If the direction ratio of two lines are given by $3 l m-4 l n+m n=0$ and $l+2 m+3 n=0$, then the angle…
If the direction ratio of two lines are given by $3 l m-4 l n+m n=0$ and $l+2 m+3 n=0$, then the angle between the lines, is :
- $\frac{\pi}{6}$
- $\frac{\pi}{4}$
- $\frac{\pi}{6}$
- $\frac{\pi}{2}$
Solution
We have,
$3 l m-4 l n+m n=0$ $\ldots$ (i)
and $l+2 m+3 n=0$ $\ldots$ (ii)
From Eq. (ii), $l=-(2 m+3 n)$
Using in Eq. (i), we get
$-3(2 m+3 n) m+4(2 m+3 n) n+m n=0$
$\Rightarrow \quad-6 m^2-9 m n+8 m n+12 n^2+m n=0$
$\Rightarrow \quad-6 m^2+12 n^2=0$
Now, $m^2=2 n^2 \Rightarrow m= \pm \sqrt{2} n$
Now, $\quad m=\sqrt{2} n$
$\Rightarrow \quad l=-(2 \sqrt{2} n+3 n)=-(2 \sqrt{2}+3) n$
$\therefore l: m: n=-(3+2 \sqrt{2}) n: \sqrt{2} n: n$
$=-(3+2 \sqrt{2}): \sqrt{2}: 1$
Also, $\quad m=-\sqrt{2} n \Rightarrow l=-(-2 \sqrt{2}+3) n$
$l: m: n=-(3-2 \sqrt{2}) n:-\sqrt{2} n: n$
$=-(3-2 \sqrt{2}):-\sqrt{2}: 1$
If $\theta$ is the angle between the lines, then
$\cos \theta=\frac{l_1 l_2+m_1 m_2+n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}}$
$=\frac{(3+2 \sqrt{2})(3-2 \sqrt{2})+(\sqrt{2})(-\sqrt{2})+1.1}{\sqrt{(3+2 \sqrt{2})^2+(\sqrt{2})^2+1^2}{\sqrt{(3-2 \sqrt{2})^2+(-\sqrt{2})^2+1^2}}}$
$=\frac{9-8-2+1}{\sqrt{9+8+12 \sqrt{2}+2+1} \sqrt{9+8-12 \sqrt{2}+2+1}}=0$
$\Rightarrow \cos \theta=0 \Rightarrow \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}$
Asked in: AP EAMCET 2003
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