The distance between the parallel lines $\frac{x-2}{2}=\frac{y-4}{5}=\frac{z-1}{2}$ and…
The distance between the parallel lines $\frac{x-2}{2}=\frac{y-4}{5}=\frac{z-1}{2}$ and $\frac{x-1}{3}=\frac{y+1}{5}=\frac{z+3}{2}$ is
- $\frac{1}{\sqrt{38}}$ units
- $\sqrt{\frac{333}{38}}$ units
- $\sqrt{\frac{300}{37}}$ units
- $\sqrt{\frac{300}{35}}$ units
Solution
Let $\overline{\mathrm{a}}_1=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{a}}_2=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$\therefore \overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1=-\hat{\mathrm{i}}-6 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ and let $\overline{\mathrm{b}}=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\begin{aligned}
& \overline{\mathrm{b}}\left|\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right|=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & 5 & 2 \\
-1 & -6 & -4
\end{array}\right|=\hat{\mathrm{i}}(-20+12)-\hat{\mathrm{j}}(-12+2)+\hat{\mathrm{k}}(-18+5) \\
& =-8 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\
& \therefore\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|=\sqrt{64+100+169}=\sqrt{133}
\end{aligned}$
Also $|\overline{\mathrm{b}}|=\sqrt{9+25+4}=\sqrt{38}$
$\therefore$ Distance between given lines $=\sqrt{\frac{333}{38}}$ units
Asked in: MHT CET 2021 (23 Sep Shift 2)
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