An angle between the lines whose direction cosines are given by the equations, l+3 m+5 n= 0 and 5 l m-2 m n+6 n l=0, is

Given $$\begin{aligned} &l+3 m+5 n=0 \\ &\text { and } 5 l m-2 m n+6 n l=0 \end{aligned}$$​l+3m+5n=0 and 5lm−2mn+6nl=0​ From eq. (1) we have $$l=-3 m-5 n$$l=−3m−5n Put the value of $l$l in eq. (2), we get; $$\begin{aligned} &5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0 \\ &\Rightarrow 15 m^2+45 m n+30 n^2=0 \\ &\Rightarrow m^2+3 m n+2 n^2=0 \\ &\Rightarrow m^2+2 m n+m n+2 n^2=0 \\ &\Rightarrow(m+n)(m+2 n)=0 \\ &\therefore m=-n \text { or } m=-2 n \\ &\text { For } m=-n, l=-2 n \\ &\text { And for } m=-2 n, l=n \end{aligned}$$​5(−3m−5n)m−2mn+6n(−3m−5n)=0⇒15m2+45mn+30n2=0⇒m2+3mn+2n2=0⇒m2+2mn+mn+2n2=0⇒(m+n)(m+2n)=0∴m=−n or m=−2n For m=−n,l=−2n And for m=−2n,l=n​ $$\begin{aligned} &\therefore(l, m, n)=(-2 n,-n, n) \text { Or }(l, m, n) \\ &=(n,-2 n, n) \\ &\Rightarrow(l, m, n)=(-2,-1,1) \text { Or }(l, m, n) \\ &=(1,-2,1) \end{aligned}$$​∴(l,m,n)=(−2n,−n,n) Or (l,m,n)=(n,−2n,n)⇒(l,m,n)=(−2,−1,1) Or (l,m,n)=(1,−2,1)​ Therefore, angle between the lines is given as: $$\begin{aligned} &\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}} \\ &\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right) \end{aligned}$$​cos(θ)=6​⋅6​(−2)(1)+(−1)⋅(−2)+(1)(1)​⇒cos(θ)=61​⇒θ=cos−1(61​)​