Q99 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
The direction cosines of two lines are connected by the relation $l-m+n=0$ and $2 l m-3 m n+n l=0$. If $\theta$ is the angle between these two lines, then $\cos \theta=$
- A$\frac{1}{4}$
- Correct answerB$\frac{1}{\sqrt{19}}$
- C$\frac{1}{3}$
- D$\frac{1}{3 \sqrt{2}}$
Solution
Given : $l-m+n=0$
$\Rightarrow m=l+n$ $\qquad ....\mathrm{(i)}$
Substituting in $2 l m-3 mn+n l=0$
$\begin{aligned}
& \Rightarrow 2 l^2-3 n^2=0 \qquad ....\mathrm{(ii)} \\
& \Rightarrow l=\sqrt{\frac{3}{2}} n, l=-\sqrt{\frac{3}{2}} n
\end{aligned}$
From (i), $m=\left(1+\sqrt{\frac{3}{2}}\right) \mathrm{n}, \mathrm{m}=\left(1-\frac{\sqrt{3}}{2}\right) \mathrm{n}$
$\therefore$ d.r.'s of line (i) are $\left(\sqrt{\frac{3}{2}},\left(1+\frac{\sqrt{3}}{2}\right), 1\right)$
d.r.'s of line (ii) are $\left(-\sqrt{\frac{3}{2}}, 1-\sqrt{\frac{3}{2}}, 1\right)$
$\cos \theta=\left|\frac{-\frac{3}{2}+1+1-\frac{3}{2}}{\left.\sqrt{\frac{3}{2}+1+2 \sqrt{\frac{3}{2}}+\frac{3}{2}+1} \times \sqrt{\frac{3}{2}+1+1+\frac{3}{2}-2 \sqrt{\frac{3}{2}}} \right\rvert\,}\right|$
$=\frac{1}{\sqrt{19}}$.
Hence, the correct answer is option: 'B'.