JEE Main 2026 (6 Apr, Shift 1) Maths Q16: Let a line L be…

Question

Let a line L be perpendicular to both the lines $\mathrm{L}_1:\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\mathrm{L}_2:\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$.
If $	heta$ is the acute angle between the lines L and $\mathrm{L}_3:\frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$, then $	an	heta$ is equal to:

Accepted Answer

Option 2: $\frac{5}{2}\sqrt{2}$

Answer

$\frac{3}{2}\sqrt{2}$

Answer

$\frac{5}{2}\sqrt{2}$

Answer

$\frac{5}{3}\sqrt{2}$

Answer

$\frac{4}{3}\sqrt{2}$

JEE Main 6 April 2026, Shift 1, Maths Q16

Same topic in other shifts