在棱长为 1 的正方体 ABCD - A 1 B 1 C 1 D 1 中，求直线 BD 1 与平面 ABCD 所成角的正弦值。 3 3 线面角线面角法向量正弦值新课标I卷 2023

3 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"/> 以 A 为原点建系： B ( 1 , 0 , 0 ) ， D 1 ( 0 , 1 , 1 ) ，故 B D 1 <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"/> = ( − 1 , 1 , 1 ) 。平面 A B C D 就是 x O y 平面，取法向量 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"/> = ( 0 , 0 , 1 ) 。线面角的正弦等于方向向量与法向量夹角余弦的绝对值： sin θ = ∣ B D 1 <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"/> ∣ ∣ B D 1 <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： 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"/> × 1 ∣1∣ = 3 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"/> 。

2023 新课标I卷｜空间向量与空间角·线面角

在棱长为

1

的正方体

A B C D - A_{1} B_{1} C_{1} D_{1}

中，求直线

B D_{1}

与平面

A B C D

所成角的正弦值。

答案

\frac{3}{3}

分步解析

1以 $A$ 为原点建系： $B (1, 0, 0)$ ， $D_{1} (0, 1, 1)$ ，故 $B D_{1} = (- 1, 1, 1)$ 。
2平面 $A B C D$ 就是 $x O y$ 平面，取法向量 $n = (0, 0, 1)$ 。
3线面角的正弦等于方向向量与法向量夹角余弦的绝对值： $sin θ = \frac{∣ B D _{1} \cdot n ∣}{∣ B D _{1} ∣∣ n ∣}$ 。
4计算： $\frac{∣1∣}{3 \times 1} = \frac{3}{3}$ 。

点评线面角用「正弦」、二面角用「余弦」，这是空间向量法最容易记混的一处；写成

sin θ = ∣ cos ⟨ a, n ⟩ ∣

就不会错。

#线面角#法向量#正弦值

2023 年真题新课标I卷

2023 年新课标I卷·第18题

分步解析

同考点推荐空间向量与空间角 · 线面角