Q34 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$. Then the cosine of the acute angle between these lines is :
- A$\frac{20}{3 \sqrt{38}}$
- Correct answerB$\frac{10}{3 \sqrt{38}}$
- C$\frac{10}{7 \sqrt{38}}$
- D$\frac{10}{\sqrt{38}}$
Solution
From $n = 4l + m$, substitute in $2mn + 10nl + 3lm = 0$:
$40l^2 + 21lm + 2m^2 = 0$. Let $t = l/m$: $40t^2 + 21t + 2 = 0$.
$t = \frac{-21 \pm 11}{80}$, giving $t = -1/8$ or $t = -2/5$.
$l/m = -1/8$: direction $(l,m,n) = (-1, 8, 4)$.
$l/m = -2/5$: direction $(l,m,n) = (-2, 5, -3)$.
$\cos\theta = \frac{|2+40-12|}{\sqrt{81}\sqrt{38}} = \frac{30}{9\sqrt{38}} = \frac{10}{3\sqrt{38}}$.
$40l^2 + 21lm + 2m^2 = 0$. Let $t = l/m$: $40t^2 + 21t + 2 = 0$.
$t = \frac{-21 \pm 11}{80}$, giving $t = -1/8$ or $t = -2/5$.
$l/m = -1/8$: direction $(l,m,n) = (-1, 8, 4)$.
$l/m = -2/5$: direction $(l,m,n) = (-2, 5, -3)$.
$\cos\theta = \frac{|2+40-12|}{\sqrt{81}\sqrt{38}} = \frac{30}{9\sqrt{38}} = \frac{10}{3\sqrt{38}}$.
Hence, the correct answer is option: 'B'.