JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026mcqmedium

Let a line $L$ be perpendicular to both the lines $L_1: \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7}$ and $L_2: \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7}$ . If $\theta$ is the acute angle between the lines $L$ and $L_3: \dfrac{x - \dfrac{8}{7}}{2} = \dfrac{y - \dfrac{4}{7}}{1} = \dfrac{z}{2}$ , then $\tan\theta$ is equal to:

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{3}{2}\sqrt{2}$
B
$\dfrac{5}{2}\sqrt{2}$
C
$\dfrac{5}{3}\sqrt{2}$
D
$\dfrac{4}{3}\sqrt{2}$

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Solution

The direction vector of line $L_1$ is $\vec{d_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$ .

The direction vector of line $L_2$ is $\vec{d_2} = \hat{i} + 4\hat{j} + 7\hat{k}$ .

Since line $L$ is perpendicular to both $L_1$ and $L_2$ , its direction vector $\vec{d}$ is given by the cross product of $\vec{d_1}$ and $\vec{d_2}$ :

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$

$\vec{d} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$

The direction ratios of $L$ are proportional to $1, -2, 1$ .

The direction vector of line $L_3$ is $\vec{d_3} = 2\hat{i} + \hat{j} + 2\hat{k}$ .

The acute angle $\theta$ between $L$ and $L_3$ is given by:

$\cos\theta = \dfrac{|\vec{d} \cdot \vec{d_3}|}{|\vec{d}| |\vec{d_3}|}$

$\cos\theta = \dfrac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}}$

$\cos\theta = \dfrac{|2 - 2 + 2|}{\sqrt{6} \sqrt{9}} = \dfrac{2}{3\sqrt{6}}$

From $\cos\theta = \dfrac{2}{3\sqrt{6}}$ , considering a right-angled triangle with adjacent side $2$ and hypotenuse $3\sqrt{6}$ :

The opposite side is $\sqrt{(3\sqrt{6})^2 - 2^2} = \sqrt{54 - 4} = \sqrt{50} = 5\sqrt{2}$ .

Therefore, $\tan\theta = \dfrac{5\sqrt{2}}{2}$ .

Answer: $\dfrac{5}{2}\sqrt{2}$

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Solution

The direction vector of line $L_1$ is $\vec{d_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$ .

The direction vector of line $L_2$ is $\vec{d_2} = \hat{i} + 4\hat{j} + 7\hat{k}$ .

Since line $L$ is perpendicular to both $L_1$ and $L_2$ , its direction vector $\vec{d}$ is given by the cross product of $\vec{d_1}$ and $\vec{d_2}$ :

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$

$\vec{d} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$

The direction ratios of $L$ are proportional to $1, -2, 1$ .

The direction vector of line $L_3$ is $\vec{d_3} = 2\hat{i} + \hat{j} + 2\hat{k}$ .

The acute angle $\theta$ between $L$ and $L_3$ is given by:

$\cos\theta = \dfrac{|\vec{d} \cdot \vec{d_3}|}{|\vec{d}| |\vec{d_3}|}$

$\cos\theta = \dfrac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}}$

$\cos\theta = \dfrac{|2 - 2 + 2|}{\sqrt{6} \sqrt{9}} = \dfrac{2}{3\sqrt{6}}$

From $\cos\theta = \dfrac{2}{3\sqrt{6}}$ , considering a right-angled triangle with adjacent side $2$ and hypotenuse $3\sqrt{6}$ :

The opposite side is $\sqrt{(3\sqrt{6})^2 - 2^2} = \sqrt{54 - 4} = \sqrt{50} = 5\sqrt{2}$ .

Therefore, $\tan\theta = \dfrac{5\sqrt{2}}{2}$ .

Answer: $\dfrac{5}{2}\sqrt{2}$

JEE Main 2026 — Mathematics Vectors & 3D Geometry

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JEE Main 2026 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry