Q100 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
If the direction cosines of the two lines satisfy the equations $l+m+n=0,2 l m+2 l n-m n=0$, then the acute angle between these lines is
- A$\cos ^{-1}\left(\frac{1}{3}\right)$
- B$30^{\circ}$
- C$\cos ^{-1}\left(\frac{2}{3}\right)$
- Correct answerD$60^{\circ}$
Solution
The direction cosines of two lines satisfy the equations
$$
\begin{aligned}
& l+m+n=0 \Rightarrow m=-(l+n) \\
& \text { and } \quad 2 l m+2 l n-m n=0 \\
& \Rightarrow \quad 2 l n+m(2 l-n)=0 \\
& \Rightarrow \quad 2 \ln -(l+n)(2 l-n)=0 \\
& \Rightarrow 2 l n-2 l^2+n l-2 \ln +n^2=0 \\
& \Rightarrow \quad-2 l^2+n l+n^2=0 \\
& \Rightarrow \quad 2 l^2-n l-n^2=0 \\
& \Rightarrow \quad 2 l^2-2 n l+n l-n^2=0 \\
& \Rightarrow \quad 2 l(l-n)+n(l-n)=0 \\
& \Rightarrow \quad(l-n)(2 l+n)=0 \\
& \Rightarrow \quad l=n \text { and } l=-\frac{n}{2} \\
&
\end{aligned}
$$
If $l=n$ and $l=-\frac{n}{2}$
Substituting in Eq. (i),
$$
\begin{gathered}
m=-(n+n) \text { and } m=-\left(-\frac{n}{2}+n\right) \\
\frac{m}{-2}=n \text { and } m=-\frac{n}{2}
\end{gathered}
$$
From Eqs. (iii) and (iv),
$$
\frac{l}{1}=\frac{m}{-2}=\frac{n}{1} \text { and } \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}
$$
So, $l: m: n=1:-2: 1$ and $l: m: n=1: 1:-2$
So, acute angle between lines is
$$
\begin{aligned}
& \cos \theta=\left|\frac{1 \times 1+(-2) \times 1+1 \times(-2)}{\sqrt{1^2+2^2+1^2} \sqrt{1^2+1^2+2^2}}\right|=\left|\frac{1-2-2}{\sqrt{6} \sqrt{6}}\right|=\left|\frac{-3}{6}\right| \\
& \cos \theta=\frac{1}{2} \\
& \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta=\frac{\pi}{3} \Rightarrow \theta=60^{\circ}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.