Question

Find the angle between the lines whose direction cosines are given by the equation $3l+m+5n=0$ and $6mn-2nl+5lm=0$.

Answer 1

Given equation $3l+m+5n=0$ & $6mn-2nl+5lm=0$-----(2)

$m=-(3l+5n)----\left(1\right)$

Sub $\left(1\right)$ in $\left(2\right)$ $6[-(3l+5n\left)\right]n-2ln+5l[-(3l+5n\left)\right]=0$

$(-18l-30n)n-2nl-15l^2-25ln=0$

$-18ln-30n^2-2ln-15l^2-25ln=0$

$-15l^2-45ln-30n^2=0\Rightarrow l^2+3ln+2n^2=0$

$l(l+2n)+n(l+2n)=0\Rightarrow (l+n)(l+2n)=0$

$\therefore l=-n$ & $l=-2n$

from $\left(1\right)$ $m=-(3l+5n)$ $m=-(3l+5n)$

$l=-n\Rightarrow m=-(-3n+5n)$ $l=-2n\Rightarrow m=-\left(3\right(-2n)+5n)$

$m=-2n$ $m=n$

$l=-n$ & $m=-2n$ $l=-2n$ & $m=n$

$\frac{l}{-1}=\frac{n}{1}$ $\frac{m}{-2}=\frac{n}{1}$ $\frac{l}{-2}=\frac{n}{1}$ $\frac{m}{1}=\frac{n}{1}$

$(l_1,m_1,n_1)=(-1,-2,1)$ $(l_2,m_2,n_2)=(-2,1,1)$

Angle between direction cosines are

$\cos \theta =\frac{l_1{l}_2+m_1{m}_2+n_1{n}_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}$

$=\frac{(-1)(-2)+(-2)\left(1\right)+1\left(1\right)}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}\sqrt{(-2{)}^2+1^2+1^2}}$

$=\frac{+2-1+1}{\sqrt{1+4+1}4+1+1}=\frac{1}{\sqrt{6}.\sqrt{6}}=\frac{1}{6}$

$\cos \theta =\frac{1}{6}$

$\theta ={\cos }^{-1}\left(\frac{1}{6}\right)$.