Q100 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l+m+n$ $=0$ and $2 l m+2 n l-m n=0$, then $\cos \theta=$

  • Correct answer
    A
    $\frac{1}{2}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\frac{5}{6}$
  • D
    $\frac{3}{5}$

Solution

$l+m+n=0$ $\begin{aligned} & 2 l m+2 n l-m n=0 \\ & l=-(m+n) \\ & \Rightarrow-2 m(m+n)-2 n(m+n)-m n=0 \\ & \Rightarrow-2 m^2-2 m n-2 m n-2 n^2-m n=0 \\ & \Rightarrow-2 m^2-2 n^2-5 m n=0 \\ & \Rightarrow 2 m^2+5 m n+2 n^2=0 \\ & \Rightarrow \quad(2 m+n)(m+2 n)=0 \\ & \Rightarrow n=-2 m, m=-2 n \\ & \text { when } n=-2 m \\ & l=-(m+n)=-(m-2 m)=m \\ & \therefore \quad l=m=\frac{-n}{2} \\ & \Rightarrow \quad \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\end{aligned}$ $\therefore \quad$ D.R. $\equiv(1,1,-2)$ when $n=-2 m$ $\begin{aligned} & l=-(m+n)=-(m-2 m)=m \\ & \therefore \quad l=m=\frac{-n}{2} \\ & \Rightarrow \quad \frac{l}{1}=\frac{m}{1}=\frac{n}{-2} \\ & \therefore \quad \text { D.R. } \equiv(1,1,-2)\end{aligned}$ when $m=-2 n$ $\begin{aligned} & l=-(m+n)=-(-2 n+n)=n \\ & \therefore \quad l=n=\frac{m}{-2} \\ & \Rightarrow \quad \frac{l}{1}=\frac{m}{-2}=\frac{n}{1} \\ & \therefore \quad \text { D.R. } \equiv(1,-2,1) \\ & \therefore \quad \cos \theta=\left|\frac{1-2-2}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right|=\frac{3}{6}=\frac{1}{2} .\end{aligned}$

Hence, the correct answer is option: 'A'.

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