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  • Find the shortest distance between the lines <img alt="\frac{x-1}{2}= \frac{y-2}{3}= \frac{z-3}{4}\, \ and \ \ \, \frac{x-2}{3}= \frac{y-4}{4}= \frac{z-5}{5}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7Bx-1%7D%7B2%7D%3D%20%5Cfrac%7By-2

Find the shortest distance between the lines \frac{x-1}{2}= \frac{y-2}{3}= \frac{z-3}{4}\, \ and \ \ \, \frac{x-2}{3}= \frac{y-4}{4}= \frac{z-5}{5}.

 

 

 

 
 
 
 
 

Answers (1)

writing lines in the vector form:
\vec{r}= \hat{i}+2\hat{j}+3\hat{k}+\lambda \left ( 2\hat{i}+3\hat{j}+4\hat{k}\right )
\vec{r}= 2\hat{i}+4\hat{j}+5\hat{k}+\mu \left ( 3\hat{i}+4\hat{j}+5\hat{k}\right )
so, we have \vec{a_{1}}= \hat{i}+2\hat{j}+3\hat{k},\vec{b_{1}}= 2\hat{i}+3\hat{j}+4\hat{k}
                   \vec{a_{2}}= 2\hat{i}+4\hat{j}+5\hat{k},\vec{b_{2}}= 3\hat{i}+4\hat{j}+5\hat{k}
\therefore \vec{a_{2}}- \vec{a_{1}}= \hat{i}+2\hat{j}+2\hat{k}
and \vec{b_{1}}\times \vec{b_{2}}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 2 & 3&4 \\ 3& 4 &5 \end{vmatrix}= -\hat{i}+2\hat{j}-\hat{k}\, also
\left | \vec{b_{1}}\times \vec{b_{2}} \right |= \sqrt{1+4+1}\Rightarrow \sqrt{6}
Hence S.D = \frac{\left | \left ( \vec{b_{1}}\times \vec{b_{2}} \right )\left ( \vec{a_{2}} - \vec{a_{1}} \right ) \right |}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |}=\frac{\left | -1+4-2 \right |}{\sqrt{6}}
    = \frac{1}{\sqrt{6}}\: or\; \frac{\sqrt{6}}{6}\, units
 

Posted by

Ravindra Pindel

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