1.4.2(3) 用空间向量研究距离问题
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【基础过关系列】2022-2023学年高二数学上学期同步知识点剖析精品讲义(人教A版2019)
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选择性必修第一册同步巩固,难度3颗星!
基础知识
点A、B间的距离
\(A B=|\overrightarrow{A B}|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}\).
【例】若\(P(0,1,-2)\),\(Q(1,3,-1)\),则\(|PQ|=\)\(\underline{\quad \quad}\) .
解析 \(|P Q|=\sqrt{1+4+1}=\sqrt{6}\).
点Q到直线l 距离
若\(Q\)为直线\(l\)外的一点,\(P\)在直线上,\(\vec{a}\)为直线\(l\)的方向向量, \(\vec{b}=\overrightarrow{P Q}\),则点\(Q\)到直线\(l\)距离为
\(d=\dfrac{1}{|\vec{a}|} \sqrt{(|\vec{a}||\vec{b}|)^{2}-(\vec{a} \cdot \vec{b})^{2}}\)
公式推导
如图, \(d=|\vec{b}| \sin \theta=|\vec{b}| \sqrt{1-\cos ^{2} \theta}=|\vec{b}| \sqrt{1-\left(\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)^{2}}=\dfrac{1}{|\vec{a}|} \sqrt{(|\vec{a}||\vec{b}|)^{2}-(\vec{a} \cdot \vec{b})^{2}}\) .
注 ① 不要死记公式,而要理解其公式推导过程;
② 也可先求\(PQ\)在直线\(l\)上的投影,再用勾股定理求出距离\(d\).
【例】已知直线\(l\)的方向向量为 \(\vec{a}=(-1,0,1)\),点\(A(1,2,-1)\)在\(l\)上,则点\(P(2,-1,2)\)到\(l\)的距离为\(\underline{\quad \quad}\) .
解析 根据题意,得 \(\overrightarrow{P A}=(-1,3,-3)\), \(\vec{a}=(-1,0,1)\),
\(\therefore \cos <\vec{a}, \overrightarrow{P A}>=\dfrac{1+0-3}{\sqrt{2} \times \sqrt{19}}=-\sqrt{\dfrac{2}{19}}\),
\(\therefore \sin <\vec{a}, \overrightarrow{P A}>=\sqrt{\dfrac{17}{19}}\);
又 \(\because|\overrightarrow{P A}|=\sqrt{19}\),
\(∴\)点\(P(2,-1,2)\)到直线\(l\)的距离为 \(|\overrightarrow{P A}| \sin <\vec{a}, \overrightarrow{P A}>=\sqrt{19} \times \sqrt{\dfrac{17}{19}}=\sqrt{17}\).
平行线m与平行线n的距离
在直线\(m\)上任取一点\(Q\),再求点\(Q\)到直线\(n\)的距离便可.
直线点Q到平面α的距离
若点\(Q\)为平面\(α\)外一点,点\(M\)为平面\(α\)内任一点,平面\(α\)的法向量为 \(\vec{n}\),则Q到平面\(α\)的距离等于 \(\overrightarrow{M Q}\)在法向量 \(\vec{n}\)方向上的投影的绝对值,即 \(d=\dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}|}\) .
公式推导
如图, \(d=|\overrightarrow{M Q}| \sin \alpha=|\overrightarrow{M Q}||\cos \langle\vec{n}, \overrightarrow{M Q}\rangle|=|\overrightarrow{M Q}| \cdot \dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}||\overrightarrow{M Q}|}=\dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}|}\) .
【例】 已知平面\(α\)的一个法向量 \(\vec{n}=(-2,-2,1)\),点\(A(-1,3,0)\)在\(α\)内,则\(P(-2,1,4)\)到\(α\)的距离为\(\underline{\quad \quad}\).
解析 根据题意,可得\(A(-1,3,0)\),\(P(-2,1,4)\),
\(\therefore \overrightarrow{P A}=(-1,-2,4)\),
又\(∵\)平面\(α\)的一个法向量 \(\vec{n}=\left(-2, -2,1\right)\),点\(A\)在\(α\)内,
\(∴P(-2,1,4)\)到\(α\)的距离等于向量 \(\overrightarrow{P A}\)在 \(\vec{n}\)上的投影的绝对值,
即 \(d=\dfrac{|\overrightarrow{P A} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{|-1 \times(-2)+(-2) \times(-2)+4 \times 1|}{\sqrt{4+4+1}}=\dfrac{10}{3}\).
直线 a平面α之间的距离
当一条直线和一个平面平行时,直线上的各点到平面的距离相等.由此可知,直线到平面的距离可转化为求直线上任一点到平面的距离,即转化为点面距离.
平面间的距离
利用两平行平面间的距离处处相等,可将两平行平面间的距离转化为求点面距离.
基本方法
【题型1】点到点的距离
【典题1】 如图,在空间直角坐标系中,有一棱长为\(2\)的正方体\(ABCD-A_1 B_1 C_1 D_1\),\(A_1 C\)的中点\(E\)到\(AB\)的中点\(F\)的距离为 ( )
A.\(2\sqrt{2}\) \(\qquad\qquad \qquad \qquad\) B. \(\sqrt{2}\) \(\qquad\qquad \qquad \qquad\) C.\(2\)\(\qquad \qquad \qquad\qquad\) D.\(1\)
解析 在空间直角坐标系中,有一棱长为\(2\)的正方体\(ABCD-A_1 B_1 C_1 D_1\)
\(∴A_1 (2,0,2)\),\(C(0,2,0)\),\(A_1 C\)的中点\(E(1,1,1)\),
\(A(2,0,0)\),\(B(2,2,0)\),\(AB\)的中点\(F(2,1,0)\),
\(∴A_1 C\)的中点\(E\)到\(AB\)的中点\(F\)的距离为 \(|E F|=\sqrt{(2-1)^{2}+(1-1)^{2}+(0-1)^{2}}=\sqrt{2}\).
故选:\(B\).
巩固练习
1若\(O\)为坐标原点, \(\overrightarrow{O A}=(1,1,-2)\), \(\overrightarrow{O B}=(3,2,8)\), \(\overrightarrow{O C}=(0,1,0)\),则线段\(AB\)的中点\(P\)到点\(C\)的距离为( )
A. \(\dfrac{\sqrt{165}}{2}\) \(\qquad \qquad\) B. \(2 \sqrt{14}\) \(\qquad \qquad\) C. \(\sqrt{53}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{53}}{2}\)
2如图,在正四棱柱\(ABCD-A_1 B_1 C_1 D_1\)中,\(AA_1=2\),\(AB=BC=1\),动点\(P\),\(Q\)分别在线段\(C_1 D\)、\(AC\)上,则线段\(PQ\)长度的最小值是\(\underline{\quad \quad}\).
参考答案
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答案 \(D\)
解析 \(\because \overrightarrow{O P}=\dfrac{\overrightarrow{O A}+\overrightarrow{O B}}{2}=\left(2, \dfrac{3}{2}, 3\right)\)
\(\therefore \overrightarrow{P C}=\overrightarrow{O C}-\overrightarrow{O P}=\left(-2,-\dfrac{1}{2},-3\right)\),
\(\therefore|P C|=\dfrac{\sqrt{53}}{2}\), 答案:\(D\) -
答案 \(\dfrac{2}{3}\)
解析 建立如图所示的空间直角坐标系,则\(A(1,0,0)\),\(B(1,1,0)\),\(C(0,1,0)\),\(C_1 (0,1,2)\),
设 \(\overrightarrow{D P}=\lambda \overrightarrow{D C_{1}}\), \(\overrightarrow{A Q}=\mu \overrightarrow{A C}\),\((λ,μ∈[0,1])\).
\(\therefore \overrightarrow{D P}=\lambda(0,1,2)=(0, \lambda, 2 \lambda)\),
\(\overrightarrow{D Q}=\overrightarrow{D A}+\mu(\overrightarrow{D C}-\overrightarrow{D A})=(1,0,0)+\mu(-1,1,0)=(1-\mu, \mu, 0)\).
\(\therefore|\overrightarrow{P Q}|=\sqrt{(1-\mu)^{2}+(\mu-\lambda)^{2}+4 \lambda^{2}}\)\(=\sqrt{5\left(\lambda-\dfrac{\mu}{5}\right)^{2}+\dfrac{9}{5}\left(\mu-\dfrac{5}{9}\right)^{2}+\dfrac{4}{9}} \geqslant \sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\),
当且仅当 \(\lambda=\dfrac{\mu}{5}, \mu=\dfrac{5}{9}\),即 \(\lambda=\dfrac{1}{9}, \mu=\dfrac{5}{9}\)时取等号.
\(∴\)线段\(PQ\)长度的最小值为 \(\dfrac{2}{3}\).
【题型2】点到线的距离
【典题1】 \(P\)为矩形\(ABCD\)所在平面外一点,\(PA⊥\)平面\(ABCD\),若已知\(AB=3\),\(AD=4\),\(PA=1\),则点\(P\)到\(BD\)的距离为\(\underline{\quad \quad}\).
解析 方法一 \(∵\)矩形\(ABCD\)中,\(AB=3\),\(AD=4\),
\(\therefore B D=\sqrt{9+16}=5\),
过\(A\)作\(AE⊥BD\),交\(BD\)于\(E\),连结\(PE\),
\(∵PA⊥\)平面\(ABCD\),\(∴PA⊥BD\),
又 \(AE⊥BD\) \(∴BD⊥\)平面\(PAE\),
\(∴PE⊥BD\),即\(PE\)是点\(P\)到\(BD\)的距离,
\(\because \dfrac{1}{2} \times A B \times A D=\dfrac{1}{2} \times B D \times A E\),
\(\therefore A E=\dfrac{A B \times A D}{B D}=\dfrac{12}{5}\),
\(\therefore P E=\sqrt{P A^{2}+E^{2}}=\sqrt{1+\dfrac{144}{25}}=\dfrac{13}{5}\),
\(∴\)点\(P\)到\(BD\)的距离为 \(\dfrac{13}{5}\).
方法二 依题意可知,\(PA\)、\(AB\)、\(AD\)三线两两垂直,
如图建立空间直角坐标系
\(∴P(0,0,1)\),\(B(3,0,0)\),\(D(0,4,0)\),
\(\therefore \overrightarrow{B P}=(-3,0,1)\), \(\overrightarrow{B D}=(-3,4,0)\),
\(\therefore \cos <\overrightarrow{B P}, \overrightarrow{B D}>=\dfrac{\overrightarrow{B P} \cdot \overrightarrow{B D}}{|\overrightarrow{B P}||\overrightarrow{B D}|}=\dfrac{9}{5 \sqrt{10}}\),
\(∴\)点\(P\)到\(BD\)的距离为 \(d=|\overrightarrow{B P}| \sqrt{1-\cos ^{2}<\overrightarrow{B P}, \overrightarrow{B D}>}=\sqrt{10} \cdot \sqrt{1-\dfrac{81}{250}}=\dfrac{13}{5}\).
巩固练习
1已知空间直角坐标系中的点\(P(1,1,1)\),\(A(1,0,1)\),\(B(0,1,0)\),则点\(P\)到直线\(AB\)的距离为\(\underline{\quad \quad}\) .
2 在棱长为\(1\)的正方体\(ABCD-A_1 B_1 C_1 D_1\)中,求点\(B_1\)到直线\(A_1 C\)的距离\(\underline{\quad \quad}\).
3 已知正方体\(ABCD-EFGH\)的棱长为\(1\),若\(P\)点在正方体的内部且满足 \(\overrightarrow{A P}=\dfrac{3}{4} \overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}+\dfrac{2}{3} \overrightarrow{A E}\),则点\(P\)到直线\(AB\)的距离为\(\underline{\quad \quad}\).
参考答案
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答案 \(\dfrac{\sqrt{6}}{3}\)
解析 \(\overrightarrow{A P}=(0,1,0)\), \(\overrightarrow{A B}=(-1,1,-1)\),
则点\(P\)到直线\(AB\)的距离 \(d=|\overrightarrow{A P}| \sqrt{1-[\cos \langle\overrightarrow{A P}, \overrightarrow{A B}\rangle]^{2}}=1 \times \sqrt{1-\left(\dfrac{1}{\sqrt{3}}\right)^{2}}=\dfrac{\sqrt{6}}{3}\). -
答案 \(\dfrac{\sqrt{6}}{3}\)
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,
则\(B_1 (1,1,1)\),\(A_1 (1,0,1)\),\(C(0,1,0)\),
\(\therefore \overrightarrow{A_{1} C}=(-1,1,-1)\), \(\overrightarrow{A_{1} B_{1}}=(0,1,0)\),
\(\therefore \cos <\overrightarrow{A_{1} C}, \overrightarrow{A_{1} B_{1}}>=\dfrac{\overrightarrow{A_{1} C} \cdot \overrightarrow{A_{1} B_{1}}}{\left|\overrightarrow{A_{1} C}\right| \cdot\left|\overrightarrow{A_{1} B_{1}}\right|}=\dfrac{1}{\sqrt{3} \cdot 1}=\dfrac{\sqrt{3}}{3}\),
\(\therefore \sin <\overrightarrow{A_{1} C}, \overrightarrow{A_{1} B_{1}}>=\dfrac{\sqrt{6}}{3}\),
则点\(B_1\)到直线\(A_1 C\)的距离 \(\left.d=\left|\overrightarrow{A_{1} B_{1}}\right| \sin <\overrightarrow{A_{1} C}, \overrightarrow{A_{1} B_{1}}\right\rangle=\dfrac{\sqrt{6}}{3}\). -
答案 \(\dfrac{5}{6}\)
解析 分别以\(AB\),\(AD\),\(AE\)为\(x\)轴,\(y\)轴,\(z\)轴作出空间直角坐标系,
\(∵\)正方体\(ABCD-EFGH\)的棱长为\(1\), \(\therefore \overrightarrow{A B}=(1,0,0)\),
\(\because \overrightarrow{A P}=\dfrac{3}{4} \overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}+\dfrac{2}{3} \overrightarrow{A E}\),
\(\therefore \overrightarrow{A P}=\left(\dfrac{3}{4}, \dfrac{1}{2}, \dfrac{2}{3}\right)\),
可得 \(\mid \overrightarrow{|A P|}=\sqrt{\left(\dfrac{3}{4}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}+\left(\dfrac{2}{3}\right)^{2}}=\dfrac{\sqrt{181}}{12}\),
\(\because \overrightarrow{A B} \cdot \overrightarrow{A P}=1 \times \dfrac{3}{4}+0 \times \dfrac{1}{2}+0 \times \dfrac{2}{3}=\dfrac{3}{4}\),
\(\overrightarrow{A B} \cdot \overrightarrow{A P}=\overrightarrow{|A B|} \cdot \overrightarrow{|A P|} \cos \angle P A B\),
\(\therefore \sin \angle P A B=\sqrt{1-\cos ^{2} \angle P A B}=\dfrac{10}{\sqrt{181}}\),
\(∴\)点\(P\)到直线\(AB\)的距离为 \(\overrightarrow{|A P|} \sin \angle P A B=\dfrac{\sqrt{181}}{12} \cdot \dfrac{10}{\sqrt{181}}=\dfrac{5}{6}\).
【题型3】点到面的距离
【典题1】在正三棱柱\(ABC-A_1 B_1 C_1\)中,若\(AB=AA_1=4\),点\(D\)是\(AA_1\)的中点,求点\(A_1\)到平面\(DBC_1\)的距离.
解析 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,
\(A_1 (0,0,4)\),\(D(0,0,2)\), \(B(2 \sqrt{3}, 2,0)\),\(C_1 (0,4,4)\),
\(\overrightarrow{D A}_{1}=(0,0,2)\), \(\overrightarrow{D B}=(2 \sqrt{3}, 2,-2)\), \(\overrightarrow{D C}_{1}=(0,4,2)\),
设平面\(DBC_1\)的法向量 \(\vec{n}=(x, y, z)\),
则 \(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{D B}=2 \sqrt{3} x+2 y-2 z=0 \\
\vec{n} \cdot \overrightarrow{D C_{1}}=4 y+2 z=0
\end{array}\right.\),取\(x= \sqrt{3}\),得\(\vec{n}=( \sqrt{3},-1,2)\),
\(∴\)点\(A_1\)到平面\(DBC_1\)的距离 \(d=\dfrac{\left|\overrightarrow{D A_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{4}{\sqrt{8}}=\sqrt{2}\).
【典题2】已知\(E\),\(F\)分别是正方形\(ABCD\)边\(AD\),\(AB\)的中点,\(EF\)交\(AC\)于\(P\),\(GC\)垂直于\(ABCD\)所在平面.
(1)求证:\(EF⊥\)平面\(GPC\).
(2)若\(AB=4\),\(GC=2\),求点\(B\)到平面\(EFG\)的距离.
解析 (1)连接\(BD\)交\(AC\)于\(O\),
\(∵E,F\)是正方形\(ABCD\)边\(AD\),\(AB\)的中点,\(AC⊥BD\),
\(∴EF⊥AC\).
\(∵GC\)垂直于\(ABCD\)所在平面,\(EF⊂\)平面\(ABCD\),
\(∴EF⊥GC\)
\(∵AC∩GC=C\), \(∴EF⊥\)平面\(GPC\).
(2) 方法一 间接法
由题意可知 \(P C=\dfrac{3}{4} A C=3 \sqrt{2}\), \(P G=\sqrt{P C^{2}+G C^{2}}=\sqrt{22}\),
\(∵PC=3OP\),
\(∴C\)到面\(GEF\)的距离是\(O\)到面\(GEF\)距离的3倍,
在\(∆GPC\)中,点\(C\)到边\(PG\)的高为\(CM\),
又\(∵EF⊥\)平面\(GPC\),\(∴CM⊥\)平面\(EFG\),
\(∴CM\)为\(C\)到面\(GEF\)距离,
在\(∆GPC\)中,可得 \(P G \cdot C M=G C \cdot P C \Rightarrow C M=\dfrac{2 \times 3 \sqrt{2}}{\sqrt{22}}=\dfrac{6}{\sqrt{11}}\),
又\(BD∥EF\),可得\(BD∥\)平面\(GEF\),
可得\(B\)到面\(GEF\)的距离等于\(O\)到面\(GEF\)的距离: \(\dfrac{1}{3} C M=\dfrac{2}{\sqrt{11}}=\dfrac{2 \sqrt{11}}{11}\).
故答案为: \(\dfrac{2 \sqrt{11}}{11}\).
方法二 向量法
建立空间直角坐标系\(C-xyz\),则\(G(0,0,2)\),\(E(4,2,0)\),\(F(2,4,0)\),\(B(4,0,0)\)
\(∴\)向量 \(\overrightarrow{G E}=(4,2,-2)\),向量 \(\overrightarrow{E F}=(-2,2,0)\)
设面\(GEF\)的法向量\(\vec{n}=(x,y,z)\)
则 \(\overrightarrow{G E} \cdot \vec{n}=0\)且 \(\overrightarrow{E F} \cdot \vec{n}=0\)
即\(4x+2y-2z=0\)且\(-2x+2y=0\)
取\(x=1\)时,向量\(\vec{n}=(1,1,3)\)
又\(∵\)向量 \(\overrightarrow{B E}=(0,2,0)\)
则\(B\)到面\(GEF\)的距离 \(d==\dfrac{|\vec{n} \cdot \overrightarrow{B E}|}{|\vec{n}|}=\dfrac{2 \sqrt{11}}{11}\).
方法三 等积法
同方法一可得 \(P G=\sqrt{22}\), \(\therefore S_{\Delta E F G}=\dfrac{1}{2} \times P G \times E F=2 \sqrt{11}\),
易得 \(S_{\triangle E F B}=\dfrac{1}{2} \times A F \times E B=2\)
\(\because V_{B-E F G}=V_{G-E F B}\),
\(\therefore \dfrac{1}{3} \times h \times S_{\triangle E F G}=\dfrac{1}{3} \times G C \times S_{\triangle E F B}\)
\(\therefore h=\dfrac{G C \times S_{\triangle E F B}}{S_{\triangle E F G}}=\dfrac{2 \times 2}{2 \sqrt{11}}=\dfrac{2 \sqrt{11}}{11}\).
巩固练习
1 在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AB=AD=2\),\(AA_1=1\),则点\(B\)到平面\(D_1 AC\)的距离等于( )
A.\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\) B.\(\dfrac{\sqrt{6}}{3}\) \(\qquad \qquad\) C.\(1\) \(\qquad \qquad\) D.\(\sqrt{2}\)
2 如图,在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AD=AA_1=1\),\(AB=2\),点\(E\)在棱\(AB\)上移动.
(1)证明:\(D_1 E⊥A_1 D\);
(2)当\(E\)为\(AB\)的中点时,求点\(E\)到面\(ACD_1\)的距离;
3 如图所示,四棱锥\(S-ABCD\)中,\(SA⊥\)底面\(ABCD\);\(∠ABC=90^°\),\(∠ACD=60^°\),\(AC=AD\),\(SA=2\), \(A B=\sqrt{3}\),\(BC=1\).
(1)求证:\(BC∥\)平面\(SAD\);(2)求顶点\(A\)到平面\(SCD\)的距离.
4 如图,在正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(E\)为\(BB_1\)的中点.
(1)证明:\(BC_1∥\)平面\(AD_1E\);(2)求直线\(BC_1\)到平面\(AD_1 E\)的距离.
参考答案
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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,
\(B(2,2,0)\),\(A(2,0,0)\),\(C(0,2,0)\),\(D_1 (0,0,1)\),
\(\overrightarrow{A B}=(0,2,0)\), \(\overrightarrow{A C}=(-2,2,0)\), \(\overrightarrow{A D}_{1}=(-2,0,1)\),
设平面\(D_1 AC\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A C}=-2 x+2 y=0 \\ \vec{n} \cdot A D_{1}=-2 x+z=0 \end{array}\right.\),取\(x=1\),得\(\vec{n}=(1,1,2)\),
\(∴\)点\(B\)到平面\(D_1 AC\)的距离: \(d=\dfrac{|\overrightarrow{A B} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}\).
故选:\(B\).
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答案 (1)略 (2) \(\dfrac{1}{3}\)
解析 以\(D\)为坐标原点,直线\(DA\),\(DC\),\(DD_1\)分别为\(x\),\(y\),\(z\)轴,建立空间直角坐标系,
设\(AE=x\),则\(A_1 (1,0,1)\),\(D_1 (0,0,1)\),\(E(1,x,0)\),\(A(1,0,0)\),\(C(0,2,0)\)
(1)因为 \(\overrightarrow{D A_{1}} \cdot \overrightarrow{D_{1} E}=(1,0,1) \cdot(1, x,-1)=0\),
所以 \(\overrightarrow{D A_{1}} \perp \overrightarrow{D_{1} E}\).
(2)因为\(E\)为\(AB\)的中点,则\(E(1,1,0)\),
从而 \(\overrightarrow{D_{1} E}=(1,1,-1)\), \(\overrightarrow{A C}=(-1,2,0)\), \(\overrightarrow{A D}_{1}=(-1,0,1)\),
设平面\(ACD_1\)的法向量为\(\vec{n}=(a,b,c)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A C}=0 \\ \vec{n} \cdot \overrightarrow{A D_{1}}=0 \end{array}\right.\),也即 \(\left\{\begin{array}{l} -a+2 b=0 \\ -a+c=0 \end{array}\right.\),得 \(\left\{\begin{array}{l} a=2 b \\ a=c \end{array}\right.\),从而\(\vec{n}=(2,1,2)\),
所以点\(E\)到平面\(AD_1 C\)的距离为 \(h=\dfrac{\left|\overrightarrow{D_{1} E} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{2+1-2}{3}=\dfrac{1}{3}\).
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答案 (1)略 (2) \(\dfrac{2 \sqrt{21}}{7}\)
解析 (1)证明:\(∵∠ABC=90^∘\),\(∠ACD=60^∘\),\(AC=AD\),\(SA=2\), \(A B=\sqrt{3}\),\(BC=1\).
\(∴△ADC\)是等边三角形, \(A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{3+1}=2\),
\(∴∠DAC=∠ACB=60^∘\),\(∴BC∥AD\),
\(∵BC⊄\)平面\(SAD\),\(AD⊂\)平面\(SAD\),
\(∴BC∥\)平面\(SAD\).
(2)解:\(∵\)四棱锥\(S-ABCD\)中,\(SA⊥\)底面\(ABCD\),\(∠ABC=90^∘\),\(BC∥AD\),
\(∴AD⊥AB\),
以\(A\)为原点,\(AB\)为\(x\)轴,\(AD\)为\(y\)轴,\(AS\)为\(z\)轴,建立空间直角坐标系,
\(A(0,0,0)\),\(S(0,0,2)\), \(C(\sqrt{3}, 1,0)\),\(D(0,2,0)\),
\(\overrightarrow{S A}=(0,0,-2)\), \(\overrightarrow{S C}=(\sqrt{3}, 1,-2)\), \(\overrightarrow{S D}=(0,2,-2)\),
设平面\(SCD\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{S C}=\sqrt{3} x+y-2 z=0 \\ \vec{n} \cdot \overrightarrow{S D}=2 y-2 z=0 \end{array}\right.\),取\(z=1\),得 \(\vec{n}=\left(\dfrac{\sqrt{3}}{3}, 1,1\right)\),
\(∴\)顶点\(A\)到平面\(SCD\)的距离为 \(d=\dfrac{|\vec{n} \cdot \overrightarrow{S A}|}{|\vec{n}|}=\dfrac{2}{\sqrt{\dfrac{7}{3}}}=\dfrac{2 \sqrt{21}}{7}\).
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答案 (1)略 (2) \(\dfrac{2}{3}\)
解析 证明:(1)\(∵D_1 C_1∥AB\),\(D_1 C_1=AB\),
\(∴\)四边形\(D_1 ABC_1\)为平行四边形,
\(∴D_1 A∥C_1 B\),
\(∵D_1 A⊂\)面\(AD_1 E\),\(C_1 B⊄\)面\(AD_1 E\),
\(∴BC_1∥\)平面\(AD_1 E\).
解:(2)如图建立空间直角坐标系\(A-xyz\),设正方体的棱长为\(2\),
则\(A(0,0,0)\),\(B(0,2,0)\),\(D_1 (2,0,2)\),\(C_1 (2,2,2)\),\(E(0,2,1)\),
\(∵BC_1∥\)平面\(AD_1 E\),
\(∴\)直线\(BC_1\)到平面\(AD_1 E\)的距离即为点\(B\)到平面\(AD_1 E\)的距离,
\(\overrightarrow{A B}=(0,2,0)\), \(\overrightarrow{A D}_{1}=(2,0,2)\), \(\overrightarrow{A E}=(0,2,1)\),
设平面\(AD_1 E\)的一个法向量为\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A D_{1}}=2 x+2 z=0 \\ \vec{n} \cdot \overrightarrow{A E}=2 y+z=0 \end{array}\right.\),取\(z=-1\),得 \(\vec{n}=\left(1, \dfrac{1}{2},-1\right)\),
\(\therefore d=\dfrac{|\overrightarrow{A B} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{\left|(0,2,0) \cdot\left(1, \dfrac{1}{2},-1\right)\right|}{\sqrt{1+1+\dfrac{1}{4}}}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}\),
\(∴\)直线\(BC_1\)到平面\(AD_1 E\)的距离为 \(\dfrac{2}{3}\);
分层练习
【A组---基础题】
1 已知\(M\)为\(z\)轴上一点,且点\(M\)到点\(A(-1,0,1)\)与点\((1,-3,2)\)的距离相等,则点\(M\)的坐标为( )
A.\((3,0,0)\) \(\qquad \qquad\) B.\((0,-2,0)\) \(\qquad \qquad\) C.\((0,0,6)\) \(\qquad \qquad\) D.\((0,0,-3)\)
2 已知\(A(0,0,2)\),\(B(1,0,2)\),\(C(0,2,0)\),则点\(A\)到直线\(BC\)的距离为( )
A. \(\dfrac{2 \sqrt{2}}{3}\) \(\qquad \qquad\) B.\(1\) \(\qquad \qquad\) C. \(\sqrt{2}\) \(\qquad \qquad\) D. \(2 \sqrt{2}\)
3 已知平面\(α\)的法向量为 \(\vec{n}=(-2,-2,1)\),点\(A(x,3,0)\)在平面\(α\)内,则点\(P(-2,1,4)\)到平面\(α\)的距离为 \(\dfrac{10}{3}\),则\(x=\)( )
A.\(-1\)\(\qquad \qquad\) B.\(-11\) \(\qquad \qquad\) C.\(-1\)或\(-11\)\(\qquad \qquad\) D.\(-21\)
4 在直三棱柱\(ABC-A_1 B_1 C_1\)中,\(AB=AC=AA_1=2\),\(∠BAC=90^∘\),\(M\)为\(BB_1\)的中点,\(N\)为\(BC\)的中点.
(1)求点\(M\)到直线\(AC_1\)的距离;(2)求点\(N\)到平面\(MA_1 C_1\)的距离.
5 如图,在四棱锥\(P-ABCD\)中,底面\(ABCD\)为矩形,侧棱\(PA⊥\)底面\(ABCD\),\(AB=\sqrt{3}\),\(BC=1\),\(PA=2\),\(E\)为\(PD\)的中点.
(1)求点\(C\)到平面\(PBD\)的距离;
(2)在侧面\(PAB\)内找一点\(N\),使\(NE⊥\)面\(PAC\),并求出\(N\)点到\(AB\)和\(AP\)的距离.
6 如图,已知斜三棱柱\(ABC-A_1 B_1 C_1\),\(∠BCA=90°\),\(AC=BC=2\),\(A_1\)在底面\(ABC\)上的射影恰为\(AC\)的中点\(D\),又知\(BA_1⊥AC_1\).
(1)求证:\(AC_1⊥\)平面\(A_1 BC\); (2)求\(CC_1\)到平面\(A_1 AB\)的距离.
参考答案
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答案 \(C\)
解析 \(∵M\)为\(z\)轴上一点,\(∴\)设\(M(0,0,t)\),
\(∵\)点\(M\)到点\(A(-1,0,1)\)与点\((1,-3,2)\)的距离相等,
\(\therefore \sqrt{(0+1)^{2}+(0-0)^{2}+(t-1)^{2}}\)\(=\sqrt{(0-1)^{2}+(0+3)^{2}+(t-2)^{2}}\),
解得\(t=6\),
\(∴\)点\(M\)的坐标为\(M(0,0,6)\).
故选:\(C\). -
答案 \(A\)
解析 \(∵A(0,0,2)\),\(B(1,0,2)\),\(C(0,2,0)\), \(\overrightarrow{A B}=(1,0,0)\), \(\overrightarrow{B C}=(-1,2,-2)\),
\(∴\)点\(A\)到直线\(BC\)的距离为: \(d=|\overrightarrow{A B}| \sqrt{1-(\cos \langle\overrightarrow{A B}, \overrightarrow{B C}\rangle)^{2}}=1 \times \sqrt{1-\left(\dfrac{-1}{1 \times 3}\right)^{2}}=\dfrac{2 \sqrt{2}}{3}\).
故选:\(A\). -
答案 \(C\)
解析 \(\overrightarrow{A P}=(-2-x,-2,4)\), \(|\overrightarrow{A P}|=\sqrt{(-2-x)^{2}+(-2)^{2}+4^{2}}=\sqrt{x^{2}+4 x+24}\),
\(|\vec{n}|=\sqrt{4+4+1}=3\), \(\overrightarrow{A P} \cdot \vec{n}=-2(-2-x)+4+4=2 x+12\),
\(\therefore \cos <\overrightarrow{A P}, \vec{n}>=\dfrac{\overrightarrow{A P} \cdot \vec{n}}{|\overrightarrow{A P}||\vec{n}|}=\dfrac{2 x+12}{\sqrt{x^{2}+4 x+24} \times 3}\),
设\(AP\)与平面\(α\)所成角为\(θ\),则 \(\sin \theta=\dfrac{|2 x+12|}{3 \sqrt{x^{2}+4 x+24}}\),
\(∴P\)到平面\(α\)的距离为 \(|A P| \cdot \sin \theta=\dfrac{|2 x+12|}{3}=\dfrac{10}{3}\),解得\(x=-1\)或\(x=-11\).
故选:\(C\). -
答案 (1) \(\dfrac{3 \sqrt{2}}{2}\)(2) \(\dfrac{3 \sqrt{5}}{5}\)
解析 连接\(AM\),建立如图的空间直角坐标系,
则\(A(0,0,0)\),\(A_1 (0,0,2)\),\(M(2,0,1)\),\(C_1 (0,2,2)\),
直线\(AC_1\)的一个单位方向向量为 \(\overrightarrow{\boldsymbol{s}_{0}}=\left(0, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)\), \(\overrightarrow{A M}=(2,0,1)\),
故点\(M\)到直线\(AC_1\)的距离 \(d=\sqrt{|\overrightarrow{\mathrm{AM}}|^{2}-\left|\overrightarrow{\mathrm{AM}} \cdot s_{0}\right|^{2}}=\sqrt{5-\dfrac{1}{2}}=\dfrac{3 \sqrt{2}}{2}\);
连接\(MN\),设平面\(MA_1 C_1\)的法向量为\(\vec{n}=(x,y,z)\),
则 \(\vec{n} \cdot \overrightarrow{A_{1} C_{1}}=0\),且 \(\vec{n} \cdot \overrightarrow{A_{1} M}=0\),
即\((x,y,z)⋅(0,2,0)=0\)且\((x,y,z)⋅(2,0,-1)=0\) ,
即\(y=0\)且\(2x-z=0\),取\(x=1\),得\(z=2\),故\(\vec{n}=(1,0,2)\),
与 \(\vec{n}\)同向的单位向量为 \(n_{0}=\left(\dfrac{\sqrt{5}}{5}, 0, \dfrac{2 \sqrt{5}}{5}\right)\),
因为\(N(1,1,0)\),所以 \(\overrightarrow{M N}=(-1,1,-1)\),
故求点\(N\)到平面\(MA_1 C_1\)的距离 \(d=\left|\overrightarrow{M N} \cdot n_{0}\right|=\dfrac{3 \sqrt{5}}{5}\). -
答案 (1)( \(\dfrac{2 \sqrt{57}}{19}\)
(2)\(N\)点的坐标为 \(\left(\dfrac{\sqrt{3}}{6}, 0,1\right)\),从而\(N\)点到\(AB,AP\)的距离分别为\(1\), \(\dfrac{\sqrt{3}}{6}\)
解析 (1)建立如图所示的空间直角坐标系,则\(A(0,0,0)\)、\(B(\sqrt{3},0,0)\)、\(C(\sqrt{3},1,0)\)、\(D(0,1,0)\)、\(P(0,0,2)\)、 \(E\left(0, \dfrac{1}{2}, 1\right)\),
从而 \(\overrightarrow{P D}=(0,1,-2)\), \(\overrightarrow{P B}=(\sqrt{3}, 0,-2)\), \(\overrightarrow{P C}=(\sqrt{3}, 1,-2)\)
设平面\(PBD\)的一个法向量为\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{P D}=0 \\ \vec{n} \cdot \overrightarrow{P B}=0 \end{array}\right.\),即 \(\left\{\begin{array}{l} y-2 z=0 \\ \sqrt{3} x-2 z=0 \end{array}\right.\),令\(z=1\),得 \(\vec{n}=\left(\dfrac{2}{\sqrt{3}}, 2,1\right)\),
所以点\(C\)到平面\(PBD\)的距离 \(d=\dfrac{|\overrightarrow{P C} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{2 \sqrt{57}}{19}\).
(2)由于\(N\)点在侧面\(PAB\)内,故可设\(N\)点坐标为\((x,0,z)\),则 \(\overrightarrow{N E}=\left(-x, \dfrac{1}{2}, 1-z\right)\),
由\(NE⊥\)面\(PAC\)可得, \(\left\{\begin{array}{l} \overrightarrow{N E} \cdot \overrightarrow{A P}=0 \\ \overrightarrow{N E} \cdot \overrightarrow{A C}=0 \end{array}\right.\),即 \(\left\{\begin{array}{l} z-1=0 \\ -\sqrt{3} x+\dfrac{1}{2}=0 \end{array}\right.\)
\(\therefore x=\dfrac{\sqrt{3}}{6}\),\(z=1\).
即\(N\)点的坐标为 \(\left(\dfrac{\sqrt{3}}{6}, 0,1\right)\),从而\(N\)点到\(AB,AP\)的距离分别为\(1\), \(\dfrac{\sqrt{3}}{6}\).
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答案 (1) 略 (2) \(\dfrac{2 \sqrt{21}}{7}\)
解析 (1)\(∵A_1\)在底面\(ABC\)上的射影为\(AC\)的中点\(D\),
\(∴\)平面\(A_1 ACC_1⊥\)平面\(ABC\),
\(∵BC⊥AC\)且平面\(A_1 ACC_1∩\)平面\(ABC=AC\),
\(∴BC⊥\)平面\(A_1 ACC_1\),
\(∴BC⊥AC_1\),
\(∵AC_1⊥BA_1\)且\(BC∩BA_1=B\),
\(∴AC_1⊥\)平面\(A_1 BC\).
(2)如图所示,以\(C\)为坐标原点建立空间直角坐标系,
\(∵AC_1⊥\)平面\(A_1 BC\),\(∴AC_1⊥A_1 C\),
\(∴\)四边形\(A_1 ACC_1\)是菱形,
\(∵D\)是\(AC\)的中点,\(A_1 D⊥AC\)
\(∴∠A_1 AD=60°\),
\(∴A(2,0,0)\), \(A_{1}(1,0, \sqrt{3})\),\(B(0,2,0)\), \(C_{1}(-1,0, \sqrt{3})\),
\(\therefore \overrightarrow{A_{1} A}=(1,0,-\sqrt{3})\), \(\overrightarrow{A B}=(-2,2,0)\),
设平面\(A_1 AB\)的法向量\(\vec{n}=(x,y,z)\),则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A_{1}} A=0 \\ \vec{n} \cdot \overrightarrow{A B}=0 \end{array}\right.\),
\(\therefore\left\{\begin{array}{l} x-\sqrt{3} z=0 \\ -2 x+2 y=0 \end{array}\right.\),取 \(\vec{n}=(\sqrt{3}, \quad \sqrt{3}, 1)\),
\(\because \overrightarrow{C_{1} A_{1}}=(2,0,0)\),
\(∴C_1\)到平面\(A_1 AB\)的距离 \(d=\dfrac{\left|\overrightarrow{C_{1} A_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{2 \sqrt{21}}{7}\).
\(∵CC_1//AA_1\),\(AA_1⊂\)平面\(A_1 AB\),\(CC_1⊄\)平面\(A_1 AB\)
\(∴CC_1//\)平面\(A_1 AB\),
\(∴CC_1\)到平面\(A_1 AB\)的距离等于\(C_1\)到平面\(A_1 AB\)的距离 \(\dfrac{2 \sqrt{21}}{7}\).
【B组---提高题】
1已知三棱锥\(S-ABC\),满足\(SA\),\(SB\),\(SC\)两两垂直,且\(SA=SB=SC=2\),\(Q\)是三棱锥\(S-ABC\)外接球上一动点,则点\(Q\)到平面\(ABC\)的距离的最大值为\(\underline{\quad \quad}\).
2 如图,四棱锥\(P-ABCD\)中,底面\(ABCD\)为菱形,\(∠ABC=60^°\),\(PA⊥\)平面\(ABCD\),\(AB=2\), \(P A=\dfrac{2 \sqrt{3}}{3}\),\(E\)为\(BC\)中点,\(F\)在棱\(PD\)上,\(AF⊥PD\),点\(B\)到平面\(AEF\)的距离为\(\underline{\quad \quad}\).
3 如图,三棱锥\(A-BCD\)中,\(E\),\(F\)分别是棱\(BC\),\(CD\)上的点,且\(EF∥\)平面\(ABD\).
(1)求证:\(BD∥\)平面\(AEF\);
(2)若\(AE⊥\)平面\(BCD\),\(DE⊥BC\),\(BE=DE=2AE=4\),\(P\)为线段\(DE\)的中点,求\(P\)到直线\(AB\)的距离.
参考答案
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答案 \(\dfrac{4 \sqrt{3}}{3}\)
解析 \(∵\)锥\(S-ABC\),满足\(SA\),\(SB\),\(SC\)两两垂直,且\(SA=SB=SC=2\),
\(∴\)如图,\(SA\),\(SB\),\(SC\)是棱长为\(2\)的正方体\(MNPB-ADCS\)上具有公共顶点\(S\)的三条棱,
以\(B\)为原点,\(BM\)、\(BP\)、\(BS\)分别为\(x\)轴,\(y\)轴,\(z\)轴,建立空间直角坐标系,
则\(B(0,0,0)\),\(A(2,0,2)\),\(C(0,2,2)\),\(S(0,0,2)\),\(N(2,2,0)\),
\(\overrightarrow{B A}=(2,0,2)\), \(\overrightarrow{B C}=(0,2,2)\), \(\overrightarrow{B N}=(2,2,0)\),
设平面\(ABC\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{B A}=2 x+2 z=0 \\ \vec{n} \cdot \overrightarrow{B C}=2 y+2 z=0 \end{array}\right.\),取\(x=1\),得\(\vec{n}=(1,1,-2)\),
三棱锥\(S-ABC\)外接球就是棱长为\(2\)的正方体\(MNPB-ADCS\)的外接球,
\(∵Q\)是三棱锥\(S-ABC\)外接球上一动点,
\(∴\)点\(Q\)与\(N\)重合时,点\(Q\)到平面\(ABC\)的距离的最大值,
\(∴\)点\(Q\)到平面\(ABC\)的距离的最大值为: \(d=\dfrac{|\overrightarrow{B N} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{|2+2+0|}{\sqrt{3}}=\dfrac{4 \sqrt{3}}{3}\). -
答案 \(\dfrac{\sqrt{3}}{2}\)
解析 \(∵\)四棱锥\(P-ABCD\)中,底面\(ABCD\)为菱形,\(∠ABC=60^°\),\(PA⊥\)平面\(ABCD\),
\(∴\)以\(A\)为原点,\(AE\)为\(x\)轴,\(AD\)为\(y\)轴,\(AP\)为\(z\)轴,建立空间直角坐标系,
\(∵AB=2\), \(P A=\dfrac{2 \sqrt{3}}{3}\),\(E\)为\(BC\)中点,\(F\)在棱\(PD\)上,\(AF⊥PD\),
\(∴A(0,0,0)\), \(B(\sqrt{3},-1,0)\), \(E(\sqrt{3}, 0,0)\), \(P\left(0,0, \dfrac{2 \sqrt{3}}{3}\right)\),\(D(0,2,0)\),
设\(F(a,b,c)\), \(\overrightarrow{P F}=\lambda \overrightarrow{P D}\),
则 \(\left(a, b, c-\dfrac{2 \sqrt{3}}{3}\right)=\left(0,2 \lambda,-\dfrac{2 \sqrt{3}}{3} \lambda\right)\),解得\(a=0\),\(b=2λ\), \(c=\dfrac{2 \sqrt{3}}{3}-\dfrac{2 \sqrt{3}}{3} \lambda\),
\(\therefore \overrightarrow{A F}=\left(0,2 \lambda, \dfrac{2 \sqrt{3}}{3}-\dfrac{2 \sqrt{3}}{3} \lambda\right)\), \(\overrightarrow{P D}=\left(0,2,-\dfrac{2 \sqrt{3}}{3}\right)\),
\(∵AF⊥PD\), \(\therefore \overrightarrow{A F} \cdot \overrightarrow{P D}=4 \lambda-\dfrac{4}{3}+\dfrac{4}{3} \lambda=0\),
解得 \(\lambda=\dfrac{1}{4}\), \(\therefore \overrightarrow{A B}=(\sqrt{3},-1,0)\), \(\overrightarrow{A E}=(\sqrt{3}, 0,0)\), \(\overrightarrow{A F}=\left(0, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),
设平面\(AEF\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{A E}=\sqrt{3} x=0 \\ \vec{n} \cdot \overrightarrow{A F}=\dfrac{1}{2} y+\dfrac{\sqrt{3}}{2} z=0 \end{array}\right.\),取 \(y=\sqrt{3}\),得 \(\vec{n}=(0, \sqrt{3},-1) \text {, }\),
\(∴\)点\(B\)到平面\(AEF\)的距离为: \(d=\dfrac{|\vec{n} \cdot \overrightarrow{A B}|}{|\vec{n}|}=\dfrac{\sqrt{3}}{2}\). -
答案 (1) 略 (2) \(\dfrac{6 \sqrt{5}}{5}\)
解析 (1)证明:\(∵EF∥\)平面\(ABD\),\(EF⊂\)平面\(BCD\),平面\(BCD∩\)平面\(ABD=BD\),
\(∴EF∥BD\),
又\(BD⊄\)平面\(AEF\),\(EF⊂\)平面\(AEF\),
\(∴BD∥\)平面\(AEF\).
(2)因为\(AE⊥\)平面\(BCD\),\(BE⊂\)平面\(BCD\),\(DE⊂\)平面\(BCD\),
所以有\(AE⊥BE\),\(AE⊥DE\),又有\(DE⊥BC\),
所以可以建立以\(E\)为坐标原点,\(ED\)为\(x\)轴,\(EB\)为\(y\)轴,\(EA\)为\(z\)轴建立如下图所示的空间直角坐标系,
又由\(BE=DE=2AE=4\),\(P\)为线段\(DE\)的中点
可得各点坐标为\(E(0,0,0)\),\(B(0,4,0)\),\(D(4,0,0)\),\(A(0,0,2)\),\(P(2,0,0)\),
即 \(\overrightarrow{A P}=(2,0,-2)\), \(\overrightarrow{A B}=(0,4,-2)\)
过\(P\)点作\(PH\)垂直于\(AB\)交\(AB\)于\(H\),
所以\(A\)到垂足\(H\)的距离 \(d=\dfrac{|\overrightarrow{A P} \cdot \overrightarrow{A B}|}{|\overrightarrow{A B}|}=\dfrac{|2 \times 0+0 \times 4+(-2) \times(-2)|}{\sqrt{0^{2}+4^{2}+(-2)^{2}}}=\dfrac{2 \sqrt{5}}{5}\),
所以\(P\)到直线\(AB\)的距离为 \(\sqrt{|\overrightarrow{A P}|^{2}-d^{2}}=\sqrt{8-\left(\dfrac{2 \sqrt{5}}{5}\right)^{2}}=\dfrac{6 \sqrt{5}}{5}\).
【C组---拓展题】
1 正方体\(ABCD-A_1 B_1 C_1 D_1\)的棱长为\(a\),则平面\(AB_1 D_1\)与平面\(BDC_1\)的距离为\(\underline{\quad \quad}\).
2 如图在四棱锥\(P-ABCD\)中,侧面\(PAD⊥\)底面\(ABCD\),侧棱 \(P A=P D=\sqrt{2}\),底面\(ABCD\)为直角梯形,其中\(BC∥AD\),\(AB⊥AD\),\(AD=2AB=2BC=2\),\(O\)为\(AD\)的中点.
(1)求证\(PO⊥\)平面\(ABCD\);
(2)求二面角\(C-PD-A\)夹角的正弦值;
(3)线段\(AD\)上是否存在\(Q\),使得它到平面\(PCD\)的距离为 \(\dfrac{\sqrt{3}}{2}\)?若存在,求出 \(\dfrac{A Q}{Q D}\)的值;若不存在,说明理由.
参考答案
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答案 \(\dfrac{\sqrt{3}}{3} a\)
解析 建立空间直角坐标系如图.
则\(A(a,0,0)\),\(B(a,a,0)\),\(D(0,0,0)\),\(C_1 (0,a,a)\),\(D_1 (0,0,a)\),\(B_1 (a,a,a)\),
\(\therefore \overrightarrow{A B_{1}}=(0, a, a)\), \(\overrightarrow{A D_{1}}=(-a, 0, a)\), \(\overrightarrow{B C_{1}}=(-a, 0, a)\), \(\overrightarrow{D C_{1}}=(0, a, a)\)
设\(\vec{n}=(x,y,z)\)为平面\(AB_1 D_1\)的法向量,
则 \(\left\{\begin{array}{c} \vec{n} \cdot \overrightarrow{A B_{1}}=a(y+z)=0 \\ \vec{n} \cdot \overrightarrow{A D_{1}}=a(-x+z)=0 \end{array}\right.\) 得 \(\left\{\begin{array}{c} y=-z \\ x=z \end{array}\right.\) 令\(z=1\),则\(\vec{n}=(1,-1,1)\)
\(\because \overrightarrow{A D_{1}} / / \overrightarrow{B C_{1}}\), \(\overrightarrow{A B_{1}} / / \overrightarrow{D C_{1}}\),
\(∴AD_1∥BC_1\),\(AB_1∥DC_1\),\(AD_1∩AB_1=A\),\(DC_1∩BC_1=C_1\),
\(∴\)平面\(AB_1 D_1∥\)平面\(BDC_1\).
\(∴\)平面\(AB_1 D_1\)与平面\(BDC_1\)的距离可转化为点\(C_1\)到平面\(AB_1 D_1\)的距离\(d\).
\(\because \overrightarrow{C_{1} B_{1}}=(a, 0,0)\),平面\(AB_1 D_1\)的法向量为\(\vec{n}=(1,-1,1)\)
\(\therefore d=\dfrac{\left|\overrightarrow{C_{1} B_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{\sqrt{3}}{3} a\). -
答案 (1) 略 (2) \(\dfrac{\sqrt{6}}{3}\) (3) \(\dfrac{1}{3}\)
解析 证明:(1)\(∵\)侧棱 \(P A=P D=\sqrt{2}\),\(O\)为\(AD\)的中点,
\(∴PO⊥AD\),
\(∵\)侧面\(PAD⊥\)底面\(ABCD\),侧面\(PAD∩\)底面\(ABCD=AD\),
\(∴PO⊥\)平面\(ABCD\).
解:(2)\(∵\)底面\(ABCD\)为直角梯形,其中\(BC∥AD\),\(AB⊥AD\),\(AD=2AB=2BC=2\),
\(∴OC⊥AD\),又\(PO⊥\)平面\(ABCD\),
\(∴\)以\(O\)为原点,\(OC\)为\(x\)轴,\(OD\)为\(y\)轴,\(OP\)为\(z\)轴,建立空间直角坐标系,
平面\(PAD\)的法向量\(\vec{m}=(1,0,0)\),
\(C(1,0,0)\),\(D(0,1,0)\),\(P(0,0,1)\), \(\overrightarrow{P C}=(1,0,-1)\), \(\overrightarrow{P D}=(0,1,-1)\),
设平面\(PCD\)的法向量\(\vec{n}=(x,y,z)\),
则 \(\left\{\begin{array}{l} \vec{n} \cdot \overrightarrow{P C}=x-z=0 \\ \vec{n} \cdot \overrightarrow{P D}=y-z=0 \end{array}\right.\),取\(x=1\),得\(\vec{n}=(1,1,1)\),
设二面角\(C-PD-A\)夹角为\(θ\),
则 \(\cos \theta=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{3}}\),
\(\therefore \sin \theta=\sqrt{1-\left(\dfrac{1}{\sqrt{3}}\right)^{2}}=\dfrac{\sqrt{6}}{3}\),
\(∴\)二面角\(C-PD-A\)夹角的正弦值为 \(\dfrac{\sqrt{6}}{3}\).
(3)设线段\(AD\)上存在\(Q(0,m,0)\),\(m∈[-1,1]\),使得它到平面\(PCD\)的距离为 \(\dfrac{\sqrt{3}}{2}\),
\(\therefore \overrightarrow{P Q}=(0, m,-1)\),
\(∴Q\)到平面\(PCD\)的距离 \(d=\dfrac{|\overrightarrow{P Q} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{|m-1|}{\sqrt{3}}=\dfrac{\sqrt{3}}{2}\),
解得 \(m=-\dfrac{1}{2}\)或 \(m=\dfrac{5}{2}\)(舍去),
\(\therefore Q\left(0,-\dfrac{1}{2}, 0\right)\),则 \(\dfrac{A Q}{Q D}=\dfrac{\dfrac{1}{2}}{\dfrac{3}{2}}=\dfrac{1}{3}\).