JEE Main 2024 — 6 April, Shift 1 — Question 9

$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \& \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$

S.D $=\frac{\left|\left(\overline{\mathrm{a}}_{2} \cdot \overline{\mathrm{a}}_{1}\right) \cdot\left(\overline{\mathrm{b}}_{1} \cdot \overline{\mathrm{~b}}_{2}\right)\right|}{\left|\overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}\right|}$

$\mathrm{a}_{1}=3,-15,9$

$\mathrm{b}_{1}=2,-7,5$

$\mathrm{a}_{2}=-1,1,9$

$\mathrm{b}_{2}=2,1,-3$

$\mathrm{a}_{2}-\mathrm{a}_{1}=-4,16,0$

${{\overset{}{\mathop{b}}\,}_{1}}\times {{\overset{}{\mathop{b}}\,}_{2}}=\left| \begin{matrix}\overset{}{\mathop{i}}\, & \overset{}{\mathop{j}}\, & \overset{}{\mathop{k}}\, \\2 & -7 & 5 \\2 & 1 & -3 \\\end{matrix} \right|$=$\widehat i\left(16\right)-\widehat j\left(-16\right)+\widehat k\left(16\right)$

=$16(\hat{i}+\hat{j}+\hat{k})$ $\left|\overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}\right|=16 \sqrt{3}$

$\therefore\left(\overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}\right) \cdot\left(\overline{\mathrm{b}}_{1}-\overline{\mathrm{b}}_{2}\right)=16[-4+16]=(16)(12)$

S.D. $=\frac{(16)(12)}{16 \sqrt{3}}=4 \sqrt{3}$