题目
Problem
With respect to the origin O O O A A A O A → = i + 5 j + 6 k \overrightarrow{OA}=\mathbf{i}+5\mathbf{j}+6\mathbf{k} O A = i + 5 j + 6 k l l l
r = 4 i + k + λ ( − i + 2 j + 3 k ) . \mathbf{r}=4\mathbf{i}+\mathbf{k}+\lambda(-\mathbf{i}+2\mathbf{j}+3\mathbf{k}). r = 4 i + k + λ ( − i + 2 j + 3 k ) . (a) Find in degrees the acute angle between the directions of O A OA O A l l l
(3)
(b) Find the position vector of the foot of the perpendicular from A A A l l l
(4)
(c) Hence find the position vector of the reflection of A A A l l l
(2)
题目中文翻译
相对于原点 O O O A A A O A → = i + 5 j + 6 k \overrightarrow{OA}=\mathbf{i}+5\mathbf{j}+6\mathbf{k} O A = i + 5 j + 6 k l l l
r = 4 i + k + λ ( − i + 2 j + 3 k ) . \mathbf{r}=4\mathbf{i}+\mathbf{k}+\lambda(-\mathbf{i}+2\mathbf{j}+3\mathbf{k}). r = 4 i + k + λ ( − i + 2 j + 3 k ) . (a) 求 O A OA O A l l l
(3)
(b) 求从 A A A l l l
(4)
(c) 由此求 A A A l l l
(2)
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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使用位置向量 O A → \overrightarrow{OA} O A
答题过程
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A direction vector of l l l d = ( − 1 , 2 , 3 ) \mathbf d=(-1,2,3) d = ( − 1 , 2 , 3 ) O A → = ( 1 , 5 , 6 ) \overrightarrow{OA}=(1,5,6) O A = ( 1 , 5 , 6 )
Their scalar product is
O A → ⋅ d = − 1 + 10 + 18 = 27. \overrightarrow{OA}\cdot\mathbf d=-1+10+18=27. O A ⋅ d = − 1 + 10 + 18 = 27. Their lengths are 62 \sqrt{62} 62 14 \sqrt{14} 14
θ = cos − 1 27 62 14 ≈ 23.6 ∘ . \theta=\cos^{-1}\frac{27}{\sqrt{62}\sqrt{14}}
\approx\boxed{23.6^\circ}. θ = cos − 1 62 14 27 ≈ 23. 6 ∘ .
(b)
解法一
思路
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设垂足 P P P A P → \overrightarrow{AP} A P
答题过程
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A general point on l l l P = ( 4 − λ , 2 λ , 1 + 3 λ ) P=(4-\lambda,2\lambda,1+3\lambda) P = ( 4 − λ , 2 λ , 1 + 3 λ )
A P → = ( 3 − λ , 2 λ − 5 , 3 λ − 5 ) . \overrightarrow{AP}=(3-\lambda,2\lambda-5,3\lambda-5). A P = ( 3 − λ , 2 λ − 5 , 3 λ − 5 ) . For the perpendicular foot,
A P → ⋅ ( − 1 , 2 , 3 ) = 0. \overrightarrow{AP}\cdot(-1,2,3)=0. A P ⋅ ( − 1 , 2 , 3 ) = 0. Therefore
− ( 3 − λ ) + 2 ( 2 λ − 5 ) + 3 ( 3 λ − 5 ) = 0 , -(3-\lambda)+2(2\lambda-5)+3(3\lambda-5)=0, − ( 3 − λ ) + 2 ( 2 λ − 5 ) + 3 ( 3 λ − 5 ) = 0 , so 14 λ − 28 = 0 14\lambda-28=0 14 λ − 28 = 0 λ = 2 \lambda=2 λ = 2
O P → = 2 i + 4 j + 7 k . \boxed{\overrightarrow{OP}=2\mathbf i+4\mathbf j+7\mathbf k}. O P = 2 i + 4 j + 7 k .
解法二
思路
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也可把点到直线上任意点的距离平方配方，最小值对应的点就是垂足。
答题过程
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For P = ( 4 − λ , 2 λ , 1 + 3 λ ) P=(4-\lambda,2\lambda,1+3\lambda) P = ( 4 − λ , 2 λ , 1 + 3 λ )
A P 2 = ( 3 − λ ) 2 + ( 2 λ − 5 ) 2 + ( 3 λ − 5 ) 2 = 14 λ 2 − 56 λ + 59 = 14 ( λ − 2 ) 2 + 3. \begin{align*}
AP^2
=&\,(3-\lambda)^2+(2\lambda-5)^2\\[4mm]
&\,\hspace{2pt}+(3\lambda-5)^2\\[4mm]
=&\,14\lambda^2-56\lambda+59\\[4mm]
=&\,14(\lambda-2)^2+3.
\end{align*} A P 2 = = = ( 3 − λ ) 2 + ( 2 λ − 5 ) 2 + ( 3 λ − 5 ) 2 14 λ 2 − 56 λ + 59 14 ( λ − 2 ) 2 + 3. Equivalently,
d ( A P 2 ) d λ = 28 λ − 56 \dfrac{\mathrm{d}(AP^2)}{\mathrm{d}\lambda}=28\lambda-56 d λ d ( A P 2 ) = 28 λ − 56 28 > 0 28>0 28 > 0 λ = 2 \lambda=2 λ = 2
O P → = 2 i + 4 j + 7 k . \boxed{\overrightarrow{OP}=2\mathbf i+4\mathbf j+7\mathbf k}. O P = 2 i + 4 j + 7 k .
(c)
解法一
思路
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垂足是点 A A A A A A
答题过程
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Let A ′ A' A ′ A A A l l l P P P A A ′ AA' A A ′
O A ′ → = 2 O P → − O A → = 2 ( 2 , 4 , 7 ) − ( 1 , 5 , 6 ) = 3 i + 3 j + 8 k . \begin{align*}
\overrightarrow{OA'}
=&\,2\overrightarrow{OP}-\overrightarrow{OA}\\[4mm]
=&\,2(2,4,7)-(1,5,6)\\[4mm]
=&\,\boxed{3\mathbf i+3\mathbf j+8\mathbf k}.
\end{align*} O A ′ = = = 2 O P − O A 2 ( 2 , 4 , 7 ) − ( 1 , 5 , 6 ) 3 i + 3 j + 8 k . 