题目
Problem
The points A A A B B B C C C
4 i − 4 j + k , − 4 i + 3 j − 4 k , 4 i − j − 2 k , 4\mathbf i-4\mathbf j+\mathbf k,\quad -4\mathbf i+3\mathbf j-4\mathbf k,\quad 4\mathbf i-\mathbf j-2\mathbf k, 4 i − 4 j + k , − 4 i + 3 j − 4 k , 4 i − j − 2 k , respectively, relative to the origin O O O
(a) Find the equation of the plane A B C ABC A B C a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(5)
(b) Find the perpendicular distance from O O O A B C ABC A B C
(2)
(c) The point D D D 2 i + 3 j − 3 k 2\mathbf i+3\mathbf j-3\mathbf k 2 i + 3 j − 3 k
Find the coordinates of the point of intersection of the line O D OD O D A B C ABC A B C
(3)
题目中文翻译
点 A A A B B B C C C O O O
4 i − 4 j + k , − 4 i + 3 j − 4 k , 4 i − j − 2 k . 4\mathbf i-4\mathbf j+\mathbf k,\quad -4\mathbf i+3\mathbf j-4\mathbf k,\quad 4\mathbf i-\mathbf j-2\mathbf k. 4 i − 4 j + k , − 4 i + 3 j − 4 k , 4 i − j − 2 k . (a) 求平面 A B C ABC A B C a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(5)
(b) 求 O O O A B C ABC A B C
(2)
(c) 点 D D D 2 i + 3 j − 3 k 2\mathbf i+3\mathbf j-3\mathbf k 2 i + 3 j − 3 k
求直线 O D OD O D A B C ABC A B C
(3)
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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平面内两条不平行向量的叉积给出法向量，再代入已知点求常数。
答题过程
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Two vectors in the plane are
A B → = ( − 8 7 − 5 ) , A C → = ( 0 3 − 3 ) . \overrightarrow{AB}=\begin{pmatrix}-8\\7\\-5\end{pmatrix},\qquad
\overrightarrow{AC}=\begin{pmatrix}0\\3\\-3\end{pmatrix}. A B = − 8 7 − 5 , A C = 0 3 − 3 . Their cross product is
A B → × A C → = ( 7 ( − 3 ) − ( − 5 ) ( 3 ) ( − 5 ) ( 0 ) − ( − 8 ) ( − 3 ) ( − 8 ) ( 3 ) − 7 ( 0 ) ) = ( − 6 − 24 − 24 ) . \begin{align*}
\overrightarrow{AB}\times\overrightarrow{AC}
=&\,\begin{pmatrix}7(-3)-(-5)(3)\\(-5)(0)-(-8)(-3)\\(-8)(3)-7(0)\end{pmatrix}\\[4mm]
=&\,\begin{pmatrix}-6\\-24\\-24\end{pmatrix}.
\end{align*} A B × A C = = 7 ( − 3 ) − ( − 5 ) ( 3 ) ( − 5 ) ( 0 ) − ( − 8 ) ( − 3 ) ( − 8 ) ( 3 ) − 7 ( 0 ) − 6 − 24 − 24 . So a normal vector is ( 1 , 4 , 4 ) (1,4,4) ( 1 , 4 , 4 ) A = ( 4 , − 4 , 1 ) A=(4,-4,1) A = ( 4 , − 4 , 1 )
x + 4 y + 4 z = 4 + 4 ( − 4 ) + 4 ( 1 ) , x+4y+4z=4+4(-4)+4(1), x + 4 y + 4 z = 4 + 4 ( − 4 ) + 4 ( 1 ) , giving x + 4 y + 4 z = − 8 \boxed{x+4y+4z=-8} x + 4 y + 4 z = − 8
解法二
思路
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也可直接设平面的一般方程，把三个点代入后相减，解出三个系数的比例。
答题过程
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Let the plane be p x + q y + r z = d px+qy+rz=d p x + q y + r z = d ( p , q , r ) ≠ ( 0 , 0 , 0 ) (p,q,r)\ne(0,0,0) ( p , q , r ) = ( 0 , 0 , 0 ) A , B , C A,B,C A , B , C
4 p − 4 q + r = d , − 4 p + 3 q − 4 r = d , 4 p − q − 2 r = d . \begin{align*}
4p-4q+r=&\,d,\\[4mm]
-4p+3q-4r=&\,d,\\[4mm]
4p-q-2r=&\,d.
\end{align*} 4 p − 4 q + r = − 4 p + 3 q − 4 r = 4 p − q − 2 r = d , d , d . Subtracting the first equation from the third gives 3 q − 3 r = 0 3q-3r=0 3 q − 3 r = 0 r = q r=q r = q
− 8 p + 7 q − 5 r = 0 ⟹ q = 4 p . -8p+7q-5r=0\quad\Longrightarrow\quad q=4p. − 8 p + 7 q − 5 r = 0 ⟹ q = 4 p . Thus r = 4 p r=4p r = 4 p p ≠ 0 p\ne0 p = 0 p p p A A A
x + 4 y + 4 z = − 8 . \boxed{x+4y+4z=-8}. x + 4 y + 4 z = − 8 .
(b)
解法一
思路
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用点到平面的距离公式；分母必须是完整法向量的模。
答题过程
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The plane is x + 4 y + 4 z + 8 = 0 x+4y+4z+8=0 x + 4 y + 4 z + 8 = 0 O = ( 0 , 0 , 0 ) O=(0,0,0) O = ( 0 , 0 , 0 )
∣ 8 ∣ 1 2 + 4 2 + 4 2 = 8 33 . \boxed{\frac{|8|}{\sqrt{1^2+4^2+4^2}}=\frac8{\sqrt{33}}}. 1 2 + 4 2 + 4 2 ∣8∣ = 33 8 .
(c)
解法一
思路
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直线经过原点，可用 D D D
答题过程
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On O D OD O D
( x , y , z ) = ( 2 t , 3 t , − 3 t ) , t ∈ R . (x,y,z)=(2t,3t,-3t),\qquad t\in\mathbb R. ( x , y , z ) = ( 2 t , 3 t , − 3 t ) , t ∈ R . Substituting into the plane,
2 t + 4 ( 3 t ) + 4 ( − 3 t ) = − 8 ⟹ t = − 4. 2t+4(3t)+4(-3t)=-8\quad\Longrightarrow\quad t=-4. 2 t + 4 ( 3 t ) + 4 ( − 3 t ) = − 8 ⟹ t = − 4. Hence the intersection is ( − 8 , − 12 , 12 ) \boxed{(-8,-12,12)} ( − 8 , − 12 , 12 )
