JEE Main 2026 — 4 April, Evening Shift — Question 37

$\vec{a}_{1}=\frac{1}{3} \hat{i}+2 \hat{j}+\frac{8}{3} \hat{k}$ $\overrightarrow{\mathrm{a}}_{2}=\frac{-2}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{b}}_{1}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}_{2}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ $\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} 2 & -5 & 6 0 & 1 & -1\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ $\vec{a}_{1}-\vec{a}_{2}=\hat{i}+2 \hat{j}+3 \hat{k}$ S.D $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$ $S . D=\left|\frac{(\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})}{3}\right|$ $=\frac{9}{3}=3$