The shortest distance between the lines $$\overrightarrow{\mathrm{r}}=\left(\frac{1}{3} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\frac{8}{3} \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$$ and $\vec{r}=\left(-\frac{2}{3} \hat{i}-\frac{1}{3} \hat{k}\right)+\mu(\hat{j}-\hat{k}), \lambda, \mu \in \mathbb{R}$, is :

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Three Dimensional Geometry

The shortest distance between the lines

r = (\frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 6 \hat{k})

and

r = (- \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}) + μ (\hat{j} - \hat{k}), λ, μ \in R

, is :

Answer: (B)

The lines pass through

a_{1} = \frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}

with direction

b_{1} = 2 \hat{i} - 5 \hat{j} + 6 \hat{k}

, and through

a_{2} = - \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}

with direction

b_{2} = \hat{j} - \hat{k}

. Their common normal is

b_{1} \times b_{2} = \hat{i} 20 \hat{j} - 5 1 \hat{k} 6 - 1 = - \hat{i} + 2 \hat{j} + 2 \hat{k}

, with

∣ b_{1} \times b_{2} ∣ = 1 + 4 + 4 = 3

. Also

a_{1} - a_{2} = \hat{i} + 2 \hat{j} + 3 \hat{k}

, so

(a_{1} - a_{2}) \cdot (b_{1} \times b_{2}) = - 1 + 4 + 6 = 9

. The shortest distance is

\frac{∣9∣}{3} = 3

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option