Question

The shortest distance between the skew lines $\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1},\frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}$ is

• A. $6$
• B. $7$
• C. $3\sqrt{5}$
• D. $\sqrt{35}$

Answer 1

The correct option is D $\sqrt{35}$

We have,

$\begin{array}{c}\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}\overrightarrow{r_1}=3\hat{i}+4\hat{j}-2\hat{k}+\lambda \left(-\hat{i}+2\hat{j}+\hat{k}\right)\\ \frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}\overrightarrow{r_2}=\hat{i}-7\hat{j}-2\hat{k}+\lambda \left(\hat{i}+3\hat{j}+2\hat{k}\right)\\ d=\frac{\left|\left(2\hat{i}+11\hat{j}\right).\left(-\hat{i}+2\hat{j}+\hat{k}\right)\times \left(\hat{i}+3\hat{j}+2\hat{k}\right)\right|}{\left|\left(-\hat{i}+2\hat{j}+\hat{k}\right)\times \left(\hat{i}+3\hat{j}+2\hat{k}\right)\right|}\\ =\frac{\left|\left(2\hat{i}+11\hat{j}\right).\left(\hat{i}+3\hat{j}-5\hat{k}\right)\right|}{\left|\left(\hat{i}+3\hat{j}-5\hat{k}\right)\right|}\\ =\frac{\left|35\right|}{\sqrt{35}}\\ =\sqrt{35}\end{array}$

Then,

Option $D$ is correct answer.