题目
Problem
The plane Π \Pi Π r = 3 i − 2 j + k + λ ( − i + 2 j + k ) \mathbf r=3\mathbf i-2\mathbf j+\mathbf k+\lambda(-\mathbf i+2\mathbf j+\mathbf k) r = 3 i − 2 j + k + λ ( − i + 2 j + k ) r = 4 i + 4 j + 2 k + μ ( 3 i + 2 j − k ) \mathbf r=4\mathbf i+4\mathbf j+2\mathbf k+\mu(3\mathbf i+2\mathbf j-\mathbf k) r = 4 i + 4 j + 2 k + μ ( 3 i + 2 j − k )
(a) Find a Cartesian equation of Π \Pi Π a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
The line l l l P P P 2 i + 3 j + k 2\mathbf i+3\mathbf j+\mathbf k 2 i + 3 j + k j + k \mathbf j+\mathbf k j + k
(b) Find the acute angle between l l l Π \Pi Π
(3)
(c) Find the position vector of the foot of the perpendicular from P P P Π \Pi Π
(4)
题目中文翻译
平面 Π \Pi Π r = 3 i − 2 j + k + λ ( − i + 2 j + k ) \mathbf r=3\mathbf i-2\mathbf j+\mathbf k+\lambda(-\mathbf i+2\mathbf j+\mathbf k) r = 3 i − 2 j + k + λ ( − i + 2 j + k ) r = 4 i + 4 j + 2 k + μ ( 3 i + 2 j − k ) \mathbf r=4\mathbf i+4\mathbf j+2\mathbf k+\mu(3\mathbf i+2\mathbf j-\mathbf k) r = 4 i + 4 j + 2 k + μ ( 3 i + 2 j − k )
(a) 求平面 Π \Pi Π a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
直线 l l l 2 i + 3 j + k 2\mathbf i+3\mathbf j+\mathbf k 2 i + 3 j + k P P P j + k \mathbf j+\mathbf k j + k
(b) 求 l l l Π \Pi Π
(3)
(c) 求从 P P P Π \Pi Π
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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平面包含两条直线的方向，其叉积给法向量；代入一点求常数，并核对另一条直线也满足。
答题过程
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The line directions are u = ( − 1 , 2 , 1 ) \mathbf u=(-1,2,1) u = ( − 1 , 2 , 1 ) v = ( 3 , 2 , − 1 ) \mathbf v=(3,2,-1) v = ( 3 , 2 , − 1 )
u × v = ( − 4 , 2 , − 8 ) = − 2 ( 2 , − 1 , 4 ) . \mathbf u\times\mathbf v=(-4,2,-8)=-2(2,-1,4). u × v = ( − 4 , 2 , − 8 ) = − 2 ( 2 , − 1 , 4 ) . Taking normal ( 2 , − 1 , 4 ) (2,-1,4) ( 2 , − 1 , 4 ) ( 3 , − 2 , 1 ) (3,-2,1) ( 3 , − 2 , 1 )
2 x − y + 4 z = 12 . \boxed{2x-y+4z=12}. 2 x − y + 4 z = 12 . The other point ( 4 , 4 , 2 ) (4,4,2) ( 4 , 4 , 2 ) 8 − 4 + 8 = 12 8-4+8=12 8 − 4 + 8 = 12
(b)
解法一
思路
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直线方向与法向量的点积绝对值除以两模之积，给线面锐角的正弦。
答题过程
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The direction of l l l d = ( 0 , 1 , 1 ) \mathbf d=(0,1,1) d = ( 0 , 1 , 1 ) Π \Pi Π n = ( 2 , − 1 , 4 ) \mathbf n=(2,-1,4) n = ( 2 , − 1 , 4 ) ϕ \phi ϕ
sin ϕ = ∣ d ⋅ n ∣ ∣ d ∣ ∣ n ∣ = ∣ − 1 + 4 ∣ 2 21 = 3 42 . \begin{align*}
\sin\phi=&\,\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}\\[4mm]
=&\,\frac{|-1+4|}{\sqrt2\sqrt{21}}\\[4mm]
=&\,\frac3{\sqrt{42}}.
\end{align*} sin ϕ = = = ∣ d ∣∣ n ∣ ∣ d ⋅ n ∣ 2 21 ∣ − 1 + 4∣ 42 3 . Therefore ϕ = 27.6 ∘ \boxed{\phi=27.6^\circ} ϕ = 27. 6 ∘
(c)
解法一
思路
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垂足位于经过P、方向为平面法向量的直线上，将参数式代入平面方程。
答题过程
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The foot F F F
O F → = ( 2 , 3 , 1 ) + t ( 2 , − 1 , 4 ) = ( 2 + 2 t , 3 − t , 1 + 4 t ) . \begin{align*}
\overrightarrow{OF}=&\,(2,3,1)+t(2,-1,4)\\[4mm]
=&\,(2+2t,3-t,1+4t).
\end{align*} O F = = ( 2 , 3 , 1 ) + t ( 2 , − 1 , 4 ) ( 2 + 2 t , 3 − t , 1 + 4 t ) . Since F F F Π \Pi Π
2 ( 2 + 2 t ) − ( 3 − t ) + 4 ( 1 + 4 t ) = 12. 2(2+2t)-(3-t)+4(1+4t)=12. 2 ( 2 + 2 t ) − ( 3 − t ) + 4 ( 1 + 4 t ) = 12. Thus 5 + 21 t = 12 5+21t=12 5 + 21 t = 12 t = 1 / 3 t=1/3 t = 1/3
O F → = 1 3 ( 8 i + 8 j + 7 k ) . \boxed{\overrightarrow{OF}=\frac13(8\mathbf i+8\mathbf j+7\mathbf k)}. O F = 3 1 ( 8 i + 8 j + 7 k ) . 