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Let the direction cosines of two lines satisfy the equations $4 \ell+m-n=0$ and $2 \mathrm{mn}+10 \mathrm{n} \ell+3 \ell \mathrm{~m}=0$. Then the cosine of the acute angle between these lines is
[JEE MAIN 23 Jan 2026 S I]
(1) $\frac{10}{\sqrt{38}}$
(2) $\frac{20}{3 \sqrt{38}}$
(3) $\frac{10}{7 \sqrt{38}}$
(4) $\frac{10}{3 \sqrt{38}}$Solution
$4 \ell+\mathrm{m}-\mathrm{n}=0 \Rightarrow \mathrm{n}=4 \ell+\mathrm{m}$
$\mathrm{n}(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$
$\Rightarrow(4 \ell+\mathrm{m})(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$
$\Rightarrow 8 \ell \mathrm{~m}+40 \ell^{2}+2 \mathrm{~m}^{2}+10 \ell \mathrm{~m}+3 \ell \mathrm{~m}=0$
$\Rightarrow 40 \ell^{2}+16 \ell \mathrm{~m}+5 \ell \mathrm{~m}+2 \mathrm{~m}^{2}=0$
$\Rightarrow 8 \ell(5 \ell+2 \mathrm{~m})+\mathrm{m}(5 \ell+2 \mathrm{~m})=0$
$\Rightarrow(5 \ell+2 \mathrm{~m})(8 \ell+\mathrm{m})=0$
$5 \mathrm{l}+2 \mathrm{~m}=0 \quad \& \quad 8 \ell+\mathrm{m}=0$
$\Rightarrow \mathrm{m}=-\frac{5 \ell}{2}$
$\Rightarrow \mathrm{m}=-8 \ell$
$\mathrm{n}=4 \ell+\mathrm{m}$
$\mathrm{n}=4 \ell+\mathrm{m}=4 \ell-8 \ell$
$\mathrm{n}=4 \ell-\frac{5 \ell}{2}=\frac{3 \ell}{2}$
$\Rightarrow \mathrm{n}=-4 \ell$
$\ell, \mathrm{m}, \mathrm{m} \rightarrow \ell,-\frac{5 \ell}{2}, \frac{3 \ell}{2}$
$\ell, \mathrm{m}, \mathrm{n} \rightarrow \ell,-8 \ell,-4 \ell$
or line $\mathrm{L}_{1} \mathrm{drs} 2,-5,3 \mid$ line $\mathrm{L}_{2} \mathrm{drs}-1,8,4$\[ \cos \theta=\left|\frac{2(-1)-5(8)+3(4)}{\sqrt{4+25+9} \sqrt{1+64+16}}\right|=\frac{30}{\sqrt{38} \times 9}=\frac{10}{3 \sqrt{38}} \text { Answer }(4) \]
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