题目
Problem
The points A A A B B B ( 4 − 1 2 ) \begin{pmatrix}4\\-1\\2\end{pmatrix} 4 − 1 2 ( − 1 − 3 5 ) \begin{pmatrix}-1\\-3\\5\end{pmatrix} − 1 − 3 5 t t t A A A B B B
(a) Find a vector equation for t t t
(2)
The line s s s
r = ( 2 1 5 ) + μ ( − 3 − 4 5 ) . \mathbf{r}=\begin{pmatrix}2\\1\\5\end{pmatrix}+\mu\begin{pmatrix}-3\\-4\\5\end{pmatrix}. r = 2 1 5 + μ − 3 − 4 5 . (b) Find the acute angle between s s s t t t
(3)
(c) Find the length of the perpendicular from A A A s s s
(5)
题目中文翻译
点 A A A B B B ( 4 − 1 2 ) \begin{pmatrix}4\\-1\\2\end{pmatrix} 4 − 1 2 ( − 1 − 3 5 ) \begin{pmatrix}-1\\-3\\5\end{pmatrix} − 1 − 3 5 t t t A A A B B B
(a) 求直线 t t t
(2)
直线 s s s
r = ( 2 1 5 ) + μ ( − 3 − 4 5 ) . \mathbf{r}=\begin{pmatrix}2\\1\\5\end{pmatrix}+\mu\begin{pmatrix}-3\\-4\\5\end{pmatrix}. r = 2 1 5 + μ − 3 − 4 5 . (b) 求直线 s s s t t t
(3)
(c) 求从点 A A A s s s
(5)
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
取 A B → \overrightarrow{AB} A B A A A
答题过程
展开
The direction is A B → = ( − 5 , − 2 , 3 ) \overrightarrow{AB}=(-5,-2,3) A B = ( − 5 , − 2 , 3 )
t : r = ( 4 − 1 2 ) + λ ( − 5 − 2 3 ) , λ ∈ R . t:\quad\mathbf r=\begin{pmatrix}4\\-1\\2\end{pmatrix}
+\lambda\begin{pmatrix}-5\\-2\\3\end{pmatrix},
\qquad \lambda\in\mathbb R. t : r = 4 − 1 2 + λ − 5 − 2 3 , λ ∈ R .
(b)
解法一
思路
展开
两直线的夹角由方向向量决定。用数量积除以模长之积，再取反余弦，并选择锐角。
答题过程
展开
The direction vectors are d = ( − 5 , − 2 , 3 ) \mathbf d=(-5,-2,3) d = ( − 5 , − 2 , 3 ) e = ( − 3 , − 4 , 5 ) \mathbf e=(-3,-4,5) e = ( − 3 , − 4 , 5 ) 38 38 38 38 \sqrt{38} 38 50 \sqrt{50} 50 ϕ \phi ϕ
cos ϕ = 38 38 50 , ϕ = 29.33387 … ∘ = 29.3 ∘ (1 decimal place) . \begin{align*}
\cos\phi=&\,\frac{38}{\sqrt{38}\sqrt{50}},\\[4mm]
\phi=&\,29.33387\ldots^\circ\\[4mm]
=&\,29.3^\circ\quad\text{(1 decimal place)}.
\end{align*} cos ϕ = ϕ = = 38 50 38 , 29.33387 … ∘ 29. 3 ∘ (1 decimal place) .
(c)
解法一
思路
展开
设垂足为直线 s s s
答题过程
展开
A general point Q Q Q s s s
A Q → = ( − 2 − 3 μ , 2 − 4 μ , 3 + 5 μ ) . \overrightarrow{AQ}=(-2-3\mu,\ 2-4\mu,\ 3+5\mu). A Q = ( − 2 − 3 μ , 2 − 4 μ , 3 + 5 μ ) . At the foot of the perpendicular, A Q → ⋅ ( − 3 , − 4 , 5 ) = 0 \overrightarrow{AQ}\cdot(-3,-4,5)=0 A Q ⋅ ( − 3 , − 4 , 5 ) = 0
− 3 ( − 2 − 3 μ ) − 4 ( 2 − 4 μ ) + 5 ( 3 + 5 μ ) = 0 , 6 + 9 μ − 8 + 16 μ + 15 + 25 μ = 0 , 13 + 50 μ = 0. \begin{align*}
&\,-3(-2-3\mu)-4(2-4\mu)+5(3+5\mu)=0,\\[4mm]
&\,6+9\mu-8+16\mu+15+25\mu=0,\\[4mm]
&\,13+50\mu=0.
\end{align*} − 3 ( − 2 − 3 μ ) − 4 ( 2 − 4 μ ) + 5 ( 3 + 5 μ ) = 0 , 6 + 9 μ − 8 + 16 μ + 15 + 25 μ = 0 , 13 + 50 μ = 0. Hence μ = − 13 / 50 \mu=-13/50 μ = − 13/50 A Q → = ( − 61 , 152 , 85 ) / 50 \overrightarrow{AQ}=(-61,152,85)/50 A Q = ( − 61 , 152 , 85 ) /50
A Q 2 = 61 2 + 152 2 + 85 2 50 2 = 34050 2500 = 681 50 . \begin{align*}
AQ^2=&\,\frac{61^2+152^2+85^2}{50^2}\\[4mm]
=&\,\frac{34050}{2500}=\frac{681}{50}.
\end{align*} A Q 2 = = 5 0 2 6 1 2 + 15 2 2 + 8 5 2 2500 34050 = 50 681 . Thus the perpendicular length is 681 / 50 ≈ 3.69 \sqrt{681/50}\approx3.69 681/50 ≈ 3.69
解法二
思路
展开
也可将 A A A
答题过程
展开
For Q Q Q s s s
A Q 2 = 50 μ 2 + 26 μ + 17 = 50 ( μ + 13 50 ) 2 + 681 50 . \begin{align*}
AQ^2=&\,50\mu^2+26\mu+17\\[4mm]
=&\,50\bigg(\mu+\frac{13}{50}\bigg)^2+\frac{681}{50}.
\end{align*} A Q 2 = = 50 μ 2 + 26 μ + 17 50 ( μ + 50 13 ) 2 + 50 681 . The minimum occurs at μ = − 13 / 50 \mu=-13/50 μ = − 13/50 681 / 50 ≈ 3.69 \sqrt{681/50}\approx3.69 681/50 ≈ 3.69
解法三
思路
展开
选取直线上的固定点，先求该点到 A A A
答题过程
展开
Let P = ( 2 , 1 , 5 ) P=(2,1,5) P = ( 2 , 1 , 5 ) s s s P A → = ( 2 , − 2 , − 3 ) \overrightarrow{PA}=(2,-2,-3) P A = ( 2 , − 2 , − 3 ) 17 17 17 ( − 3 , − 4 , 5 ) (-3,-4,5) ( − 3 , − 4 , 5 )
( 2 , − 2 , − 3 ) ⋅ ( − 3 , − 4 , 5 ) 50 = − 13 50 . \frac{(2,-2,-3)\cdot(-3,-4,5)}{\sqrt{50}}
=-\frac{13}{\sqrt{50}}. 50 ( 2 , − 2 , − 3 ) ⋅ ( − 3 , − 4 , 5 ) = − 50 13 . By Pythagoras, the perpendicular distance is
17 − 169 50 = 681 50 ≈ 3.69. \sqrt{17-\frac{169}{50}}=\sqrt{\frac{681}{50}}\approx3.69. 17 − 50 169 = 50 681 ≈ 3.69.
解法四
思路
展开
还可用向量积表示固定点到 A A A
答题过程
展开
With P = ( 2 , 1 , 5 ) P=(2,1,5) P = ( 2 , 1 , 5 ) e = ( − 3 , − 4 , 5 ) \mathbf e=(-3,-4,5) e = ( − 3 , − 4 , 5 )
P A → × e = ( ( − 2 ) 5 − ( − 3 ) ( − 4 ) , ( − 3 ) ( − 3 ) − 2 ( 5 ) , 2 ( − 4 ) − ( − 2 ) ( − 3 ) ) = ( − 22 , − 1 , − 14 ) . \begin{align*}
\overrightarrow{PA}\times\mathbf e
=&\,\bigl((-2)5-(-3)(-4),\\[4mm]
&\,\hspace{2pt}(-3)(-3)-2(5),\\[4mm]
&\,\hspace{4pt}2(-4)-(-2)(-3)\bigr)\\[4mm]
=&\,(-22,-1,-14).
\end{align*} P A × e = = ( ( − 2 ) 5 − ( − 3 ) ( − 4 ) , ( − 3 ) ( − 3 ) − 2 ( 5 ) , 2 ( − 4 ) − ( − 2 ) ( − 3 ) ) ( − 22 , − 1 , − 14 ) . The perpendicular distance is
∣ P A → × e ∣ ∣ e ∣ = 22 2 + 1 + 14 2 50 = 681 50 ≈ 3.69. \begin{align*}
\frac{|\overrightarrow{PA}\times\mathbf e|}{|\mathbf e|}
=&\,\frac{\sqrt{22^2+1+14^2}}{\sqrt{50}}\\[4mm]
=&\,\sqrt{\frac{681}{50}}\approx3.69.
\end{align*} ∣ e ∣ ∣ P A × e ∣ = = 50 2 2 2 + 1 + 1 4 2 50 681 ≈ 3.69. 