题目
Problem
The points A A A B B B O A → = 2 i + 2 j − k \overrightarrow{OA}=2\mathbf{i}+2\mathbf{j}-\mathbf{k} O A = 2 i + 2 j − k O B → = 4 i + 2 j + 4 k \overrightarrow{OB}=4\mathbf{i}+2\mathbf{j}+4\mathbf{k} O B = 4 i + 2 j + 4 k O O O
(a) Show that the perpendicular distance from A A A O O O B B B 1 3 65 \frac13\sqrt{65} 3 1 65
(5)
(b) The point C C C O C → = 3 i + p j + q k \overrightarrow{OC}=3\mathbf{i}+p\mathbf{j}+q\mathbf{k} O C = 3 i + p j + q k p p p q q q
Given that
O C → \overrightarrow{OC} O C A B → \overrightarrow{AB} A B ∠ A O C = ∠ C O B \angle AOC=\angle COB ∠ A O C = ∠ C O B
find the values of p p p q q q
(7)
题目中文翻译
相对于原点 O O O A A A B B B O A → = 2 i + 2 j − k \overrightarrow{OA}=2\mathbf{i}+2\mathbf{j}-\mathbf{k} O A = 2 i + 2 j − k O B → = 4 i + 2 j + 4 k \overrightarrow{OB}=4\mathbf{i}+2\mathbf{j}+4\mathbf{k} O B = 4 i + 2 j + 4 k
(a) 证明点 A A A O O O B B B 1 3 65 \frac13\sqrt{65} 3 1 65
(5)
(b) 点 C C C O C → = 3 i + p j + q k \overrightarrow{OC}=3\mathbf{i}+p\mathbf{j}+q\mathbf{k} O C = 3 i + p j + q k p p p q q q
已知
O C → \overrightarrow{OC} O C A B → \overrightarrow{AB} A B ∠ A O C = ∠ C O B \angle AOC=\angle COB ∠ A O C = ∠ C O B
求 p p p q q q
(7)
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
展开
设直线 O B OB O B
答题过程
展开
Let P P P A A A O B OB O B O P → = λ ( 4 , 2 , 4 ) \overrightarrow{OP}=\lambda(4,2,4) O P = λ ( 4 , 2 , 4 )
A P → = ( 4 λ − 2 , 2 λ − 2 , 4 λ + 1 ) . \overrightarrow{AP}=(4\lambda-2,\ 2\lambda-2,\ 4\lambda+1). A P = ( 4 λ − 2 , 2 λ − 2 , 4 λ + 1 ) . Using A P → ⋅ O B → = 0 \overrightarrow{AP}\cdot\overrightarrow{OB}=0 A P ⋅ O B = 0
4 ( 4 λ − 2 ) + 2 ( 2 λ − 2 ) + 4 ( 4 λ + 1 ) = 0 , 36 λ − 8 = 0 , \begin{align*}
4(4\lambda-2)+2(2\lambda-2)+4(4\lambda+1)=&\,0,\\
36\lambda-8=&\,0,
\end{align*} 4 ( 4 λ − 2 ) + 2 ( 2 λ − 2 ) + 4 ( 4 λ + 1 ) = 36 λ − 8 = 0 , 0 , giving λ = 2 / 9 \lambda=2/9 λ = 2/9
A P → = ( − 10 9 , − 14 9 , 17 9 ) . \overrightarrow{AP}=\bigg(-\frac{10}{9},-\frac{14}{9},\frac{17}{9}\bigg). A P = ( − 9 10 , − 9 14 , 9 17 ) . The perpendicular distance is
A P = 100 + 196 + 289 9 = 585 9 = 65 3 . \begin{align*}
AP=&\,\frac{\sqrt{100+196+289}}9\\[4mm]
=&\,\frac{\sqrt{585}}9=\boxed{\frac{\sqrt{65}}3}.
\end{align*} A P = = 9 100 + 196 + 289 9 585 = 3 65 .
解法二
思路
展开
先用点积求 A O B AOB A O B
答题过程
展开
Let θ = ∠ A O B \theta=\angle AOB θ = ∠ A O B
O A = 3 , O B = 6 , O A → ⋅ O B → = 8 + 4 − 4 = 8. \begin{align*}
OA=&\,3,\qquad OB=6,\\[4mm]
\overrightarrow{OA}\cdot\overrightarrow{OB}=&\,8+4-4=8.
\end{align*} O A = O A ⋅ O B = 3 , O B = 6 , 8 + 4 − 4 = 8. Thus
cos θ = 8 3 ( 6 ) = 4 9 , \cos\theta=\frac8{3(6)}=\frac49, cos θ = 3 ( 6 ) 8 = 9 4 , and the perpendicular distance from A A A O B OB O B
d = O A sin θ = 3 1 − 16 81 = 65 3 . d=OA\sin\theta
=3\sqrt{1-\frac{16}{81}}
=\boxed{\frac{\sqrt{65}}3}. d = O A sin θ = 3 1 − 81 16 = 3 65 .
解法三
思路
展开
以 B B B B A BA B A B O BO B O
答题过程
展开
Let P P P A A A O B OB O B
B A → = ( − 2 , 0 , − 5 ) , B O → = ( − 4 , − 2 , − 4 ) . \overrightarrow{BA}=(-2,0,-5),\qquad
\overrightarrow{BO}=(-4,-2,-4). B A = ( − 2 , 0 , − 5 ) , B O = ( − 4 , − 2 , − 4 ) . Therefore
B A = 29 , B O = 6 , B A → ⋅ B O → = 8 + 20 = 28. BA=\sqrt{29},\quad BO=6,\quad
\overrightarrow{BA}\cdot\overrightarrow{BO}=8+20=28. B A = 29 , B O = 6 , B A ⋅ B O = 8 + 20 = 28. Writing ϕ = ∠ O B A \phi=\angle OBA ϕ = ∠ O B A
cos ϕ = 28 6 29 , B P = B A cos ϕ = 14 3 . \begin{align*}
\cos\phi=&\,\frac{28}{6\sqrt{29}},\\[4mm]
BP=&\,BA\cos\phi=\frac{14}{3}.
\end{align*} cos ϕ = B P = 6 29 28 , B A cos ϕ = 3 14 . Hence
A P = 29 − ( 14 3 ) 2 = 65 9 = 65 3 . AP=\sqrt{29-\bigg(\frac{14}{3}\bigg)^2}
=\sqrt{\frac{65}{9}}
=\boxed{\frac{\sqrt{65}}3}. A P = 29 − ( 3 14 ) 2 = 9 65 = 3 65 .
(b)
解法一
思路
展开
垂直条件先给出 q q q O C OC O C
答题过程
展开
The direction of A B AB A B
A B → = ( 4 , 2 , 4 ) − ( 2 , 2 , − 1 ) = ( 2 , 0 , 5 ) . \overrightarrow{AB}=(4,2,4)-(2,2,-1)=(2,0,5). A B = ( 4 , 2 , 4 ) − ( 2 , 2 , − 1 ) = ( 2 , 0 , 5 ) . Since O C OC O C A B AB A B
( 3 , p , q ) ⋅ ( 2 , 0 , 5 ) = 0 ⟹ 6 + 5 q = 0. (3,p,q)\cdot(2,0,5)=0
\quad\Longrightarrow\quad
6+5q=0. ( 3 , p , q ) ⋅ ( 2 , 0 , 5 ) = 0 ⟹ 6 + 5 q = 0. Thus q = − 6 / 5 q=-6/5 q = − 6/5
Also O A = 3 OA=3 O A = 3 O B = 6 OB=6 O B = 6 ∠ A O C \angle AOC ∠ A O C ∠ C O B \angle COB ∠ C O B
6 + 2 p − q 3 9 + p 2 + q 2 = 12 + 2 p + 4 q 6 9 + p 2 + q 2 . \frac{6+2p-q}{3\sqrt{9+p^2+q^2}}
=\frac{12+2p+4q}{6\sqrt{9+p^2+q^2}}. 3 9 + p 2 + q 2 6 + 2 p − q = 6 9 + p 2 + q 2 12 + 2 p + 4 q . Cancelling the positive common length,
2 ( 6 + 2 p − q ) = 12 + 2 p + 4 q , 2 p = 6 q . \begin{align*}
2(6+2p-q)=&\,12+2p+4q,\\
2p=&\,6q.
\end{align*} 2 ( 6 + 2 p − q ) = 2 p = 12 + 2 p + 4 q , 6 q . Hence
p = − 18 5 , q = − 6 5 . \boxed{p=-\frac{18}{5},\qquad q=-\frac65}. p = − 5 18 , q = − 5 6 . For these values both dot products with O C → \overrightarrow{OC} O C π / 2 \pi/2 π /2
