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 -8/7 2 =frac y -4/7 1 =frac z 2 , then tan θ is equal to:

Mathematics · 3D Geometry

JEE Main 2026 — 6 April, Morning Shift — Question 36

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:

Option A:
$\frac{3}{2} \sqrt{2}$
Option B:
$\frac{5}{2} \sqrt{2}$
Correct
Option C:
$\frac{5}{3} \sqrt{2}$
Option D:
$\frac{4}{3} \sqrt{2}$

Answer: B

Step-by-step solution

$\overrightarrow{\mathrm{r}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ 3 & 5 & 7\\ 1 & 4 & 7\end{array}\right|=7 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$ $\equiv 7(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ Now,

$\because \cos \theta=\frac{2-2+2}{\sqrt{1+4+1} \sqrt{4+1+4}}$ $\Rightarrow \cos \theta=\frac{2}{3 \sqrt{6}} \Rightarrow \tan \theta=\frac{5 \sqrt{2}}{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Angle Between lines using D.C'S ,D.R.'S

Let the image of the point P(1,6, a) in the line L: frac x1=frac y-12=frac z- a+1 b, b 0 , be (frac a3, 0, a+ c ) . If S(alpha, beta…

Answer: B

Step-by-step solution

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