JEE Main 2026 — Mathematics Vectors & 3D Geometry

The given lines are $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$r=a1​+λb1​ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$r=a2​+μb2​, where:

$$\vec{a}_1 = \dfrac{1}{3}\hat{i} + 2\hat{j} + \dfrac{8}{3}\hat{k}$$a1​=31​i^+2j^​+38​k^

$$\vec{b}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k}$$b1​=2i^−5j^​+6k^

$$\vec{a}_2 = -\dfrac{2}{3}\hat{i} - \dfrac{1}{3}\hat{k}$$a2​=−32​i^−31​k^

$$\vec{b}_2 = \hat{j} - \hat{k}$$b2​=j^​−k^

The shortest distance 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​)∣​.

First, finding the vector $\vec{a}_2 - \vec{a}_1$a2​−a1​:

$$\vec{a}_2 - \vec{a}_1 = \left(-\dfrac{2}{3} - \dfrac{1}{3}\right)\hat{i} + (0 - 2)\hat{j} + \left(-\dfrac{1}{3} - \dfrac{8}{3}\right)\hat{k} = -\hat{i} - 2\hat{j} - 3\hat{k}$$a2​−a1​=(−32​−31​)i^+(0−2)j^​+(−31​−38​)k^=−i^−2j^​−3k^

Next, finding the cross product $\vec{b}_1 \times \vec{b}_2$b1​×b2​:

$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(5 - 6) - \hat{j}(-2 - 0) + \hat{k}(2 - 0) = -\hat{i} + 2\hat{j} + 2\hat{k}$$b1​×b2​=​i^20​j^​−51​k^6−1​​=i^(5−6)−j^​(−2−0)+k^(2−0)=−i^+2j^​+2k^

The magnitude of the cross product is:

$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$$∣b1​×b2​∣=(−1)2+22+22​=1+4+4​=3

Now, finding the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$(a2​−a1​)⋅(b1​×b2​):

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k})$$(a2​−a1​)⋅(b1​×b2​)=(−i^−2j^​−3k^)⋅(−i^+2j^​+2k^)

$$= (-1)(-1) + (-2)(2) + (-3)(2) = 1 - 4 - 6 = -9$$=(−1)(−1)+(−2)(2)+(−3)(2)=1−4−6=−9

Finally, the shortest distance is:

$$d = \dfrac{|-9|}{3} = \dfrac{9}{3} = 3$$d=3∣−9∣​=39​=3

Answer: $3$3