题目
Problem
4.
The plane Π 1 \Pi_1 Π 1 A ( 2 , 4 , − 5 ) A(2,4,-5) A ( 2 , 4 , − 5 )
( − 1 3 3 ) \begin{pmatrix}
-1\\
3\\
3
\end{pmatrix} − 1 3 3 The plane Π 2 \Pi_2 Π 2 B ( 3 , 6 , − 2 ) B(3,6,-2) B ( 3 , 6 , − 2 )
( 2 0 − 5 ) \begin{pmatrix}
2\\
0\\
-5
\end{pmatrix} 2 0 − 5 The line l l l Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(a) Determine a vector equation for l l l
(7)
The points C C C D D D l l l
Given that C C C D D D
(b) determine the exact volume of the tetrahedron A B C D ABCD A B C D
(5)
题目中文翻译
4.
平面 Π 1 \Pi_1 Π 1 A ( 2 , 4 , − 5 ) A(2,4,-5) A ( 2 , 4 , − 5 )
( − 1 3 3 ) \begin{pmatrix}
-1\\
3\\
3
\end{pmatrix} − 1 3 3 平面 Π 2 \Pi_2 Π 2 B ( 3 , 6 , − 2 ) B(3,6,-2) B ( 3 , 6 , − 2 )
( 2 0 − 5 ) \begin{pmatrix}
2\\
0\\
-5
\end{pmatrix} 2 0 − 5 直线 l l l Π 1 \Pi_1 Π 1 Π 2 \Pi_2 Π 2
(a) 求 l l l
(7)
点 C C C D D D l l l
已知 C C C D D D
(b) 求四面体 A B C D ABCD A B C D
(5)
解答
(a)
解法一：联立两个平面方程
思路
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先由已知点和法向量分别写出两个平面的笛卡尔方程，再把其中一个坐标设为参数，联立求出另外两个坐标。这样可直接读出交线上的一点和方向向量。
答题过程
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The equations of the two planes are
Π 1 : − x + 3 y + 3 z = − 2 + 12 − 15 = − 5 , Π 2 : 2 x − 5 z = 6 + 10 = 16. \begin{align*}
\Pi_1:\quad
-x+3y+3z
=&\,-2+12-15=-5, \\
\Pi_2:\quad
2x-5z
=&\,6+10=16.
\end{align*} Π 1 : − x + 3 y + 3 z = Π 2 : 2 x − 5 z = − 2 + 12 − 15 = − 5 , 6 + 10 = 16. Let y = λ y=\lambda y = λ 2 x − 5 z = 16 2x-5z=16 2 x − 5 z = 16
z = 2 x − 16 5 . z=\frac{2x-16}{5}. z = 5 2 x − 16 . Substituting this into − x + 3 y + 3 z = − 5 -x+3y+3z=-5 − x + 3 y + 3 z = − 5
− x + 3 λ + 3 ( 2 x − 16 5 ) = − 5 , − 5 x + 15 λ + 6 x − 48 = − 25 , x = 23 − 15 λ . \begin{align*}
-x+3\lambda
+3\biggl(\frac{2x-16}{5}\biggr)
=&\,-5, \\
-5x+15\lambda+6x-48
=&\,-25, \\
x=&\,23-15\lambda.
\end{align*} − x + 3 λ + 3 ( 5 2 x − 16 ) = − 5 x + 15 λ + 6 x − 48 = x = − 5 , − 25 , 23 − 15 λ . Hence
z = 2 ( 23 − 15 λ ) − 16 5 = 6 − 6 λ . z=\frac{2(23-15\lambda)-16}{5}
=6-6\lambda. z = 5 2 ( 23 − 15 λ ) − 16 = 6 − 6 λ . Therefore, a vector equation of l l l
r = ( 23 0 6 ) + λ ( − 15 1 − 6 ) , λ ∈ R . \boxed{
\mathbf r=
\begin{pmatrix}
23\\
0\\
6
\end{pmatrix}
+\lambda
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}
},
\qquad \lambda\in\mathbb R. r = 23 0 6 + λ − 15 1 − 6 , λ ∈ R .
解法二：叉积求方向向量
思路
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交线同时位于两个平面内，所以它的方向向量同时垂直于两个平面的法向量，可由两法向量的叉积得到。再令一个坐标为零，联立平面方程找出交线上的一点。
答题过程
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The equations of the planes are
− x + 3 y + 3 z = − 5 , 2 x − 5 z = 16. -x+3y+3z=-5,
\qquad
2x-5z=16. − x + 3 y + 3 z = − 5 , 2 x − 5 z = 16. A direction vector of l l l
( − 1 3 3 ) × ( 2 0 − 5 ) = ( 3 ( − 5 ) − 3 ( 0 ) 3 ( 2 ) − ( − 1 ) ( − 5 ) ( − 1 ) ( 0 ) − 3 ( 2 ) ) = ( − 15 1 − 6 ) . \begin{align*}
\begin{pmatrix}
-1\\
3\\
3
\end{pmatrix}
\times
\begin{pmatrix}
2\\
0\\
-5
\end{pmatrix}
=&\,
\begin{pmatrix}
3(-5)-3(0)\\
3(2)-(-1)(-5)\\
(-1)(0)-3(2)
\end{pmatrix} \\
=&\,
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}.
\end{align*} − 1 3 3 × 2 0 − 5 = = 3 ( − 5 ) − 3 ( 0 ) 3 ( 2 ) − ( − 1 ) ( − 5 ) ( − 1 ) ( 0 ) − 3 ( 2 ) − 15 1 − 6 . To find a point on l l l z = 0 z=0 z = 0
2 x = 16 ⟹ x = 8 , 2x=16
\quad\Longrightarrow\quad
x=8, 2 x = 16 ⟹ x = 8 , and
− 8 + 3 y = − 5 ⟹ y = 1. -8+3y=-5
\quad\Longrightarrow\quad
y=1. − 8 + 3 y = − 5 ⟹ y = 1. Hence ( 8 , 1 , 0 ) (8,1,0) ( 8 , 1 , 0 ) l l l
r = ( 8 1 0 ) + λ ( − 15 1 − 6 ) , λ ∈ R . \boxed{
\mathbf r=
\begin{pmatrix}
8\\
1\\
0
\end{pmatrix}
+\lambda
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}
},
\qquad \lambda\in\mathbb R. r = 8 1 0 + λ − 15 1 − 6 , λ ∈ R .
(b)
解法一：标量三重积
思路
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取 (a) 中交线上的方便点作为 C C C C D → \overrightarrow{CD} C D
答题过程
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Sliding C C C D D D l l l C D = 5 CD=5 C D = 5 △ A C D \triangle ACD △ A C D A A A l l l B B B Π 1 \Pi_1 Π 1
Take C = ( 8 , 1 , 0 ) C=(8,1,0) C = ( 8 , 1 , 0 ) l l l
( − 15 ) 2 + 1 2 + ( − 6 ) 2 = 262 . \sqrt{(-15)^2+1^2+(-6)^2}=\sqrt{262}. ( − 15 ) 2 + 1 2 + ( − 6 ) 2 = 262 . Since C C C D D D
C D → = 5 262 ( − 15 1 − 6 ) . \overrightarrow{CD}
=\frac{5}{\sqrt{262}}
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}. C D = 262 5 − 15 1 − 6 . Also,
C A → = ( − 6 3 − 5 ) , C B → = ( − 5 5 − 2 ) . \overrightarrow{CA}
=
\begin{pmatrix}
-6\\
3\\
-5
\end{pmatrix},
\qquad
\overrightarrow{CB}
=
\begin{pmatrix}
-5\\
5\\
-2
\end{pmatrix}. C A = − 6 3 − 5 , C B = − 5 5 − 2 . Now
C A → × C B → = ( 3 ( − 2 ) − ( − 5 ) ( 5 ) ( − 5 ) ( − 5 ) − ( − 6 ) ( − 2 ) ( − 6 ) ( 5 ) − 3 ( − 5 ) ) = ( 19 13 − 15 ) . \overrightarrow{CA}\times\overrightarrow{CB}
=
\begin{pmatrix}
3(-2)-(-5)(5)\\
(-5)(-5)-(-6)(-2)\\
(-6)(5)-3(-5)
\end{pmatrix}
=
\begin{pmatrix}
19\\
13\\
-15
\end{pmatrix}. C A × C B = 3 ( − 2 ) − ( − 5 ) ( 5 ) ( − 5 ) ( − 5 ) − ( − 6 ) ( − 2 ) ( − 6 ) ( 5 ) − 3 ( − 5 ) = 19 13 − 15 . Therefore, the volume V V V
V = 1 6 ∣ C D → ⋅ ( C A → × C B → ) ∣ = 1 6 ∣ 5 262 ( − 15 1 − 6 ) ⋅ ( 19 13 − 15 ) ∣ = 1 6 ∣ 5 262 ( − 182 ) ∣ = 455 3 262 . \begin{align*}
V
=&\,\frac16
\bigg|
\overrightarrow{CD}\mathbin{\boldsymbol{\cdot}}
\bigl(
\overrightarrow{CA}\times\overrightarrow{CB}
\bigr)
\bigg| \\
=&\,\frac16\bigg|
\frac{5}{\sqrt{262}}
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}
\mathbin{\boldsymbol{\cdot}}
\begin{pmatrix}
19\\
13\\
-15
\end{pmatrix}
\bigg| \\
=&\,\frac16\bigg|
\frac{5}{\sqrt{262}}(-182)
\bigg| \\
=&\,\boxed{\frac{455}{3\sqrt{262}}}.
\end{align*} V = = = = 6 1 C D ⋅ ( C A × C B ) 6 1 262 5 − 15 1 − 6 ⋅ 19 13 − 15 6 1 262 5 ( − 182 ) 3 262 455 .
解法二：三角形面积与垂直高度
思路
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把 △ A C D \triangle ACD △ A C D Π 1 \Pi_1 Π 1 B B B Π 1 \Pi_1 Π 1
答题过程
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As in Method 1, take C = ( 8 , 1 , 0 ) C=(8,1,0) C = ( 8 , 1 , 0 )
C D → = 5 262 ( − 15 1 − 6 ) , C A → = ( − 6 3 − 5 ) . \overrightarrow{CD}
=\frac{5}{\sqrt{262}}
\begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix},
\qquad
\overrightarrow{CA}
=
\begin{pmatrix}
-6\\
3\\
-5
\end{pmatrix}. C D = 262 5 − 15 1 − 6 , C A = − 6 3 − 5 . Since
( − 15 1 − 6 ) × ( − 6 3 − 5 ) = ( 1 ( − 5 ) − ( − 6 ) ( 3 ) ( − 6 ) ( − 6 ) − ( − 15 ) ( − 5 ) ( − 15 ) ( 3 ) − 1 ( − 6 ) ) = ( 13 − 39 − 39 ) , \begin{pmatrix}
-15\\
1\\
-6
\end{pmatrix}
\times
\begin{pmatrix}
-6\\
3\\
-5
\end{pmatrix}
=
\begin{pmatrix}
1(-5)-(-6)(3)\\
(-6)(-6)-(-15)(-5)\\
(-15)(3)-1(-6)
\end{pmatrix}
=
\begin{pmatrix}
13\\
-39\\
-39
\end{pmatrix}, − 15 1 − 6 × − 6 3 − 5 = 1 ( − 5 ) − ( − 6 ) ( 3 ) ( − 6 ) ( − 6 ) − ( − 15 ) ( − 5 ) ( − 15 ) ( 3 ) − 1 ( − 6 ) = 13 − 39 − 39 , the area of △ A C D \triangle ACD △ A C D
Area ( △ A C D ) = 1 2 ∣ C D → × C A → ∣ = 1 2 ⋅ 5 262 13 2 + ( − 39 ) 2 + ( − 39 ) 2 = 65 19 2 262 . \begin{align*}
\operatorname{Area}(\triangle ACD)
=&\,\frac12
\bigg|
\overrightarrow{CD}\times\overrightarrow{CA}
\bigg| \\
=&\,\frac12\cdot
\frac{5}{\sqrt{262}}
\sqrt{13^2+(-39)^2+(-39)^2} \\
=&\,\frac{65\sqrt{19}}{2\sqrt{262}}.
\end{align*} Area ( △ A C D ) = = = 2 1 C D × C A 2 1 ⋅ 262 5 1 3 2 + ( − 39 ) 2 + ( − 39 ) 2 2 262 65 19 . The plane Π 1 \Pi_1 Π 1
− x + 3 y + 3 z + 5 = 0. -x+3y+3z+5=0. − x + 3 y + 3 z + 5 = 0. Hence the perpendicular distance from B ( 3 , 6 , − 2 ) B(3,6,-2) B ( 3 , 6 , − 2 ) Π 1 \Pi_1 Π 1
h = ∣ − 3 + 18 − 6 + 5 ∣ ( − 1 ) 2 + 3 2 + 3 2 = 14 19 . \begin{align*}
h
=&\,\frac{|-3+18-6+5|}
{\sqrt{(-1)^2+3^2+3^2}} \\
=&\,\frac{14}{\sqrt{19}}.
\end{align*} h = = ( − 1 ) 2 + 3 2 + 3 2 ∣ − 3 + 18 − 6 + 5∣ 19 14 . Therefore,
V = 1 3 Area ( △ A C D ) h = 1 3 ⋅ 65 19 2 262 ⋅ 14 19 = 455 3 262 . \begin{align*}
V
=&\,\frac13
\operatorname{Area}(\triangle ACD)h \\
=&\,\frac13\cdot
\frac{65\sqrt{19}}{2\sqrt{262}}
\cdot\frac{14}{\sqrt{19}} \\
=&\,\boxed{\frac{455}{3\sqrt{262}}}.
\end{align*} V = = = 3 1 Area ( △ A C D ) h 3 1 ⋅ 2 262 65 19 ⋅ 19 14 3 262 455 . 