Q91 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
The distance between parallel lines $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$ and $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ is
- A$\sqrt{2}$
- B$\frac{1}{3}$
- C$\frac{1}{\sqrt{3}}$ units
- Correct answerD$\frac{\sqrt{2}}{3}$ units
Solution
The distance between given parallel lines
$$
\begin{aligned}
& =\left|\frac{[(\hat{\mathrm{i}}-2 \hat{\mathrm{i}})+(-\hat{\mathrm{j}}+\hat{\mathrm{j}})+(2 \hat{\mathrm{k}}-\hat{\mathrm{k}})] \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{|2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}|}\right|=\left|\frac{(-\hat{\mathrm{i}}+\hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{(2)^2+(1)^2+(-2)^2}}\right| \\
& (-\hat{\mathrm{i}}+\hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})=\left|\begin{array}{ccc}
\mathrm{i} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-1 & 0 & 1 \\
2 & 1 & -2
\end{array}\right|=\hat{\mathrm{i}}(0-1)-\hat{\mathrm{j}}(2-2)+\hat{\mathrm{k}}(-1-0)=-\hat{\mathrm{i}}-\hat{\mathrm{k}} \\
& \therefore \mathrm{d}=\left|\frac{\mid-\hat{\mathrm{i}}-\hat{\mathrm{k}}}{\sqrt{9}}\right|=\frac{\sqrt{2}}{3} \text { units }
\end{aligned}
$$
Hence, the correct answer is option: 'D'.