Question

Determine whether the following pair of lines intersect or not:
(i) $\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+\lambda \left(2\hat{i}+\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(2\hat{i}-\hat{j}\right)+\mathrm{\mu }\left(\hat{i}+\hat{j}-\hat{k}\right)$

(ii) $\frac{x-1}{2}=\frac{y+1}{3}=z\mathrm{and}\frac{x+1}{5}=\frac{y-2}{1};z=2$

(iii) $\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\mathrm{and}\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}$

(iv) $\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\mathrm{and}\frac{x-8}{7}=\frac{y-4}{1}=\frac{3-5}{3}$

Answer 1

(i) $\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+\lambda \left(2\hat{i}+\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(2\hat{i}-\hat{j}\right)+\mathrm{\mu }\left(\hat{i}+\hat{j}-\hat{k}\right)$

The position vectors of two arbitrary points on the given lines are

$$\begin{gathered} \left(\hat{i}-\hat{j}\right)+\lambda \left(2\hat{i}+\hat{k}\right)=\left(1+2\lambda \right)\hat{i}-\hat{j}+\lambda \hat{k} \\ \left(2\hat{i}-\hat{j}\right)+\mu \left(\hat{i}+\hat{j}-\hat{k}\right)=\left(2+\mu \right)\hat{i}+\left(-1+\mu \right)\hat{j}-\mu \hat{k} \end{gathered}$$

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$\left(1+2\lambda \right)\hat{i}+-\hat{j}+\lambda \hat{k}=\left(2+\mu \right)\hat{i}+\left(-1+\mu \right)\hat{j}-\mu \hat{k}$

Equating the coefficients of $\hat{i},\hat{j}\mathrm{and}\hat{k}$, we get

$$\begin{gathered} 1+2\lambda =2+\mu ...\left(1\right) \\ -1=-1+\mu ...\left(2\right) \\ \lambda =-\mu ...\left(3\right) \end{gathered}$$

Solving (2) and (3), we get
$$\begin{gathered} \lambda =0 \\ \mu =0 \end{gathered}$$.

Substituting the values $\lambda =0\mathrm{and}\mu =0$ in (1), we get
$$\begin{gathered} \mathrm{LHS}=1+2\lambda \\ =1+2\left(0\right) \\ =1 \\ \mathrm{RHS}=2+\mu \\ =2+0 \\ =2 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \\ \mathrm{Since}\lambda =0\mathrm{and}\mu =0\mathrm{do}\mathrm{not}\mathrm{satisfy}\left(1\right),\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{do}\mathrm{not}\mathrm{intersect}. \end{gathered}$$


(ii) $\frac{x-1}{2}=\frac{y+1}{3}=z\mathrm{and}\frac{x+1}{5}=\frac{y-2}{1};z=2$

The coordinates of any point on the first line are given by

$$\begin{gathered} \frac{x-1}{2}=\frac{y+1}{3}=z=\lambda \\ \Rightarrow x=2\lambda +1 \\ y=3\lambda -1 \\ z=\lambda \end{gathered}$$

The coordinates of a general point on the first line are $\left(2\lambda +1,3\lambda -1,\lambda \right)$.

Also, the coordinates of any point on the second line are given by

$$\begin{gathered} \frac{x+1}{5}=\frac{y-2}{1}=\mu ,z=2 \\ \Rightarrow x=5\mu -1 \\ y=\mu +2 \\ z=2 \end{gathered}$$

The coordinates of a general point on the second line are $\left(5\mu -1,\mu +2,2\right)$.

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$$\begin{gathered} 2\lambda +1=5\mu -1,3\lambda -1=\mu +2,\lambda =2 \\ \Rightarrow 2\lambda -5\mu =-2...\left(1\right) \\ 3\lambda -\mu =3...\left(2\right) \\ \lambda =2...\left(3\right) \\ \mathrm{Solving}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \lambda =2 \\ \mu =3 \\ \mathrm{Substituting}\lambda =2\mathrm{and}\mu =3\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=2\lambda -5\mu \\ =2\left(2\right)-5\left(3\right) \\ =4-15 \\ =-11\ne -2 \\ \Rightarrow \mathrm{LHS}\ne \mathrm{RHS} \\ \mathrm{Since}\lambda =2\mathrm{and}\mu =3\mathrm{do}\mathrm{not}\mathrm{satisfy}\left(1\right),\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{do}\mathrm{not}\mathrm{intersect}. \end{gathered}$$


(iii) $\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\mathrm{and}\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}$

The coordinates of any point on the first line are given by

$$\begin{gathered} \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda \\ \Rightarrow x=3\lambda +1 \\ y=-\lambda +1 \\ z=-1 \end{gathered}$$

The coordinates of a general point on the first line are $\left(3\lambda +1,-\lambda +1,-1\right)$.

Also, the coordinates of any point on the second line are given by

$$\begin{gathered} \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu \\ \Rightarrow x=2\mu +4 \\ y=0 \\ z=3\mu -1 \end{gathered}$$

The coordinates of a general point on the second line are $\left(2\mu +4,0,3\mu -1\right)$.

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$$\begin{gathered} 3\lambda +1=2\mu +4,-\lambda +1=0,-1=3\mu -1 \\ \Rightarrow 3\lambda -2\mu =3...\left(1\right) \\ \lambda =1...\left(2\right) \\ \mu =0...\left(3\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \lambda =1 \\ \mu =0 \\ \mathrm{Substituting}\lambda =1\mathrm{and}\mu =0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=3\lambda -2\mu \\ =3\left(1\right)-2\left(0\right) \\ =3 \\ =\mathrm{RHS} \\ \mathrm{Since}\lambda =1\mathrm{and}\mu =0\mathrm{satisfy}\left(1\right),\mathrm{the}\mathrm{lines}\mathrm{intersect}. \\ \mathrm{Substituting}\lambda =1\mathrm{and}\mu =0\mathrm{in}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{a}\mathrm{general}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{first}\mathrm{line},\mathrm{we}\mathrm{get} \\ x=4 \\ y=0 \\ z=-1 \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}\mathrm{at}\left(4,0,-1\right). \end{gathered}$$


(iv) $\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\mathrm{and}\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$

The coordinates of any point on the first line are given by

$$\begin{gathered} \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda \\ \Rightarrow x=4\lambda +5 \\ y=4\lambda +7 \\ z=-5\lambda -3 \end{gathered}$$

The coordinates of a general point on the first line are $\left(4\lambda +5,4\lambda +7,-5\lambda -3\right)$.

The coordinates of any point on the second line are given by

$$\begin{gathered} \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu \\ \Rightarrow x=7\mu +8 \\ y=\mu +4 \\ z=3\mu +5 \end{gathered}$$

The coordinates of a general point on the second line are $\left(7\mu +8,\mu +4,3\mu +5\right)$.

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$$\begin{gathered} 4\lambda +5=7\mu +8,4\lambda +7=\mu +4,-5\lambda -3=3\mu +5 \\ \Rightarrow 4\lambda -7\mu =3...\left(1\right) \\ 4\lambda -\mu =-3...\left(2\right) \\ 5\lambda +3\mu =-8...\left(3\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \lambda =-1 \\ \mu =-1 \\ \mathrm{Substituting}\lambda =-1\mathrm{and}\mu =-1\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=5\lambda +3\mu \\ =5\left(-1\right)+3\left(-1\right) \\ =-8 \\ =\mathrm{RH}S \\ \mathrm{Since}\lambda =-1\mathrm{and}\mu =-1\mathrm{satisfy}\left(3\right),\mathrm{the}\mathrm{lines}\mathrm{intersect}. \\ \mathrm{Substituting}\lambda =-1\mathrm{and}\mu =-1\mathrm{in}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{a}\mathrm{general}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{first}\mathrm{line},\mathrm{we}\mathrm{get} \\ x=1 \\ y=3 \\ z=2 \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}\mathrm{at}\left(1,3,2\right). \end{gathered}$$

Disclaimer: The question printed in the book is incorrect. Instead of z, 3 is printed.