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Find the shortest distance between two lines l1 and l2 whose vector equations is given below.\(\vec{r}=3\hat{i}-4\hat{j}+2\hat{k}+λ(4\hat{i}+\hat{j}+\hat{k})\)\(\vec{r}=5\hat{i}+\hat{j}-\hat{k}+μ(2\hat{i}-\hat{j}-3\hat{k})\) (a) \(\frac{11}{\sqrt{12}}\)(b) \(\frac{23}{\sqrt{10}}\)(c) \(\frac{18}{\sqrt{10}}\)(d) \(\frac{10}{\sqrt{11}}\)I have been asked this question in homework.The doubt is from Three Dimensional Geometry topic in portion Three Dimensional Geometry of Mathematics – Class 12

Answer»

Correct choice is (c) \(\FRAC{18}{\sqrt{10}}\)

To elaborate: The distance between two SKEW lines is given by

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

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

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

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

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

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

=\(6\hat{j}-6\hat{k}\)

d=\(\left |{6\hat{j}-6\hat{k}).(\hat{i}-3\hat{j}+3\hat{k})}{|6\hat{i}-2\hat{k}|}\right |\)

\(\left|\frac{0-18-18}{\sqrt{6^2+2^2}}\right |=\frac{36}{\sqrt{40}}=\frac{18}{\sqrt{10}}\)


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