Question

Find the shortest distance between the following pairs of lines whose cartesian equations are:
(i) $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\mathrm{and}\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}$

(ii) $\frac{x-1}{2}=\frac{y+1}{3}=z\mathrm{and}\frac{x+1}{3}=\frac{y-2}{1};z=2$

(iii) $\frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}\mathrm{and}\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}$

(iv) $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\mathrm{and}\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$

Answer 1

(i) The equations of the given lines are

$$\begin{gathered} \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}...\left(1\right) \\ \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}...\left(2\right) \end{gathered}$$

Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \\ \mathrm{Here}, \\ \overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b_1}=2\hat{i}+3\hat{j}+4\hat{k} \end{gathered}$$

Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} \\ \mathrm{Here}, \\ \overrightarrow{a_2}=2\hat{i}+3\hat{j}+5\hat{k} \\ \overrightarrow{b_2}=3\hat{i}+4\hat{j}+5\hat{k} \end{gathered}$$

Now,
$$\begin{gathered} \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}+\hat{j}+2\hat{k} \\ \mathrm{and}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right| \\ =-\hat{i}+2\hat{j}-\hat{k} \\ \Rightarrow \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{{\left(-1\right)}^2+2^2+{\left(-1\right)}^2} \\ =\sqrt{1+4+1} \\ =\sqrt{6} \\ \mathrm{and}\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)=\left(\hat{i}+\hat{j}+2\hat{k}\right).\left(-\hat{i}+2\hat{j}-\hat{k}\right) \\ =-1+2-2 \\ =-1 \end{gathered}$$

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\mathrm{and}\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by

$$\begin{gathered} d=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right| \\ =\left|\frac{-1}{\sqrt{6}}\right| \\ =\frac{1}{\sqrt{6}} \end{gathered}$$

(ii) The equations of the given lines are

$$\begin{gathered} \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}...\left(1\right) \\ \frac{x+1}{3}=\frac{y-2}{1}=\frac{z-2}{0}...\left(2\right) \end{gathered}$$

Since line (1) passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \\ \mathrm{Here}, \\ \overrightarrow{a_1}=\hat{i}-\hat{j}+0\hat{k} \\ \overrightarrow{b_1}=2\hat{i}+3\hat{j}+\hat{k} \end{gathered}$$

Also, line (2) passes through the point ($-$1, 2, 2) and has direction ratios proportional to 3, 1, 0.
Its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} \\ \mathrm{Here}, \\ \overrightarrow{a_2}=-\hat{i}+2\hat{j}+2\hat{k} \\ \overrightarrow{b_2}=3\hat{i}+\hat{j}+0\hat{k} \end{gathered}$$

Now,
$$\begin{gathered} \overrightarrow{a_2}-\overrightarrow{a_1}=-2\hat{i}+3\hat{j}+2\hat{k} \\ \mathrm{and}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 3& 1& 0\end{array}\right| \\ =-\hat{i}+3\hat{j}-7\hat{k} \\ \Rightarrow \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{{\left(-1\right)}^2+3^2+{\left(-7\right)}^2} \\ =\sqrt{1+9+49} \\ =\sqrt{59} \\ \mathrm{and}\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)=\left(-2\hat{i}+3\hat{j}+2\hat{k}\right).\left(-\hat{i}+3\hat{j}-7\hat{k}\right) \\ =2+9-14 \\ =-3 \end{gathered}$$

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\mathrm{and}\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by

$$\begin{gathered} d=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right| \\ =\left|\frac{-3}{\sqrt{59}}\right| \\ =\frac{3}{\sqrt{59}} \end{gathered}$$

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

(iii) $$\begin{gathered} \frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}...\left(1\right) \\ \frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}...\left(2\right) \end{gathered}$$

Since line (1) passes through the point (1, $-$2, 3) and has direction ratios proportional to $-$1, 1, $-$2, its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \\ \mathrm{Here}, \\ \overrightarrow{a_1}=\hat{i}-2\hat{j}+3\hat{k} \\ \overrightarrow{b_1}=-\hat{i}+\hat{j}-2\hat{k} \end{gathered}$$

Also, line (2) passes through the point (1, $-$1, $-$1) and has direction ratios proportional to 1, 2, $-$2.
Its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} \\ \mathrm{Here}, \\ \overrightarrow{a_2}=\hat{i}-\hat{j}-\hat{k} \\ \overrightarrow{b_2}=\hat{i}+2\hat{j}-2\hat{k} \end{gathered}$$

Now,
$$\begin{gathered} \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{j}-4\hat{k} \\ \mathrm{and}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& -2\\ 1& 2& -2\end{array}\right| \\ =2\hat{i}-4\hat{j}-3\hat{k} \\ \Rightarrow \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{2^2+{\left(-4\right)}^2+{\left(-3\right)}^2} \\ =\sqrt{4+16+9} \\ =\sqrt{29} \\ \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)=\left(\hat{j}-4\hat{k}\right).\left(2\hat{i}-4\hat{j}-3\hat{k}\right) \\ =-4+12 \\ =8 \end{gathered}$$

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\mathrm{and}\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by
$$\begin{gathered} d=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right| \\ =\left|\frac{8}{\sqrt{29}}\right| \\ =\frac{8}{\sqrt{29}} \end{gathered}$$


(iv) $$\begin{gathered} \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}...\left(1\right) \\ \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}...\left(2\right) \end{gathered}$$

Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, $-$2, 1, its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \\ \mathrm{Here}, \\ \overrightarrow{a_1}=3\hat{i}+5\hat{j}+7\hat{k} \\ \overrightarrow{b_1}=\hat{i}-2\hat{j}+\hat{k} \end{gathered}$$

Also, line (2) passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1.
Its vector equation is
$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} \\ \mathrm{Here}, \\ \overrightarrow{a_2}=-\hat{i}-\hat{j}-\hat{k} \\ \overrightarrow{b_2}=7\hat{i}-6\hat{j}+\hat{k} \end{gathered}$$

Now,
$$\begin{gathered} \overrightarrow{a_2}-\overrightarrow{a_1}=-4\hat{i}-6\hat{j}-8\hat{k} \\ \mathrm{and}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 1\\ 7& -6& 1\end{array}\right| \\ =4\hat{i}+6\hat{j}+8\hat{k} \\ \Rightarrow \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{4^2+6^2+8^2} \\ =\sqrt{16+36+64} \\ =\sqrt{116} \\ \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)=\left(-4\hat{i}-6\hat{j}-8\hat{k}\right).\left(4\hat{i}+6\hat{j}+8\hat{k}\right) \\ =-16-36-64 \\ =-116 \end{gathered}$$

The shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\mathrm{and}\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ is given by
$$\begin{gathered} d=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right).\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right| \\ =\left|\frac{-116}{\sqrt{116}}\right| \\ =\sqrt{116} \\ =2\sqrt{29} \end{gathered}$$