JEE Main 2026 — Mathematics Vectors & 3D Geometry

The given lines are $L_1: \dfrac{x-4}{1} = \dfrac{y-3}{2} = \dfrac{z-2}{-3}$L1​:1x−4​=2y−3​=−3z−2​ and $L_2: \dfrac{x+2}{2} = \dfrac{y-6}{4} = \dfrac{z-5}{-5}$L2​:2x+2​=4y−6​=−5z−5​.

For $L_1$L1​, a point on the line is $\vec{a}_1 = 4\hat{i} + 3\hat{j} + 2\hat{k}$a1​=4i^+3j^​+2k^ and its direction vector is $\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}$b1​=i^+2j^​−3k^.

For $L_2$L2​, a point on the line is $\vec{a}_2 = -2\hat{i} + 6\hat{j} + 5\hat{k}$a2​=−2i^+6j^​+5k^ and its direction vector is $\vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k}$b2​=2i^+4j^​−5k^.

The shortest distance $d$d between two skew lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$d=∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​.

We have $\vec{a}_2 - \vec{a}_1 = (-2 - 4)\hat{i} + (6 - 3)\hat{j} + (5 - 2)\hat{k} = -6\hat{i} + 3\hat{j} + 3\hat{k}$a2​−a1​=(−2−4)i^+(6−3)j^​+(5−2)k^=−6i^+3j^​+3k^.

The cross product of the direction vectors is:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j}$b1​×b2​=​i^12​j^​24​k^−3−5​​=i^(−10+12)−j^​(−5+6)+k^(4−4)=2i^−j^​.

The magnitude of the cross product is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$∣b1​×b2​∣=22+(−1)2+02​=5​.

The dot product is $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-6)(2) + (3)(-1) + (3)(0) = -12 - 3 = -15$(a2​−a1​)⋅(b1​×b2​)=(−6)(2)+(3)(−1)+(3)(0)=−12−3=−15.

Therefore, the shortest distance is $d = \dfrac{|-15|}{\sqrt{5}} = \dfrac{15}{\sqrt{5}} = 3\sqrt{5}$d=5​∣−15∣​=5​15​=35​.

Answer: $3\sqrt{5}$35​