Question

Find the angle between the lines whose direction cosines are given by the equations
(i) l + m + n = 0 and l² + m² − n² = 0
(ii) 2l − m + 2n = 0 and mn + nl + lm = 0
(iii) l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
(iv) 2l + 2m − n = 0, mn + ln + lm = 0

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Given}: \\ l+m+n=0...\left(1\right) \\ {l}^2+m^2-n^2=0...\left(2\right) \\ \mathrm{From}\left(1\right),\mathrm{we}\mathrm{get} \\ m=-l-n \\ \mathrm{Substituting}m=-l-n\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ {l}^2+{\left(-l-n\right)}^2-n^2 \\ \Rightarrow l^2+l^2+n^2+2\ln -n^2=0 \\ \Rightarrow 2l^2+2\ln =0 \\ \Rightarrow 2l\left(l+n\right)=0 \\ \Rightarrow l=0,l=-n \\ \mathrm{If}l=0,\mathrm{then}\mathrm{by}\mathrm{substituting}l=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}m=-n. \\ \mathrm{If}l=-n,\mathrm{then}\mathrm{by}\mathrm{substituting}l=-n\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}m=0. \\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{are}\mathrm{proportional}\mathrm{to}0,-n,n\mathrm{and}-n,0,n\mathrm{or}0,-1,1\mathrm{and}-1,0,1. \\ \mathrm{Vectors}\mathrm{parallel}\mathrm{to}\mathrm{these}\mathrm{lines}\mathrm{are} \\ \overrightarrow{a}=0\hat{i}-\hat{j}+\hat{k} \\ \overrightarrow{b}=-\hat{i}+0\hat{j}+\hat{k} \\ \mathrm{If}\theta \mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines},\mathrm{then}\theta \mathrm{is}\mathrm{also}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b}. \\ \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ =\frac{1}{\sqrt{0+1+1}\sqrt{1+0+1}} \\ =\frac{1}{2} \\ \Rightarrow \theta =\frac{\pi }{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Given}: \\ 2l-m+2n=0...\left(1\right) \\ mn+nl+lm=0...\left(2\right) \\ \mathrm{From}\left(1\right),\mathrm{we}\mathrm{get} \\ m=2l+2n \\ \mathrm{Substituting}m=2l+2n\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \left(2l+2n\right)n+nl+l\left(2l+2n\right)=0 \\ \Rightarrow 2\ln +2n^2+nl+2l^2+2\ln =0 \\ \Rightarrow 2l^2+5ln+2n^2=0 \\ \Rightarrow \left(l+2n\right)\left(2l+n\right)=0 \\ \Rightarrow l=-2n,-\frac{n}{2} \\ \mathrm{If}l=-2n,\mathrm{then}\mathrm{by}\mathrm{substituting}l=-2n\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}m=-2n. \\ \mathrm{If}l=-\frac{n}{2},\mathrm{then}\mathrm{by}\mathrm{substituting}l=-\frac{n}{2}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}m=n. \\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{are}\mathrm{proportional}\mathrm{to}-2n,-2n,n\mathrm{and}-\frac{n}{2},n,n\mathrm{or}-2,-2,1\mathrm{and}-\frac{1}{2},1,1. \\ \mathrm{Vectors}\mathrm{parallel}\mathrm{these}\mathrm{lines}\mathrm{are} \\ \overrightarrow{a}=-2\hat{i}-\widehat{2j}+\hat{k} \\ \overrightarrow{b}=-\frac{1}{2}\hat{i}+\hat{j}+\hat{k} \\ \mathrm{If}\theta \mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines},\mathrm{then}\theta \mathrm{is}\mathrm{also}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b.} \\ \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ =\frac{1-2+1}{\sqrt{4+4+1}\sqrt{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+1+1}} \\ =0 \\ \Rightarrow \theta =\frac{\pi }{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Given}: \\ l+2m+3n=0...\left(1\right) \\ 3lm-4\ln +mn=0...\left(2\right) \\ \mathrm{From}\left(1\right),\mathrm{we}\mathrm{get} \\ l=-2m-3n \\ \mathrm{Substituting}l=-2m-3n\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 3\left(-2m-3n\right)m-4\left(-2m-3n\right)n+mn=0 \\ \Rightarrow -6m^2-9mn+8mn+12n^2+mn=0 \\ \Rightarrow 12n^2-6m^2=0 \\ \Rightarrow m^2=2n^2 \\ \Rightarrow m=\sqrt{2}n,-\sqrt{2}n \\ \mathrm{If}m=\sqrt{2}n,\mathrm{then}\mathrm{by}\mathrm{substituting}m=\sqrt{2}n\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}l=n\left(-2\sqrt{2}-3\right). \\ \mathrm{If}m=-\sqrt{2}n,\mathrm{then}\mathrm{by}\mathrm{substituting}m=-\sqrt{2}n\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}l=n\left(2\sqrt{2}-3\right). \\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{are}\mathrm{proportional}\mathrm{to}n\left(-2\sqrt{2}-3\right),\sqrt{2}n,n\mathrm{and}n\left(2\sqrt{2}-3\right),-\sqrt{2}n,n\mathrm{or}\left(-2\sqrt{2}-3\right),\sqrt{2},1\mathrm{and}\left(-2\sqrt{2}-3\right),-\sqrt{2},1. \\ \mathrm{Vectors}\mathrm{parallel}\mathrm{to}\mathrm{these}\mathrm{lines}\mathrm{are} \\ \overrightarrow{a}=\left(-2\sqrt{2}-3\right)\hat{i}+\sqrt{2}\hat{j}+\hat{k} \\ \overrightarrow{b}=\left(2\sqrt{2}-3\right)\hat{i}-\sqrt{2}\hat{j}+\hat{k} \\ \mathrm{If}\theta \mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines},\mathrm{then}\theta \mathrm{is}\mathrm{also}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b}. \\ \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ =\frac{\left[\left(-2\sqrt{2}-3\right)\hat{i}+\sqrt{2}\hat{j}+\hat{k}\right].\left[\left(2\sqrt{2}-3\right)\hat{i}-\sqrt{2}\hat{j}+\hat{k}\right]}{\sqrt{8+9+12\sqrt{2}+2+1}\sqrt{8+9-12\sqrt{2}+2+1}} \\ =\frac{-\left(8-9\right)-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}} \\ =\frac{0}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}} \\ =0 \\ \Rightarrow \theta =\frac{\pi }{2} \end{gathered}$$

(iv) The given relations are

2l + 2m − n = 0 .....(1)

mn + ln + lm = 0 .....(2)

From (1), we have

n = 2l + 2m

Putting this value of n in (2), we get

$$\begin{gathered} m\left(2l+2m\right)+l\left(2l+2m\right)+lm=0 \\ \Rightarrow 2lm+2m^2+2l^2+2lm+lm=0 \\ \Rightarrow 2m^2+5lm+2l^2=0 \\ \Rightarrow \left(2m+l\right)\left(m+2l\right)=0 \\ \Rightarrow 2m+l=0\mathrm{or}m+2l=0 \\ \Rightarrow l=-2m\mathrm{or}l=-\frac{m}{2} \end{gathered}$$

When $l=-2m$, we have

$n=2\times \left(-2m\right)+2m=-4m+2m=-2m$

When $l=-\frac{m}{2}$, we have

$n=2\times \left(-\frac{m}{2}\right)+2m=-m+2m=m$

Thus, the direction ratios of two lines are proportional to

$-2m,m,-2m$ and $-\frac{m}{2},m,m$

Or $-2,1,-2$ and $-1,2,2$

So, vectors parallel to these lines are $\overrightarrow{a}=-2\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{b}=-\hat{i}+2\hat{j}+2\hat{k}$.

Let $\theta$ be the angle between these lines, then $\theta$ is also the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

$$\begin{gathered} \therefore \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ =\frac{\left(-2\hat{i}+\hat{j}-2\hat{k}\right).\left(-\hat{i}+2\hat{j}+2\hat{k}\right)}{\sqrt{4+1+4}\sqrt{1+4+4}} \\ =\frac{-2\times \left(-1\right)+1\times 2+\left(-2\right)\times 2}{3\times 3} \\ =\frac{2+2-4}{9} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Thus, the angle between the two lines whose direction cosines are given by the given relations is $\frac{\mathrm{\pi }}{2}$.