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用法向量求距离

用法向量求距离. P. C. D. O. B. A. 1 、已知正四面体 P-ABC ，棱长为 1 ，求点 P 到平面 ABC 的距离。. 解：设平面 ABC 的法向量 n ＝ (x,y,z),. AB=(2,-2,1), AC=(4,0,6). ∵ n · AB=0 2x-2y+z=0 n · AC=0 4x+6z=0. 令 x=3 ，则 y=2 ， z=-2. 即 n=(3,2,-2). 又因为 AD=(-7,-7, 7),. 17. 49. 49. n · AD. =. =. 所以 d=. 17. n.

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用法向量求距离

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  1. 用法向量求距离

  2. P C D O B A 1、已知正四面体P-ABC，棱长为1，求点P到平面ABC的距离。

  3. 解：设平面ABC的法向量n＝(x,y,z), AB=(2,-2,1), AC=(4,0,6) ∵ n·AB=0 2x-2y+z=0 n·AC=0 4x+6z=0 令x=3，则y=2，z=-2 即n=(3,2,-2) 又因为AD=(-7,-7, 7), 17 49 49 n·AD = = 所以d= 17 n 17 2、设A(2,3,1),B(4,1,2),C(6,3,7),D(-5,-4,8) 求点D到平面ABC的距离。

  4. 解：如图建立坐标系,A(a,0,0),B(a,a,0),C(0,a,0),M(0,0,a),E(a,0,a)解：如图建立坐标系,A(a,0,0),B(a,a,0),C(0,a,0),M(0,0,a),E(a,0,a) F(0,a,a),P(a/2,0,a/2) 设平面EFB的法向量为n(x,y,z) 又因为EF=(-a,a,0),EB=(a,0,-a) z -ax+ay=0 n⊥EF M ∴ F ay-az=0 n⊥EB 令x=1,则y=1,z=1 ∴n=(1,1,1) E 又∵PE=(a/2,0,a/2) P 3 a n·PE D a = = C y 所以d= 3 n 3 A B x 3、如图所示，已知四边形ABCD、EADM和MDCF都是边长为 a 的正方形，点P是ED的中点，求点P到平面EFB的距离

  5. 解：(2)如图建立坐标系，则D(0,0,0),A(1,0,0),C(0,1,0),A1(1,0,1)解：(2)如图建立坐标系，则D(0,0,0),A(1,0,0),C(0,1,0),A1(1,0,1) DA1=(1,0,1),AC=(-1,1,0) 设向量DA1与AC构成平面的法向量为n=(x,y,z) z x+z=0 n⊥DA1 ∴ D1 -x+y=0 C1 n⊥AC 令x=1,则y=1,z=-1 ∴n=(1,1,-1) A1 B1 DA=(1,0,0) 3 1 n·DA D = = C y 所以d= 3 n 3 A B x 4、已知正方体ABCD-A1B1C1D1的棱长为 1， (1)口答：直线A1D1与AB的距离；直线A1D1与AC的距离. (2)求直线DA1与AC间的距离。

  6. 解：如图建立坐标系,A(a,0,0),B(a,a,0),C(0,a,0),M(0,0,a),E(a,0,a)解：如图建立坐标系,A(a,0,0),B(a,a,0),C(0,a,0),M(0,0,a),E(a,0,a) F(0,a,a),P(a/2,0,a/2) 设平面EFB的法向量为n(x,y,z) 又因为EF=(-a,a,0),EB=(a,0,-a) z -ax+ay=0 n⊥EF M ∴ F ay-az=0 n⊥EB 令x=1,则y=1,z=1 ∴n=(1,1,1) E 又∵PE=(a/2,0,a/2) P P 3 a n·PE D a = = C y 所以d= 3 n 3 Q A B x 3、如图所示，已知四边形ABCD、EADM和MDCF都是边长为a的正方形，点P是ED的中点，求点P到平面EFB的距离 若Q为AC的中点，求异面直线PM与FQ的距离。

  7. ！ 欢 迎 指 正 ！ 谢 谢

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