Jee Main 2024 27 01 2024 Shift 1 - Question16

Answer: (4)

Solution:

$\frac{x-1}{2}=\frac{y+1}{4}=\frac{z-1}{3}$ and

$\frac{x-\frac{1}{2}}{\left(\frac{5}{2}\right)}=\frac{y-2}{3}=\frac{z}{6}$,

$\overrightarrow{a _1}-\overrightarrow{a _2}=\frac{1}{2} \hat{i}-3 \hat{j}+\hat{k}$

$\overrightarrow{n _1} \times \overrightarrow{n _2}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 3 \\ \frac{5}{2} & 3 & 6\end{array}\right|$

$=15 \hat{i}-\frac{9}{2} \hat{j}+(-4 k)$

$\vec{d}=\left|\frac{\left(\overrightarrow{a _1}-\overrightarrow{a _2}\right),\left(\overrightarrow{n _1} \times \overrightarrow{n _2}\right)}{\left|\overrightarrow{n _1} \times \overrightarrow{n _2}\right|}\right|$

$=\left|\frac{\frac{15}{2}+\frac{27}{2}-4}{\sqrt{225+\frac{81}{4}+16}}\right|$

$=\frac{\frac{34}{2}}{\sqrt{\frac{1045}{4}}}=\frac{34}{\sqrt{1045}}$ unit