JEE Main 2026 — 23 January, Morning Shift — Question 20

Direction cosines of two lines satisfy the equation $\Rightarrow 4 \ell+\mathrm{m}-\mathrm{n}=0$

$$\begin{gathered} 2 \mathrm{mn}+10 \mathrm{n} \ell+3 \ell \mathrm{~m}=0 \end{gathered}$$

and we know $\begin{gathered} \Rightarrow \ell^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=1 \end{gathered}$

$\Rightarrow \mathrm{n}=4 \ell+\mathrm{m}$ putting in eqn.

$\Rightarrow \mathrm{n}(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$

$\Rightarrow(4 \ell+\mathrm{m})(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$

$\Rightarrow 8 \ell \mathrm{~m}+40 \ell^{2}+2 \mathrm{~m}^{2}+10 \ell \mathrm{~m}+3 \ell \mathrm{~m}=0$

$\Rightarrow 40 \ell^{2}+21 \ell \mathrm{~m}+2 \mathrm{~m}^{2}=0$

$\Rightarrow(8 \ell+\mathrm{m})(5 \ell+2 \mathrm{~m})=0$

Case 1: $8 \ell+m=0 \Rightarrow m=-8 \ell$

Case 2 : $5 \ell+2 \mathrm{~m}=0 \Rightarrow \mathrm{~m}=\frac{-5}{2} \ell$

So direction ratio of $\mathrm{L}_{1}$ is $\ell,-8 \ell,-4 \ell$ & direction ratio of $\mathrm{L}_{2}$ is $\ell, \frac{-5 \ell}{2}, \frac{3 \ell}{2}$

$\cos \theta=\left|\frac{\ell^{2}+20 \ell^{2}-6 \ell^{2}}{\sqrt{\ell^{2}+64 \ell^{2}+16 \ell^{2}} \sqrt{\ell^{2}+\frac{25 \ell^{2}}{4}+\frac{9 \ell^{2}}{4}}}\right|$

$=\frac{15 \ell^{2}}{(9 \ell) \frac{\sqrt{38} \ell}{2}}=\frac{10}{3 \sqrt{38}}$

Ans. $=\frac{10}{3 \sqrt{38}}$