Question

Find the equation between the two parallel lines l1 and l2 whose equations is given below.

\(\vec{r}=3\hat{i}+2\hat{j}-\hat{k}+λ(3\hat{i}-2\hat{j}+\hat{k})\)

\(\vec{r}=2\hat{i}-\hat{j}+\hat{k}+μ(3\hat{i}-2\hat{j}+\hat{k})\)

(a) \(\sqrt{\frac{172}{14}}\)

(b) \(\sqrt{\frac{145}{14}}\)

(c) \(\sqrt{\frac{171}{14}}\)

(d) \(\sqrt{\frac{171}{134}}\)

The question was asked in an internship interview.

Enquiry is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right choice is (c) \(\sqrt{\frac{171}{14}}\)

Easy explanation: The distance between two parallel lines is given by

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

=\(\left |\frac{((3\hat{i}-2\hat{j}+\hat{k})×((3\hat{i}+2\hat{j}-\hat{k})-(2\hat{i}-\hat{j}+\hat{k})))}{|\sqrt{3^2+(-2)^2+1^2}|}\right |\)

=\(\left |\frac{(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k}))}{\sqrt{14}}\right |\)

\( (3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&1\\1&3&-2\end{vmatrix}\)

=\(\hat{i}(4-3)-\hat{j}(-6-1)+\hat{k}(9+2)\)

=\(\hat{i}+7\hat{j}+11\hat{k}\)

∴d=\(\frac{|\hat{i}+7\hat{j}+11\hat{k}|}{\sqrt{14}}=\frac{\sqrt{1+49+121}}{\sqrt{14}}=\sqrt{\frac{171}{14}}\).