题目
Problem
Relative to a fixed origin O O O l 1 l_1 l 1 l 2 l_2 l 2
l 1 : r = ( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) l_1:\ \mathbf{r}=(3\mathbf{i}+p\mathbf{j}+7\mathbf{k})+\lambda(2\mathbf{i}-5\mathbf{j}+4\mathbf{k}) l 1 : r = ( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) l 2 : r = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) l_2:\ \mathbf{r}=(8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+\mu(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) l 2 : r = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) where λ \lambda λ μ \mu μ p p p
Given that l 1 l_1 l 1 l 2 l_2 l 2
(a) find the value of p p p
(4)
(b) find the position vector of the point of intersection.
(2)
(c) Find the acute angle between l 1 l_1 l 1 l 2 l_2 l 2
Give your answer in degrees to one decimal place.
(3)
The point A A A l 1 l_1 l 1 λ = 2 \lambda=2 λ = 2
The point B B B l 2 l_2 l 2 A B → \overrightarrow{AB} A B l 2 l_2 l 2
(d) Find the coordinates of B B B
(5)
题目中文翻译
相对于固定原点 O O O l 1 l_1 l 1 l 2 l_2 l 2
l 1 : r = ( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) l_1:\ \mathbf{r}=(3\mathbf{i}+p\mathbf{j}+7\mathbf{k})+\lambda(2\mathbf{i}-5\mathbf{j}+4\mathbf{k}) l 1 : r = ( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) l 2 : r = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) l_2:\ \mathbf{r}=(8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+\mu(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) l 2 : r = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) 其中 λ , μ \lambda,\mu λ , μ p p p
已知 l 1 l_1 l 1 l 2 l_2 l 2
(a) 求 p p p
(b) 求交点的位置向量；
(c) 求 l 1 l_1 l 1 l 2 l_2 l 2
点 A A A l 1 l_1 l 1 λ = 2 \lambda=2 λ = 2
点 B B B l 2 l_2 l 2 A B → \overrightarrow{AB} A B l 2 l_2 l 2
(d) 求点 B B B
解答
(a)
解法一
思路
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两条直线相交，表示存在同一个点同时满足两条参数方程。把 i , j , k \mathbf{i},\mathbf{j},\mathbf{k} i , j , k p p p λ , μ \lambda,\mu λ , μ p p p
答题过程
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Since the two lines intersect,
( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) . (3\mathbf{i}+p\mathbf{j}+7\mathbf{k})
+\lambda(2\mathbf{i}-5\mathbf{j}+4\mathbf{k})
=(8\mathbf{i}-2\mathbf{j}+5\mathbf{k})
+\mu(4\mathbf{i}+\mathbf{j}+2\mathbf{k}). ( 3 i + p j + 7 k ) + λ ( 2 i − 5 j + 4 k ) = ( 8 i − 2 j + 5 k ) + μ ( 4 i + j + 2 k ) . Equating components gives
3 + 2 λ = 8 + 4 μ , ( 1 ) p − 5 λ = − 2 + μ , ( 2 ) 7 + 4 λ = 5 + 2 μ . ( 3 ) \begin{align*}
3+2\lambda &= 8+4\mu, \qquad (1)\\
p-5\lambda &= -2+\mu, \qquad (2)\\
7+4\lambda &= 5+2\mu. \qquad (3)
\end{align*} 3 + 2 λ p − 5 λ 7 + 4 λ = 8 + 4 μ , ( 1 ) = − 2 + μ , ( 2 ) = 5 + 2 μ . ( 3 ) From (1),
λ = 5 2 + 2 μ . \lambda=\frac52+2\mu. λ = 2 5 + 2 μ . Substitute this into (3):
7 + 4 ( 5 2 + 2 μ ) = 5 + 2 μ 17 + 8 μ = 5 + 2 μ 6 μ = − 12 μ = − 2. \begin{align*}
7+4\left(\frac52+2\mu\right)&=5+2\mu\\
17+8\mu&=5+2\mu\\
6\mu&=-12\\
\mu&=-2.
\end{align*} 7 + 4 ( 2 5 + 2 μ ) 17 + 8 μ 6 μ μ = 5 + 2 μ = 5 + 2 μ = − 12 = − 2. Then
λ = 5 2 + 2 ( − 2 ) = − 3 2 . \lambda=\frac52+2(-2)=-\frac32. λ = 2 5 + 2 ( − 2 ) = − 2 3 . Use (2):
p − 5 ( − 3 2 ) = − 2 + ( − 2 ) p + 15 2 = − 4 p = − 23 2 . \begin{align*}
p-5\left(-\frac32\right)&=-2+(-2)\\
p+\frac{15}{2}&=-4\\
p&=-\frac{23}{2}.
\end{align*} p − 5 ( − 2 3 ) p + 2 15 p = − 2 + ( − 2 ) = − 4 = − 2 23 . Therefore
p = − 23 2 . \boxed{p=-\frac{23}{2}}. p = − 2 23 .
(b)
解法一
思路
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交点已经由 (a) 的参数值确定。把 μ = − 2 \mu=-2 μ = − 2 l 2 l_2 l 2 λ = − 3 2 \lambda=-\frac32 λ = − 2 3 l 1 l_1 l 1
答题过程
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Using μ = − 2 \mu=-2 μ = − 2 l 2 l_2 l 2
r = ( 8 i − 2 j + 5 k ) − 2 ( 4 i + j + 2 k ) = 8 i − 2 j + 5 k − 8 i − 2 j − 4 k = − 4 j + k . \begin{align*}
\mathbf{r}
=&\, (8\mathbf{i}-2\mathbf{j}+5\mathbf{k})
-2(4\mathbf{i}+\mathbf{j}+2\mathbf{k})\\
=&\, 8\mathbf{i}-2\mathbf{j}+5\mathbf{k}
-8\mathbf{i}-2\mathbf{j}-4\mathbf{k}\\
=&\, -4\mathbf{j}+\mathbf{k}.
\end{align*} r = = = ( 8 i − 2 j + 5 k ) − 2 ( 4 i + j + 2 k ) 8 i − 2 j + 5 k − 8 i − 2 j − 4 k − 4 j + k . Therefore the position vector of the point of intersection is
− 4 j + k . \boxed{-4\mathbf{j}+\mathbf{k}}. − 4 j + k .
(c)
解法一
思路
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两条直线的夹角等于方向向量的夹角。用点积公式求 cos θ \cos\theta cos θ
答题过程
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The direction vectors are
a = 2 i − 5 j + 4 k , b = 4 i + j + 2 k . \mathbf{a}=2\mathbf{i}-5\mathbf{j}+4\mathbf{k},
\qquad
\mathbf{b}=4\mathbf{i}+\mathbf{j}+2\mathbf{k}. a = 2 i − 5 j + 4 k , b = 4 i + j + 2 k . Their scalar product is
a ⋅ b = 2 ( 4 ) + ( − 5 ) ( 1 ) + 4 ( 2 ) = 8 − 5 + 8 = 11. \begin{align*}
\mathbf{a}\cdot\mathbf{b}
&=2(4)+(-5)(1)+4(2)\\
&=8-5+8\\
&=11.
\end{align*} a ⋅ b = 2 ( 4 ) + ( − 5 ) ( 1 ) + 4 ( 2 ) = 8 − 5 + 8 = 11. Also,
∣ a ∣ = 2 2 + ( − 5 ) 2 + 4 2 = 45 = 3 5 , |\mathbf{a}|=\sqrt{2^2+(-5)^2+4^2}=\sqrt{45}=3\sqrt5, ∣ a ∣ = 2 2 + ( − 5 ) 2 + 4 2 = 45 = 3 5 , and
∣ b ∣ = 4 2 + 1 2 + 2 2 = 21 . |\mathbf{b}|=\sqrt{4^2+1^2+2^2}=\sqrt{21}. ∣ b ∣ = 4 2 + 1 2 + 2 2 = 21 . Therefore
cos θ = 11 ( 3 5 ) ( 21 ) = 11 3 105 . \cos\theta=\frac{11}{(3\sqrt5)(\sqrt{21})}
=\frac{11}{3\sqrt{105}}. cos θ = ( 3 5 ) ( 21 ) 11 = 3 105 11 . So
θ = cos − 1 ( 11 3 105 ) = 68.965 … ∘ . \theta=\cos^{-1}\left(\frac{11}{3\sqrt{105}}\right)=68.965\ldots^\circ. θ = cos − 1 ( 3 105 11 ) = 68.965 … ∘ . The acute angle is
69.0 ∘ . \boxed{69.0^\circ}. 69. 0 ∘ .
(d)
解法一
思路
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先由 λ = 2 \lambda=2 λ = 2 A A A B B B l 2 l_2 l 2 A B → \overrightarrow{AB} A B l 2 l_2 l 2 A B → \overrightarrow{AB} A B l 2 l_2 l 2 0 0 0 B B B μ \mu μ
答题过程
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When λ = 2 \lambda=2 λ = 2
O A → = ( 3 i + p j + 7 k ) + 2 ( 2 i − 5 j + 4 k ) = 3 i − 23 2 j + 7 k + 4 i − 10 j + 8 k = 7 i − 43 2 j + 15 k . \begin{align*}
\overrightarrow{OA}
=&\, (3\mathbf{i}+p\mathbf{j}+7\mathbf{k})
+2(2\mathbf{i}-5\mathbf{j}+4\mathbf{k})\\
=&\, 3\mathbf{i}-\frac{23}{2}\mathbf{j}+7\mathbf{k}
+4\mathbf{i}-10\mathbf{j}+8\mathbf{k}\\
=&\, 7\mathbf{i}-\frac{43}{2}\mathbf{j}+15\mathbf{k}.
\end{align*} O A = = = ( 3 i + p j + 7 k ) + 2 ( 2 i − 5 j + 4 k ) 3 i − 2 23 j + 7 k + 4 i − 10 j + 8 k 7 i − 2 43 j + 15 k . A general point B B B l 2 l_2 l 2
O B → = ( 8 + 4 μ ) i + ( − 2 + μ ) j + ( 5 + 2 μ ) k . \overrightarrow{OB}
=(8+4\mu)\mathbf{i}+(-2+\mu)\mathbf{j}+(5+2\mu)\mathbf{k}. O B = ( 8 + 4 μ ) i + ( − 2 + μ ) j + ( 5 + 2 μ ) k . Thus
A B → = O B → − O A → = ( 1 + 4 μ ) i + ( 39 2 + μ ) j + ( − 10 + 2 μ ) k . \begin{align*}
\overrightarrow{AB}
=&\, \overrightarrow{OB}-\overrightarrow{OA}\\
=&\, (1+4\mu)\mathbf{i}
+\left(\frac{39}{2}+\mu\right)\mathbf{j}
+(-10+2\mu)\mathbf{k}.
\end{align*} A B = = O B − O A ( 1 + 4 μ ) i + ( 2 39 + μ ) j + ( − 10 + 2 μ ) k . Since A B → \overrightarrow{AB} A B l 2 l_2 l 2
A B → ⋅ ( 4 i + j + 2 k ) = 0. \overrightarrow{AB}\cdot(4\mathbf{i}+\mathbf{j}+2\mathbf{k})=0. A B ⋅ ( 4 i + j + 2 k ) = 0. Hence
4 ( 1 + 4 μ ) + ( 39 2 + μ ) + 2 ( − 10 + 2 μ ) = 0 4 + 16 μ + 39 2 + μ − 20 + 4 μ = 0 21 μ + 7 2 = 0 μ = − 1 6 . \begin{align*}
4(1+4\mu)+\left(\frac{39}{2}+\mu\right)+2(-10+2\mu)&=0\\
4+16\mu+\frac{39}{2}+\mu-20+4\mu&=0\\
21\mu+\frac72&=0\\
\mu&=-\frac16.
\end{align*} 4 ( 1 + 4 μ ) + ( 2 39 + μ ) + 2 ( − 10 + 2 μ ) 4 + 16 μ + 2 39 + μ − 20 + 4 μ 21 μ + 2 7 μ = 0 = 0 = 0 = − 6 1 . Substitute μ = − 1 6 \mu=-\frac16 μ = − 6 1 l 2 l_2 l 2
O B → = ( 8 i − 2 j + 5 k ) − 1 6 ( 4 i + j + 2 k ) = 22 3 i − 13 6 j + 14 3 k . \begin{align*}
\overrightarrow{OB}
=&\, (8\mathbf{i}-2\mathbf{j}+5\mathbf{k})
-\frac16(4\mathbf{i}+\mathbf{j}+2\mathbf{k})\\
=&\, \frac{22}{3}\mathbf{i}
-\frac{13}{6}\mathbf{j}
+\frac{14}{3}\mathbf{k}.
\end{align*} O B = = ( 8 i − 2 j + 5 k ) − 6 1 ( 4 i + j + 2 k ) 3 22 i − 6 13 j + 3 14 k . Therefore
B ( 22 3 , − 13 6 , 14 3 ) . \boxed{B\left(\frac{22}{3},-\frac{13}{6},\frac{14}{3}\right)}. B ( 3 22 , − 6 13 , 3 14 ) .
解法二
思路
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也可以用几何方法。令 X X X A X B AXB A X B B B B B B B l 2 l_2 l 2
答题过程
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From part (b), the intersection point is
X = ( 0 , − 4 , 1 ) . X=(0,-4,1). X = ( 0 , − 4 , 1 ) . Also
A = ( 7 , − 43 2 , 15 ) . A=\left(7,-\frac{43}{2},15\right). A = ( 7 , − 2 43 , 15 ) . So
X A → = ( 7 , − 35 2 , 14 ) . \overrightarrow{XA}
=\left(7,-\frac{35}{2},14\right). X A = ( 7 , − 2 35 , 14 ) . A general point B B B l 2 l_2 l 2
B = ( 8 + 4 μ , − 2 + μ , 5 + 2 μ ) , B=(8+4\mu,-2+\mu,5+2\mu), B = ( 8 + 4 μ , − 2 + μ , 5 + 2 μ ) , so
X B → = ( 8 + 4 μ , 2 + μ , 4 + 2 μ ) . \overrightarrow{XB}
=(8+4\mu,2+\mu,4+2\mu). X B = ( 8 + 4 μ , 2 + μ , 4 + 2 μ ) . Since the angle between the two lines is θ \theta θ B B B
X B = X A cos θ . XB=XA\cos\theta. X B = X A cos θ . Using
cos 2 θ = ( 11 3 105 ) 2 = 121 945 , \cos^2\theta=\left(\frac{11}{3\sqrt{105}}\right)^2=\frac{121}{945}, cos 2 θ = ( 3 105 11 ) 2 = 945 121 , we get
( 8 + 4 μ ) 2 + ( 2 + μ ) 2 + ( 4 + 2 μ ) 2 = ( 7 2 + ( − 35 2 ) 2 + 14 2 ) × 121 945 . \begin{align*}
(8+4\mu)^2+(2+\mu)^2+(4+2\mu)^2
=&\, \left(7^2+\left(-\frac{35}{2}\right)^2+14^2\right)
\times\frac{121}{945}.
\end{align*} ( 8 + 4 μ ) 2 + ( 2 + μ ) 2 + ( 4 + 2 μ ) 2 = ( 7 2 + ( − 2 35 ) 2 + 1 4 2 ) × 945 121 . This simplifies to
36 μ 2 + 144 μ + 23 = 0. 36\mu^2+144\mu+23=0. 36 μ 2 + 144 μ + 23 = 0. Solving,
μ = − 1 6 or μ = − 23 6 . \mu=-\frac16
\quad\text{or}\quad
\mu=-\frac{23}{6}. μ = − 6 1 or μ = − 6 23 . At the intersection point X X X l 2 l_2 l 2 μ = − 2 \mu=-2 μ = − 2 X A → \overrightarrow{XA} X A l 2 l_2 l 2 μ > − 2 \mu>-2 μ > − 2
μ = − 1 6 . \mu=-\frac16. μ = − 6 1 . Substituting into l 2 l_2 l 2
B ( 22 3 , − 13 6 , 14 3 ) . \boxed{B\left(\frac{22}{3},-\frac{13}{6},\frac{14}{3}\right)}. B ( 3 22 , − 6 13 , 3 14 ) . 