题目
Problem
The lines l l l and m m m have vector equations
l : r = 2 i + j − 3 k + λ ( − i + 2 k ) l:\quad\mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+2\mathbf{k}) l : r = 2 i + j − 3 k + λ ( − i + 2 k )
and
m : r = 2 i + j − 3 k + μ ( 2 i − j + 5 k ) . m:\quad\mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\mu(2\mathbf{i}-\mathbf{j}+5\mathbf{k}). m : r = 2 i + j − 3 k + μ ( 2 i − j + 5 k ) .
Lines l l l and m m m intersect at the point P P P .
(a) State the coordinates of P P P .
(1)
(b) Find the exact value of the cosine of the acute angle between l l l and m m m .
(3)
(c) The point A A A on line l l l has coordinates ( 0 , 1 , 1 ) (0,1,1) ( 0 , 1 , 1 ) . The point B B B on line m m m has coordinates ( 0 , 2 , − 8 ) (0,2,-8) ( 0 , 2 , − 8 ) .
Find the exact area of triangle A P B APB A P B .
(3)
题目中文翻译
直线 l l l 和 m m m 的向量方程为
l : r = 2 i + j − 3 k + λ ( − i + 2 k ) l:\quad\mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+2\mathbf{k}) l : r = 2 i + j − 3 k + λ ( − i + 2 k )
和
m : r = 2 i + j − 3 k + μ ( 2 i − j + 5 k ) . m:\quad\mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\mu(2\mathbf{i}-\mathbf{j}+5\mathbf{k}). m : r = 2 i + j − 3 k + μ ( 2 i − j + 5 k ) .
直线 l l l 和 m m m 相交于点 P P P 。
(a) 写出 P P P 的坐标。
(1)
(b) 求直线 l l l 与 m m m 之间锐角的余弦的精确值。
(3)
(c) 直线 l l l 上的点 A A A 的坐标为 ( 0 , 1 , 1 ) (0,1,1) ( 0 , 1 , 1 ) 。直线 m m m 上的点 B B B 的坐标为 ( 0 , 2 , − 8 ) (0,2,-8) ( 0 , 2 , − 8 ) 。
求三角形 A P B APB A P B 的精确面积。
(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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两条直线相同的位置向量对应各自参数为零时的公共点。
答题过程
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Both lines pass through the point with position vector 2 i + j − 3 k 2\mathbf i+\mathbf j-3\mathbf k 2 i + j − 3 k . Hence
P = ( 2 , 1 , − 3 ) . \boxed{P=(2,1,-3)}. P = ( 2 , 1 , − 3 ) .
(b)
解法一
思路
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求两条方向向量的点积,锐角取点积绝对值。
答题过程
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The direction vectors are ( − 1 , 0 , 2 ) (-1,0,2) ( − 1 , 0 , 2 ) and ( 2 , − 1 , 5 ) (2,-1,5) ( 2 , − 1 , 5 ) . Their scalar product is
( − 1 ) ( 2 ) + 0 ( − 1 ) + 2 ( 5 ) = 8. (-1)(2)+0(-1)+2(5)=8. ( − 1 ) ( 2 ) + 0 ( − 1 ) + 2 ( 5 ) = 8.
The moduli are 5 \sqrt5 5 and 30 \sqrt{30} 30 . Thus the acute angle θ \theta θ satisfies
cos θ = 8 5 30 = 8 5 6 . \boxed{\cos\theta
=\frac8{\sqrt5\sqrt{30}}
=\frac8{5\sqrt6}}. cos θ = 5 30 8 = 5 6 8 .
解法二
思路
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在两条直线上各选距公共点一个方向向量的位置,组成三角形后用余弦定理。
答题过程
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Choose points U = P + ( − 1 , 0 , 2 ) U=P+(-1,0,2) U = P + ( − 1 , 0 , 2 ) on l l l and V = P + ( 2 , − 1 , 5 ) V=P+(2,-1,5) V = P + ( 2 , − 1 , 5 ) on m m m . Then
P U 2 = 5 , P V 2 = 30. PU^2=5,\qquad PV^2=30. P U 2 = 5 , P V 2 = 30.
Also U V → = ( 3 , − 1 , 3 ) \overrightarrow{UV}=(3,-1,3) U V = ( 3 , − 1 , 3 ) , so U V 2 = 19 UV^2=19 U V 2 = 19 .
By the cosine rule,
cos ∠ U P V = 5 + 30 − 19 2 5 30 = 8 5 6 . \cos\angle UPV
=\frac{5+30-19}{2\sqrt5\sqrt{30}}
=\boxed{\frac8{5\sqrt6}}. cos ∠ U P V = 2 5 30 5 + 30 − 19 = 5 6 8 .
This is positive, so ∠ U P V \angle UPV ∠ U P V is the acute angle between the lines.
(c)
解法一
思路
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三角形在 P P P 的夹角可能是两直线锐角的补角,但正弦相同。求实际边长后用正弦面积公式。
答题过程
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Using the coordinates,
P A → = ( − 2 , 0 , 4 ) , P B → = ( − 2 , 1 , − 5 ) . \begin{align*}
\overrightarrow{PA}=&\,(-2,0,4),\\[4mm]
\overrightarrow{PB}=&\,(-2,1,-5).
\end{align*} P A = P B = ( − 2 , 0 , 4 ) , ( − 2 , 1 , − 5 ) .
Thus P A = 2 5 PA=2\sqrt5 P A = 2 5 and P B = 30 PB=\sqrt{30} P B = 30 .
The angle ∠ A P B \angle APB ∠ A P B has the same sine as the acute angle θ \theta θ in (b). Hence
sin ∠ A P B = 1 − 64 150 = 86 150 . \sin\angle APB
=\sqrt{1-\frac{64}{150}}
=\sqrt{\frac{86}{150}}. sin ∠ A P B = 1 − 150 64 = 150 86 .
Therefore
Area ( A P B ) = 1 2 P A ⋅ P B sin ∠ A P B = 1 2 ( 2 5 ) ( 30 ) 86 150 = 86 . \begin{align*}
\operatorname{Area}(APB)
=&\,\frac12PA\cdot PB\sin\angle APB\\
=&\,\frac12(2\sqrt5)(\sqrt{30})\sqrt{\frac{86}{150}}\\
=&\,\boxed{\sqrt{86}}.
\end{align*} Area ( A P B ) = = = 2 1 P A ⋅ P B sin ∠ A P B 2 1 ( 2 5 ) ( 30 ) 150 86 86 .