Question

Find the shortest distance between the following set of parallel lines.

\vec{r}=6\hat{i}+2\hat{j}-\hat{k}+λ(\hat{i}+2\hat{j}-4\hat{k})

\vec{r}=\hat{i}+\hat{j}+\hat{k}+μ(\hat{i}+2\hat{j}-4\hat{k})

(a) d=\(\sqrt{\frac{324}{45}}\)

(b) d=\(\sqrt{\frac{405}{21}}\)

(c) d=\(\sqrt{\frac{24}{21}}\)

(d) d=\(\sqrt{\frac{21}{567}}\)

The question was posed to me in a national level competition.

This interesting question is from Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right answer is (b) d=\(\sqrt{\frac{405}{21}}\)

Explanation: The shortest distance between two parallel lines is given by:

d=\(\left |\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)

∴d=\(\left|\frac{(\hat{i}+2\hat{j}-4\hat{k})×(6\hat{i}+2\hat{j}-\hat{k})-(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1^2+2^2+(-4)^2}}\right |\)

=\(\left |\frac{(\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})}{\sqrt{21}} \right |\)

\((\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}& \hat{k}\\1&2&-4\\5&1&-2\end{vmatrix}\)

=\(\hat{i}(-4+4)-\hat{j}(-2+20)+\hat{k}(1-10)\)

=-\(18\hat{j}-9\hat{k}\)

⇒d=\(\left|\frac{\sqrt{(-18)^2+(-9)^2}}{√21}\right|\)

d=\(\sqrt{\frac{405}{21}}\)