设平面\(VEF\)的一个法向量为\(\vec{n}=(x_1,y_1,z_1)\),
由 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{E V}=\sqrt{2} z_{1}=0 \\ \vec{n} \cdot \overrightarrow{V F}=x \cdot x_{1}+(x-\sqrt{2}) y_{1}-\sqrt{2} z_{1}=0 \end{array}\right.\),取\(y_1=1\),得 \(\vec{n}=\left(\dfrac{\sqrt{2}}{x}-1,1,0\right)\).
平面\(VEF\)与平面\(VBC\)所成锐二面角的平面角为\(θ\),
则 \(\cos \theta=\dfrac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{\dfrac{\sqrt{2}}{x}}{\sqrt{3} \cdot \sqrt{\dfrac{2}{x^{2}}-\dfrac{2 \sqrt{2}}{x}+2}}=\dfrac{\sqrt{2}}{\sqrt{6 x^{2}-6 \sqrt{2} x+6}}\).
令 \(f(x)=6 x^{2}-6 \sqrt{2} x+6=6\left(x-\dfrac{\sqrt{2}}{2}\right)^{2}+3\).
当 \(x=\dfrac{\sqrt{2}}{2}\)时, \(f(x)_{\min }=3\).\(∴\cosθ\)的最大值为 \(\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}\).
故选:\(D\).
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答案 \(A\)
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\(ABC-A_1 B_1 C_1\)是棱长为\(2\)的正三棱柱,取\(D\)是\(BC\)中点,
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\(A(0,0,0)\),\(A_1 (0,0,2)\), \(B_{1}(\sqrt{3}, 1,2)\),\(C(0,2,0)\), \(D\left(\dfrac{\sqrt{3}}{2}, \dfrac{3}{2}, 0\right)\),
\(\overrightarrow{A C}=(0,2,0)\), \(\overrightarrow{B_{1} D}=\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2},-2\right)\), \(\overrightarrow{A_{1} B_{1}}=(\sqrt{3}, 1,0)\),
\(∵\)直线\(B_1 D\)与直线\(AC\)所成的角为\(θ_1\),
\(\therefore \cos \theta_{1}=\dfrac{\left|\overrightarrow{B_{1} D} \cdot \overrightarrow{A C}\right|}{\left|\overrightarrow{B_{1} D}\right| \cdot|\overrightarrow{A C}|}=\dfrac{2 \times \dfrac{1}{2}}{2 \cdot \sqrt{5}}=\dfrac{\sqrt{5}}{10}\),
\(∵\)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 \(\vec{m}=\left(\sqrt{3},-3,-\dfrac{3}{2}\right)\),
\(∵\)二面角\(C_1-A_1 B_1-D\)的平面角为\(θ_3\),
\(\therefore \cos \theta_{3}=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{\dfrac{3}{2}}{\sqrt{\dfrac{57}{4}} \cdot 1}=\dfrac{\sqrt{57}}{19}\).
且\(θ_1,θ_2\)与\(θ_3\)均为锐角,结合余弦函数在 \(\left[0, \dfrac{\pi}{2}\right]\)上为减函数,则\(θ_2<θ_1,θ_2<θ_3\),
故选:\(A\).
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答案 (1)略 (2) \(\dfrac{2 \sqrt{3}}{7}\)
解析 (1)证明:因为\(△ABC\)是等边三角形,\(E\)是\(BC\)的中点,所以\(AE⊥BC\),
因为\(AM=AN=4\),所以\(MN//BC\),所以\(MN⊥AE\),
可得\(MN⊥PF\),\(MN⊥FE\),
又\(PF\cap FE=F\),所以\(MN⊥\)平面\(PFE\),
又\(PE⊂\)平面\(PFE\),所以\(MN⊥PE\).
(2)解:因为\(MN⊥PF\),\(MN⊥FE\),
所以二面角\(P-MN-B\)的平面角为\(∠PFE\),
所以\(∠PFE=\dfrac{2}{3} \pi\),可得\(∠PFA=\dfrac{1}{3} \pi\),
由第(1)问知,\(MN⊥\)平面\(PFE\),\(MN⊂\)平面\(ABC\),
所以平面\(ABC⊥\)平面\(PFE\),
又因为平面\(PFE∩\)平面\(ABC=AE\),
所以点\(P\)在平面\(ABC\)内的射影\(Q\)在\(AE\)上,
因为 \(P F=2 \sqrt{3}\),所以 \(Q F=\sqrt{3}\),\(PQ=3\),
过\(F\)作直线\(l//PQ\)交\(PE\)于点\(K\),以\(F\)为坐标原点,
以 \(\overrightarrow{F M}, \overrightarrow{F E}, \overrightarrow{F K}\)的方向分别为\(x,y,z\)轴的正方向,
建立如图所示的空间直角坐标系,
\(P(0,-\sqrt{3},3)\),\(C(-3,\sqrt{3},0)\),\(E(0,\sqrt{3},0)\),\(Q(0,-\sqrt{3},0)\),
\(\overrightarrow{Q C}=(-3,2 \sqrt{3}, 0)\) ,\(\overrightarrow{P C}=(-3,2 \sqrt{3},-3)\), \(\overrightarrow{P E}=(0,2 \sqrt{3},-3)\),
设平面\(PBC\)的法向量为\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{c} \vec{n} \cdot \overrightarrow{P C}=-3 x+2 \sqrt{3} y-3 z=0 \\ \vec{n} \cdot \overrightarrow{P E}=2 \sqrt{3} y-3 z=0 \end{array}\right.\),令\(z=2\),可得 \(\vec{n}=(0, \sqrt{3}, 2)\),
所以 \(\cos \langle\overrightarrow{Q C}, \vec{n}\rangle=\dfrac{6}{\sqrt{7} \times \sqrt{21}}=\dfrac{2 \sqrt{3}}{7}\),
所以直线\(QC\)与平面\(PBC\)所成角的正弦值为 \(\dfrac{2 \sqrt{3}}{7}\).
【C组---拓展题】
1 如图所示,在正三棱台\(ABC-A_1 B_1 C_1\)中, \(A B=3 A A_{1}=\dfrac{3}{2} A_{1} B_{1}=3\),记侧面\(ABB_1 A_1\)与底面\(ABC\),侧面\(ABB_1 A_1\)与侧面\(BCC_1 B_1\),以及侧面\(ABB_1 A_1\)与截面\(A_1 BC\)所成的锐二面角的平面角分别为\(α,β,γ\),则( )
A.\(γ<β=α\) \(\qquad \qquad\) B.\(β=α<γ\) \(\qquad \qquad\) C.\(β<α<γ\)\(\qquad \qquad\) D.\(α<β<γ\)
2 如图,在圆锥\(SO\)中,\(A\),\(B\)是\(⊙O\)上的动点,\(BB'\)是\(⊙O\)的直径,\(M\),\(N\)是\(SB\)的两个三等分点,\(∠AOB=θ(0<θ<π)\),记二面角\(N-OA-B\),\(M-AB'-B\)的平面角分别为\(α\),\(β\),若\(α≤β\),则\(θ\)的最大值是( )
A. \(\dfrac{5 \pi}{6}\) \(\qquad \qquad\) B. \(\dfrac{2 \pi}{3}\) \(\qquad \qquad\) C. \(\dfrac{\pi}{2}\) \(\qquad \qquad\) D. \(\dfrac{\pi}{4}\)
3折纸与数学有着千丝万缕的联系,吸引了人们的广泛兴趣.因\(A4\)纸的长宽比 \(\sqrt{2}: 1\)称为白银分割比例,故\(A4\)纸有一个白银矩形的美称.现有一张如图\(1\)所示的\(A4\)纸\(EFCH\),\(EF:EH=\sqrt{2}: 1\).\(A,B,C,D\)分别为\(EF\),\(FG\),\(GH\),\(HE\)的中点,将其按折痕\(AB\),\(BC\),\(CD\),\(DA\),\(AC\)折起(如图\(2\)),使得\(E,F,G,H\)四点重合,重合后的点记为\(S\),折得到一个如图\(3\)所示的三棱锥\(D-ABC\).记\(O\)为\(AC\)的中点,在\(△SOB\)中,\(SP\)为\(BO\)边上的高.
(1)求证:\(SP∥\)平面\(ACD\);
(2)若\(M,N\)分别是棱\(AB\),\(BC\)上的动点,且\(AM=BN\).当三棱锥\(B-DMN\)的体积最大时,求平面\(DAB\)与平面\(DMN\)所成锐二面角的余弦值.
参考答案
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答案 \(B\)
解析 如图,设底面三角形\(ABC\)的外心为\(O\),
以\(O\)为坐标原点,以\(OB\)所在直线为\(x\)轴,以过\(O\)平行于\(AC\)的直线为\(y\)轴,以过\(O\)垂直于底面\(ABC\)的直线为z轴建立空间直角坐标系,
\(∵A B=3 A A_{1}=\dfrac{3}{2} A_{1} B_{1}=3\),
\(\therefore B(\sqrt{3}, 0,0)\), \(B_{1}\left(\dfrac{2 \sqrt{3}}{3}, 0, \dfrac{\sqrt{6}}{3}\right)\), \(C\left(-\dfrac{\sqrt{3}}{2}, \dfrac{3}{2}, 0\right)\), \(A_{1}\left(-\dfrac{\sqrt{3}}{3},-1, \dfrac{\sqrt{6}}{3}\right)\),
\(\overrightarrow{B B}_{1}==\left(-\dfrac{\sqrt{3}}{3}, 0, \dfrac{\sqrt{6}}{3}\right)\), \(\overrightarrow{B A_{1}}=\left(-\dfrac{4 \sqrt{3}}{3},-1, \dfrac{\sqrt{6}}{3}\right)\), \(\overrightarrow{B C}==\left(-\dfrac{3 \sqrt{3}}{2}, \dfrac{3}{2}, 0\right)\),
设平面\(ABB_1 A_1\)的一个法向量为\(\vec{n}=(x,y,z)\),
由 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{B B}_{1}=-\dfrac{\sqrt{3}}{3} x+\dfrac{\sqrt{6}}{3} z=0 \\ \vec{n} \cdot \overrightarrow{B A_{1}}=-\dfrac{4 \sqrt{3}}{3} x-y+\dfrac{\sqrt{6}}{3} z=0 \end{array}\right.\),取\(x=\sqrt{2}\),得 \(\vec{n}=(\sqrt{2},-\sqrt{6}, 1)\);
平面\(ABC\)的一个法向量为\(\vec{α}=(0,0,1)\);
设平面\(BCC_1 B_1\)的一个法向量为\(\vec{β} =(x_1,y_1,z_1)\),
由 \(\left\{\begin{array}{l} \vec{\beta} \cdot \overrightarrow{B B_{1}}=-\dfrac{\sqrt{3}}{3} x_{1}+\dfrac{\sqrt{6}}{3} z_{1}=0 \\ \vec{\beta} \cdot \overrightarrow{B C}=-\dfrac{3 \sqrt{3}}{2} x_{1}+\dfrac{3}{2} y_{1}=0 \end{array}\right.\),取 \(x_{1}=\sqrt{2}\),得 \(\vec{\beta}=(\sqrt{2}, \sqrt{6}, 1)\);
设平面\(A_1 BC\)的一个法向量为 \(\vec{\gamma}=\left(x_{2}, y_{2}, z_{2}\right)\),
由 \(\left\{\begin{array}{l} \vec{\gamma} \cdot \overrightarrow{B A_{1}}=-\dfrac{4 \sqrt{3}}{3} x_{2}-y_{2}+\dfrac{\sqrt{6}}{3} z_{2}=0 \\ \vec{\gamma} \cdot \overrightarrow{B C}=-\dfrac{3 \sqrt{3}}{2} x_{2}+\dfrac{3}{2} y_{2}=0 \end{array}\right.\),取 \(y_{2}=\sqrt{3}\),得 \(\vec{\gamma}=\left(1, \sqrt{3}, \dfrac{7 \sqrt{2}}{2}\right)\).
\(\therefore \cos \alpha=|\cos <\vec{n}, \vec{\alpha}>|=\left|\dfrac{\vec{n} \cdot \vec{\alpha}}{|\vec{n}||\vec{\alpha}|}\right|=\dfrac{1}{3 \times 1}=\dfrac{1}{3},\),
\(\cos \beta=|\cos <\vec{n}, \vec{\beta}>|=\left|\dfrac{\vec{n} \cdot \vec{\beta}}{|\vec{n}||\vec{\beta}|}\right|=\dfrac{|2-6+1|}{3 \times 3}=\dfrac{1}{3}\);
\(\cos \gamma=|\cos <\vec{n}, \vec{\gamma}>|=\left|\dfrac{\vec{n} \cdot \vec{\gamma}}{|\vec{n}| \vec{\gamma} \mid}\right|=\dfrac{\left|\sqrt{2}-3 \sqrt{2}+\dfrac{7 \sqrt{2}}{2}\right|}{3 \times \dfrac{\sqrt{114}}{2}}=\dfrac{\sqrt{57}}{57}\).
\(\because \cos \alpha=\cos \beta>\cos \gamma\),且余弦函数在 \(\left(0, \dfrac{\pi}{2}\right)\)上为减函数,
\(∴β=α<γ\).
故选:\(B\). -
答案 \(B\)
解析 设底面圆的半径为\(r\),\(OS=a\),以\(B'B\)所在直线为\(x\)轴,以垂直于\(B'B\)所在直线为\(y\)轴,以\(OS\)所在直线为\(z\)轴建立空间直角坐标系.
则由\(∠AOB=θ(0<θ<π)\),可得:\(O(0,0,0)\),\(B(r,0,0)\),\(S(0,0,a)\), \(A(r \cos \theta, r \sin \theta, 0)\), \(B^{\prime}(-r, 0,0)\),
\(∵M,N\)是\(SB\)的两个三等分点,
\(\therefore M\left(\dfrac{r}{3}, 0, \dfrac{2 a}{3}\right)\), \(N\left(\dfrac{2 r}{3}, 0, \dfrac{a}{3}\right)\),
\(\therefore \overrightarrow{O A}=(r \cos \theta, r \sin \theta, 0)\), \(\overrightarrow{O N}=\left(\dfrac{2 r}{3}, 0, \dfrac{a}{3}\right)\),
设平面\(NOA\)的一个法向量为\(\vec{m} =(x_1,y_1,z_1)\),
由 \(\left\{\begin{array}{l} \vec{m} \cdot \overrightarrow{O A}=x_{1} r \cos \theta+y_{1} r \sin \theta=0 \\ \vec{m} \cdot \overrightarrow{O N}=\dfrac{2 x_{1} r}{3}+\dfrac{a z_{1}}{3}=0 \end{array}\right.\),取\(x_1=1\),得 \(\vec{m}=\left(1,-\dfrac{\cos \theta}{\sin \theta},-\dfrac{2 r}{a}\right)\);
平面\(OAB\)的一个法向量为\(\vec{n} =(0,0,1)\).
由图可知,\(N-OA-B\)的平面角\(α\)为锐二面角, \(\therefore \cos \alpha=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{\left|\dfrac{2 r}{a}\right|}{\sqrt{1+\dfrac{\cos ^{2} \theta}{\sin ^{2} \theta}+\dfrac{4 r^{2}}{a^{2}}}}\);
设平面\(B'AM\)的一个法向量为\(\vec{k} =(x_2,y_2,z_2)\),
\(\overrightarrow{B^{\prime} A}=(r+r \cos \theta, r \sin \theta, 0)\), \(\overrightarrow{A M}=\left(\dfrac{r}{3}-r \cos \theta,-r \sin \theta, \dfrac{2 a}{3}\right)\).
由 \(\left\{\begin{array}{l} \vec{k} \cdot \overrightarrow{B^{\prime}} A=x_{2} r+x_{2} r \cos \theta+y_{2} r \sin \theta=0 \\ \vec{k} \cdot \overrightarrow{A M}=\dfrac{x_{2} r}{3}-x_{2} r \cos \theta-y_{2} r \sin \theta+\dfrac{2 a z_{2}}{3}=0 \end{array}\right.\),
取\(x_2=1\),得 \(\vec{k}=\left(1, \dfrac{-1-\cos \theta}{\sin \theta},-\dfrac{2 r}{a}\right)\);
平面\(AB'B\)的一个法向量为\(\vec{h} =(0,0,1)\),
由题意可知,\(M-AB'-B\)的平面角β为锐二面角,
\(\therefore \cos \beta=\dfrac{|\vec{k} \cdot \vec{h}|}{|\vec{k}| \cdot|\vec{h}|}=\dfrac{\left|\dfrac{2 r}{a}\right|}{\sqrt{1+\left(\dfrac{-1-\cos \theta}{\sin \theta}\right)^{2}+\dfrac{4 r^{2}}{a^{2}}}}\).
由二面角的范围可知\(0≤α≤β≤π\).
即 \(\dfrac{\left|\dfrac{2 r}{a}\right|}{\sqrt{1+\dfrac{\cos ^{2} \theta}{\sin ^{2} \theta}+\dfrac{4 r^{2}}{a^{2}}}} \geq \dfrac{\left|\dfrac{2 r}{a}\right|}{\sqrt{1+\left(\dfrac{-1-\cos \theta}{\sin \theta}\right)^{2}+\dfrac{4 r^{2}}{a^{2}}}}\),
化简可得 \(\cos \theta \leq-\dfrac{1}{2}\),且\(0<θ<π\),则 \(0<\theta \leq \dfrac{2 \pi}{3}\).
\(∴θ\)的最大值是 \(\dfrac{2 \pi}{3}\).
故选:\(B\).
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答案 (1)略 (2) \(\dfrac{3 \sqrt{5}}{10}\)
解析 (1)证明:连接\(DO\).设\(EH=4a\),则 \(E F=4 \sqrt{2} a\),翻折后的\(BD=DE+FB=4a\).
在\(△SAC\)中,\(SA=SC=2\sqrt{2} a\),\(AC=4a\),\(O\)为\(AC\)的中点,
\(∴SO=2a\).
又\(∵\)在\(△SOB\)中,\(BS=2a\),\(SP⊥BO\),
\(∴P\)为\(BO\)的中点,
\(∴SP∥DO\).
\(∵SP⊄\)平面\(ACD\),\(DO⊂\)平面\(ACD\),
\(∴SP∥\)平面\(ACD\).
(2)解: \(\because V_{B-D M N}=V_{D-B M N}\)且三棱锥\(D-BMN\)的高为定值,
\(\therefore S_{\triangle B M N}\)最大时,三棱锥\(B-DMN\)的体积取得最大值.
设\(AM=BN=x(0≤x≤2\sqrt{3} a)\),
\(\therefore S_{\triangle B M N}=\dfrac{1}{2} B M \cdot B N \cdot \sin \angle M B N=\dfrac{1}{2} x(2 \sqrt{3} a-x) \sin \angle M B N\)
\(=\dfrac{1}{2}\left[-(x-\sqrt{3} a)^{2}+3 a^{2}\right] \sin \angle M B N\)
又 \(\because \sin \angle M B N\)为定值.
\(∴\)当 \(x=\sqrt{3} a\)时, \(S_{\triangle B M N}\)最大,即三棱锥\(B-DMN\)的体积最大.
此时\(M\),\(N\)分别是\(AB\),\(BC\)上的中点,
由(1)可得\(SP∥DO\),\(SP⊥BO\),\(∴DO⊥BO\).
\(∵DA=DC\),\(BA=BC\),\(∴DO⊥AC\),\(BO⊥AC\).
以\(O\)为坐标原点, \(\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O D}\)分别为\(x,y,z\)轴的正方向建立空间直角坐标系\(O-xyz\),
则\(A(2a,0,0)\),\(B(0,2\sqrt{2} a,0)\),\(C(-2a,0,0)\),\(D(0,0,2\sqrt{2}a)\),\(M(a,\sqrt{2}a,0)\),
\(N=(-a,\sqrt{2} a,0)\), \(\overrightarrow{D M}=(a, \sqrt{2} a,-2 \sqrt{2} a)\), \(\overrightarrow{N M}=(2 a, 0,0,)\), \(\overrightarrow{D A}=(2 a, 0,-2 \sqrt{2} a)\), \(\overrightarrow{A B}=(-2 a, 2 \sqrt{2} a, 0)\).
设平面\(DMN\)的一个法向量为 \(\overrightarrow{n_{1}}=\left(x_{1}, y_{1}, z_{1}\right)\).
\(\therefore\left\{\begin{array}{l} \overrightarrow{n_{1}} \cdot \overrightarrow{D M}=0 \\ \overrightarrow{n_{1}} \cdot N M=0 \end{array}\right.\),
\(\therefore\left\{\begin{array}{l} a x_{1}+\sqrt{2} a y_{1}-2 \sqrt{2} a z_{1}=0 \\ 2 a x_{1}=0 \end{array}\right.\),取\(z_1=1\),则\(y_1=2\),\(x_1=0\),
\(∴\)平面\(DMN\)的一个法向量为 \(\overrightarrow{n_{1}}=(0,2,1)\).
设平面\(DAB\)的一个法向量为 \(\overrightarrow{n_{2}}=\left(x_{2}, y_{2}, z_{2}\right)\).
\(\therefore\left\{\begin{array}{l} \overrightarrow{n_{2}} \cdot \overrightarrow{D A}=0 \\ \overrightarrow{n_{2}} \cdot \overrightarrow{A B}=0 \end{array}\right.\),
\(\therefore\left\{\begin{array}{l} 2 a x_{2}+2 \sqrt{2} a z_{2}=0 \\ -2 a x_{2}+2 \sqrt{2} a y_{2}=0 \end{array}\right.\)取 \(x_{2}=\sqrt{2}\),则\(y_2=z_2=1\),
\(∴\)平面\(DAB\)的一个法向量为 \(\overrightarrow{n_{2}}=(\sqrt{2}, 1,1)\).
则 \(\cos <\overrightarrow{n_{1}}, \overrightarrow{n_{2}}>=\dfrac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\dfrac{3 \sqrt{5}}{10}\).
所以平面\(DAB\)与平面\(DMN\)所成锐二面角的余弦值为 \(\dfrac{3 \sqrt{5}}{10}\).