用法向量求距离
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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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4256133

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。


4256133

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。


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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


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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。


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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。


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