Q71 / 200 · TS EAMCET 2002 · Mathematics
The acute angle between the two lines whose direction ratios are given by $l+m-n=0$ and $l^2+m^2-n^2=0$, is
- A$0$
- B$\frac{\pi}{6}$
- C$\frac{\pi}{4}$
- Correct answerD$\frac{\pi}{3}$
Solution
Given that
$$
l+m-n=0
$$
and
$$
l^2+m^2-n^2=0
$$
From equaion (i) $l=-(m-n)$
Putting in equation (ii), we get
$$
\begin{aligned}
\Rightarrow & (m-n)^2+m^2-n^2 & =0 \\
\Rightarrow & m^2+n^2-2 m n+m^2-n^2 & =0 \\
\Rightarrow & 2 m^2-2 m n & =0 \\
& 2 m(m-n) & =0 \\
\Rightarrow & m=0, m & =n
\end{aligned}
$$
From equation (i), if $m=0$, then
$$
l=n \Rightarrow l: m: n=1: 0: 1
$$
If $m=n$, then $l=0 \Rightarrow l: m: n=0: 1: 1$
If $\theta$ is the acute angle, then
$$
\begin{aligned}
\cos \theta & =\left|\frac{1.0+0.1+1.1}{\sqrt{1+1} \sqrt{1+1}}\right|=\frac{1}{2}=\cos \frac{\pi}{3} \\
\Rightarrow \quad \theta & =\frac{\pi}{3}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.