在空间直角坐标系中，已知平面过点 M(1,0,0) 且法向量为 n =(1,1,1) ，求点 P(2,3,1) 到平面的距离。 5 3 3 点到平面的距离点面距离法向量投影新高考I卷 2021

3 5 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 取平面内一点 M ，作向量 M P <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"/> = ( 2 − 1 , 3 − 0 , 1 − 0 ) = ( 1 , 3 , 1 ) 。点到平面距离公式： d = ∣ n <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ ∣ M P <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"/> ⋅ n <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ 。计算： M P <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"/> ⋅ n <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> = 1 + 3 + 1 = 5 ， ∣ n <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/> ∣ = 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 。故 d = 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 5 = 3 5 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 。

高考数学考点题库/空间向量与空间角/点到平面的距离

2021 年新高考I卷·第20题

解答题★★★★☆ 较难改编题通法

在空间直角坐标系中，已知平面

α

过点

M (1, 0, 0)

且法向量为

n = (1, 1, 1)

，求点

P (2, 3, 1)

到平面

α

的距离。

答案

\frac{5 3}{3}

分步解析

1取平面内一点 $M$ ，作向量 $M P = (2 - 1, 3 - 0, 1 - 0) = (1, 3, 1)$ 。
2点到平面距离公式： $d = \frac{∣ M P \cdot n ∣}{∣ n ∣}$ 。
3计算： $M P \cdot n = 1 + 3 + 1 = 5$ ， $∣ n ∣ = 3$ 。
4故 $d = \frac{5}{3} = \frac{5 3}{3}$ 。

点评点到平面的距离就是「连接向量在法向量上的投影长度」，公式与点到直线距离的结构完全一致。

#点面距离#法向量#投影

2021 年真题新高考I卷

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