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P4 2022 Oct Q9

A Level / Edexcel / P4

IAL 2022 Oct Paper · Question 9

题目

Problem

With respect to a fixed origin OO, the equations of lines l1l_1 and l2l_2 are given by

l1: r= (2810) +λ(−123)\begin{align*} l_1:\ \mathbf{r}=&\,\begin{pmatrix}2\\8\\10\end{pmatrix}\\ &\,+\lambda\begin{pmatrix}-1\\2\\3\end{pmatrix} \end{align*} l2: r= (−4−12) +μ(548)\begin{align*} l_2:\ \mathbf{r}=&\,\begin{pmatrix}-4\\-1\\2\end{pmatrix}\\ &\,+\mu\begin{pmatrix}5\\4\\8\end{pmatrix} \end{align*}

where λ\lambda and μ\mu are scalar parameters.

Prove that lines l1l_1 and l2l_2 are skew.

(5)
题目中文翻译

相对于固定原点 OO，直线 l1l_1 和 l2l_2 的方程分别为

l1: r= (2810) +λ(−123)\begin{align*} l_1:\ \mathbf{r}=&\,\begin{pmatrix}2\\8\\10\end{pmatrix}\\ &\,+\lambda\begin{pmatrix}-1\\2\\3\end{pmatrix} \end{align*} l2: r= (−4−12) +μ(548)\begin{align*} l_2:\ \mathbf{r}=&\,\begin{pmatrix}-4\\-1\\2\end{pmatrix}\\ &\,+\mu\begin{pmatrix}5\\4\\8\end{pmatrix} \end{align*}

其中 λ,μ\lambda,\mu 为标量参数。

证明直线 l1l_1 与 l2l_2 是异面直线。

(5)

解答

解法一

思路

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证明两条直线异面需要完成两件事：先比较方向向量，证明它们不平行；再假设两线相交，联立三个坐标方程。由前两个坐标求出参数后，第三个坐标不相等，因此两线不相交。

答题过程

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The direction vectors of l1l_1 and l2l_2 are

d1= (−123),d2= (548).\begin{align*} \mathbf d_1=&\,\begin{pmatrix} -1\\ 2\\ 3 \end{pmatrix},\\ \mathbf d_2=&\,\begin{pmatrix} 5\\ 4\\ 8 \end{pmatrix}. \end{align*}

They are not scalar multiples, since, for example,

5−1≠42.\frac{5}{-1}\ne\frac{4}{2}.

Therefore, l1l_1 and l2l_2 are not parallel.

If the lines intersect, their coordinates must satisfy

2−λ= −4+5μ,8+2λ= −1+4μ,10+3λ= 2+8μ.\begin{align*} 2-\lambda=&\,-4+5\mu,\\ 8+2\lambda=&\,-1+4\mu,\\ 10+3\lambda=&\,2+8\mu. \end{align*}

Solving the first two equations,

λ+5μ= 6,2λ−4μ= −9,\begin{align*} \lambda+5\mu=&\,6,\\ 2\lambda-4\mu=&\,-9, \end{align*}

Doubling the first equation and subtracting the second gives

14μ= 21,μ= 32.\begin{align*} 14\mu=&\,21,\\ \mu=&\,\frac32. \end{align*}

Hence

λ=6−5(32)=−32.\lambda=6-5\bigg(\frac32\bigg)=-\frac32.

Substituting these values into the third coordinate,

10+3(−32)=112,10+3\bigg(-\frac32\bigg)=\frac{11}{2},

whereas

2+8(32)=14.2+8\bigg(\frac32\bigg)=14.

These values are not equal, so the lines do not intersect.

Since l1l_1 and l2l_2 are neither parallel nor intersecting, they are skew.

The same coordinate-checking method can start with the second and third equations. These give

2λ−4μ= −9,3λ−8μ= −8.\begin{align*} 2\lambda-4\mu=&\,-9,\\ 3\lambda-8\mu=&\,-8. \end{align*}

Doubling the first of these and subtracting the second gives λ=−10\lambda=-10, and then μ=−114\mu=-\frac{11}{4}. The first coordinates are

2−(−10)= 12,−4+5(−114)= −714.\begin{align*} 2-(-10)=&\,12,\\ -4+5\bigg(-\frac{11}{4}\bigg)=&\,-\frac{71}{4}. \end{align*}

They are different, so this also shows that the lines do not intersect.

Alternatively, non-parallel directions can be checked using the scalar product:

d1⋅d2= −5+8+24= 27,∣d1∣2= 14,∣d2∣2= 105.\begin{align*} \mathbf d_1\cdot\mathbf d_2=&\,-5+8+24\\ =&\,27,\\ |\mathbf d_1|^2=&\,14,\\ |\mathbf d_2|^2=&\,105. \end{align*}

Thus the angle θ\theta between the directions satisfies

cos⁡θ=2714×105≠±1,\cos\theta=\frac{27}{\sqrt{14\times105}}\ne\pm1,

because 272=729≠1470=14×10527^2=729\ne1470=14\times105. Hence the lines are not parallel.

解法二：比较参数的一致性

思路

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若两线相交，同一个参数 λ\lambda 必须同时满足三个坐标方程。分别用第一、第二坐标以及第一、第三坐标消去 μ\mu，得到两个不同的 λ\lambda，因此不存在共同交点。最后仍须说明方向向量不成比例，才能得出异面直线的结论。

答题过程

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If the lines intersect, equating their coordinates gives

λ+5μ= 6(1),2λ−4μ= −9(2),3λ−8μ= −8(3).\begin{align*} \lambda+5\mu=&\,6 &&(1),\\ 2\lambda-4\mu=&\,-9 &&(2),\\ 3\lambda-8\mu=&\,-8 &&(3). \end{align*}

Multiplying equation (1) by 44 and equation (2) by 55 gives

4λ+20μ= 24,10λ−20μ= −45.\begin{align*} 4\lambda+20\mu=&\,24,\\ 10\lambda-20\mu=&\,-45. \end{align*}

Adding these equations,

14λ= −21,λ= −32.\begin{align*} 14\lambda=&\,-21,\\ \lambda=&\,-\frac32. \end{align*}

Instead, multiplying equation (1) by 88 and equation (3) by 55 gives

8λ+40μ= 48,15λ−40μ= −40.\begin{align*} 8\lambda+40\mu=&\,48,\\ 15\lambda-40\mu=&\,-40. \end{align*}

Adding these equations,

23λ= 8,λ= 823.\begin{align*} 23\lambda=&\,8,\\ \lambda=&\,\frac{8}{23}. \end{align*}

Since −32≠823-\frac32\ne\frac{8}{23}, no single value of λ\lambda satisfies all three equations. Therefore the lines do not intersect.

Their direction vectors are not scalar multiples, since

5−1≠42.\frac{5}{-1}\ne\frac{4}{2}.

Hence the lines are not parallel. Being neither parallel nor intersecting, l1l_1 and l2l_2 are skew.