17     Find the shortest distance between the lines whose vector equations are 

              \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k} and \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}

Answers (1)

Given two equations of the line 

\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}  \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}  in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}}   and   \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

After comparing the given equations, we obtain

\vec{a_{1}} = \widehat{i}-2\widehat{j}+3\widehat{k}                \vec{b_{1}} = -\widehat{i}+\widehat{j}-2\widehat{k}

\vec{a_{2}} = \widehat{i}-\widehat{j}-\widehat{k}             \vec{b_{2}} = \widehat{i}+2\widehat{j}-2\widehat{k}

\vec{a_{2}}-\vec{a_{1}} = (\widehat{i}-\widehat{j}-\widehat{k}) - (\widehat{i}-2\widehat{j}+3\widehat{k}) = \widehat{j}-4\widehat{k} 

Then calculating the determinant value numerator.

\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ -1& 1 &-2 \\ 1& 2& -2 \end{vmatrix}

= (-2+4)\widehat{i}-(2+2)\widehat{j}+(-2-1)\widehat{k} = 2\widehat{i}-4\widehat{j}-3\widehat{k}

That implies,

\left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(2)^2+(-4)^2+(-3)^2}

= \sqrt{4+16+9} = \sqrt{29}

\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(2\widehat{i}-4\widehat{j}-3\widehat{k})(\widehat{j}-4\widehat{k}) = -4+12 = 8

Now, after substituting the value in the above formula we get,

d= \left | \frac{8}{\sqrt{29}} \right | = \frac{8}{\sqrt{29}}

Therefore,\frac{8}{\sqrt{29}} units are the shortest distance between the two given lines. 

 

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