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

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Three Dimensional Geometry

Let a line L be perpendicular to both the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

and

L_{2} : \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}

.
If

θ

is the acute angle between the lines L and

L_{3} : \frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}

, then

tan θ

is equal to:

Answer: (B)

A line perpendicular to both

L_{1}

and

L_{2}

has direction

\hat{i} 31 \hat{j} 54 \hat{k} 77 = 7 \hat{i} - 14 \hat{j} + 7 \hat{k}

, i.e.

(1, - 2, 1)

. With

L_{3}

having direction

(2, 1, 2)

cos θ = \frac{∣2 - 2 + 2∣}{6 9} = \frac{2}{3 6} = \frac{6}{9}

. Then

sin θ = 1 - \frac{6}{81} = \frac{75}{9} = \frac{5 3}{9}

, so

tan θ = \frac{5 3 /9}{6 /9} = \frac{5 3}{6} = \frac{5}{2} = \frac{5 2}{2}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option