Question

Find the distance between the lines l1 and l2 with the following vector equations.

\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\)

\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)

(a) \(\frac{57}{\sqrt{47}}\)

(b) \(\frac{57}{\sqrt{77}}\)

(c) \(\frac{7}{\sqrt{477}}\)

(d) \(\frac{57}{\sqrt{477}}\)

I got this question in semester exam.

The query is from Three Dimensional Geometry in section Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right option is (d) \(\frac{57}{\sqrt{477}}\)

To explain I would say: We know that, the shortest distance between two skew lines is given by

d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)

The vector equations of the two lines is

\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\)

\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)

∴d=\(\left|\frac{((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k}).(4\hat{i}-\hat{j}+5\hat{k})-(2\hat{i}+2\hat{j}-2\hat{k}))}{|(3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})|}\right |\)

\((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&5\\3&-2&4\end{vmatrix}\)

=\(\hat{i}(8+10)-\hat{j}(12-15)+\hat{k}(-6-6)\)

=\(18\hat{i}+3\hat{j}-12\hat{k}\)

d=\(\left|\frac{(18\hat{i}+3\hat{j}-12\hat{k}).(2\hat{i}-3\hat{j}+7\hat{k})}{\sqrt{18^2+3^2+(-12)^2}}\right |\)

d=\(\left|\frac{36-9-84}{\sqrt{477}}\right |\)=\(\frac{57}{\sqrt{477}}\).